

\documentclass[12pt]{report}

\usepackage{amssymb, amsmath, amsthm,amsfonts}
\usepackage{xypic}
\usepackage{graphicx} 
\usepackage[left=1.5in, top=1.3in, right=1.3in, bottom=1.2in]{geometry}
\usepackage{setspace}
%\singlespacing
%\onehalfspacing
\doublespacing
\usepackage{hyperref}
\usepackage{mathabx,epsfig}
\usepackage{pst-node}
\usepackage{tikz-cd} 
\usepackage {tikz}
\usetikzlibrary {positioning}
\def\acts{\mathrel{\reflectbox{$\righttoleftarrow$}}}
\usetikzlibrary{decorations.markings}

\newtheorem{thm}{Theorem}[chapter]
\newtheorem{lem}[thm]{Lemma}
\newtheorem{prop}[thm]{Proposition}
\newtheorem{cor}[thm]{Corollary}
\newtheorem{clm}{Claim}

\theoremstyle{definition}
\newtheorem{definition}[thm]{Definition}
\theoremstyle{definition}
\newtheorem{rem}[thm]{Remark}
\newtheorem{example}{Example}
\newtheorem{innercustomlem}{Lemma}
\newenvironment{customlem}[1]
{\renewcommand\theinnercustomlem{#1}\innercustomlem}
{\endinnercustomlem}
\newtheorem{innercustomthm}{Theorem}
\newenvironment{customthm}[1]
{\renewcommand\theinnercustomthm{#1}\innercustomthm}
{\endinnercustomthm}
\numberwithin{equation}{chapter}
\numberwithin{figure}{chapter}


\newcommand{\N}{\mathbb{N}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\Q}{\mathbb{Q}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\C}{\mathbb{C}}
\newcommand{\Oh}{\mathcal{O}}
\newcommand{\F}{\mathcal{F}}
\newcommand{\B}{\mathcal{B}}
\newcommand{\Id}{\mathrm{Id}}

\begin{document}
	
	\title{Representations of the Automorphism group of a RACG}
	\author{Thomas Richard Morgan}
	\date{month year}
	
	\pagenumbering{roman}
	{
		
		{\singlespacing
			
			\newpage
			\thispagestyle{empty}
			\begin{center}
				{ %\large
					UNIVERSITY OF OKLAHOMA
					\par
					\vspace{0.16in}
					GRADUATE COLLEGE
					\par
					\vspace{1.2in}
					%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%5
					REPRESENTATIONS OF THE AUTOMORPHISM GROUP
					\par
					\vspace{0.17in}
					OF A RIGHT-ANGLED COXETER GROUP
					\par
					\vspace{1.2in}
					%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
					A DISSERTATION
					\par
					\vspace{0.17in}
					SUBMITTED TO THE GRADUATE FACULTY
					\par
					\vspace{0.17in}
					in partial fulfillment of the requirements for the
					\par
					\vspace{0.17in}
					Degree of
					\par
					\vspace{0.17in}
					DOCTOR OF PHILOSOPHY
					\par
					\vfill
					%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
					By
					\par
					\vspace{0.17in}
					THOMAS RICHARD MORGAN
					\par
					%\vspace{0.05in}
					Norman, Oklahoma
					\par
					%\vspace{0.05in}
					2019
				}
			\end{center}
			
			
			%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
			%% SIGNATURE PAGE  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
			%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
			\newpage
			\thispagestyle{empty}
			\begin{center}
				{%\large
					REPRESENTATIONS OF THE AUTOMORPHISM GROUP
					\par
					%\vspace{0.17in}
					OF A RIGHT-ANGLED COXETER GROUP
					\par
					\vspace{0.9in}
					A DISSERTATION APPROVED FOR THE
					\par
					%\vspace{0.17in}
					DEPARTMENT OF MATHEMATICS
					\par
					\vspace{0.9in}
					BY
					\par
					\vfill
					\begin{flushright}
						\begin{tabular}{cr}
						
							\  \  \  \  \  \  \  & Dr. Lucy Lifschitz, Chair\\
							\\ &  \\
							& Dr. Justin Malestein, Cochair \\
							\\ & \\
							& Dr. Ralf Schmidt \\
							\\   & \\
							& Dr. Johnathan Kujawa \\
							\\ &  \\
							& Dr. David Vishanoff \\
						\end{tabular}
					\end{flushright}
				}
			\end{center}
			
			
			%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
			%% COPYRIGHT PAGE  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
			%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
			\newpage
			\thispagestyle{empty}
			\   \
			\par
			\vfill
			\begin{center}
				\copyright \    Copyright by THOMAS RICHARD MORGAN \   2019
				
				All rights reserved.
			\end{center}
		}
		
		
		
		%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
		%% ACKNOWLEDGEMENTS PAGE  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
		%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
		\chapter*{Acknowledgements}
		
		I wish to express my gratitude to my research advisors, Dr. Justin Malestein and Dr. Lucy Lifschitz, who have taught me so much about mathematics.  They guided me in getting my first paper published, spent many hours helping me hone my arguments, and gave me inspiration when I was stuck.  I am grateful for having had the opportunity to work with them.
		
		I would like to offer my special thanks to my wife, Lauren Fehr.  Whether she was taking care of our daughter so I could work on my research, helping me to relax when I was frustrated, or providing emotional support during my long commutes, Lauren was always there for me.  I feel so lucky to have a wonderful wife who loves me so well!
		
		I am particularly grateful to my parents, Rich and Colleen Morgan, for recognizing my early interest in mathematics. They lovingly and patiently provided me with the resources and support to pursue that passion, and I would not have decided to undertake this project without them.  I would also like to thank them for supporting me throughout college and graduate school.  
		
		I wish to acknowledge the help provided by my in-laws, Jim Tappan and Kris King, for always making me feel welcome in their home.
		
		And last, I would like to thank baby Marie, for being a constant source of cuteness and motivation to be the best I can be.
		%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
		%% TABLE OF CONTENTS PAGE  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
		%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
		
		{\singlespacing
			\tableofcontents
			%\listoffigures
		}
		\newpage
		
	} %% END PAGE NUMBERING ROMAN
	
	
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\chapter*{Abstract}
	
	In 2009 Grunewald and Lubotzky published a paper in which they defined a family of linear representations of the automorphism group of a free group.  In this dissertation, we will use their ideas to construct a family of linear representation of the automorphism group of a right-angled Coxeter group.  We will then use graph-theoretic properties of the defining graph to systematically decompose the image group into a group of block upper triangular matrices.
	\newpage
		
	\pagenumbering{arabic}
	
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%% INTRODUCTION %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

	
	
	\chapter{Introduction}	
	
	\section{Main Results}\label{sect main results}
	In 2009 Grunewald and Lubotzky published a paper (\cite{gl}) in which they constructed a family of linear representations of the automorphism group of a free group.  The group $\text{Aut}(F_n)$ is in some sense analagous to the mapping class group of a surface, which is the group of orientation-preserving homeomorphisms up to homotopy.  In fact, the Dehn-Nielsen-Baer Theorem implies that the extended mapping class group of a surface is isomorphic to the outer automorphism group of the fundamental group of the surface, which is a 1-relator group.  Both $\text{Aut}(F_n)$ and the mapping class group are of much interest in geometric group theory.  Grunewald and Lubotzky subsequently published a paper in conjunction with Larsen and Malestein (\cite{gllm}) in which they used the same idea to construct a family of linear representations of the mapping class group.  They used these representations to show that the mapping class group has a rich family of arithmetic quotients.  
	
	Often a comparison is made between $\text{Aut}(F_n)$/mapping class groups and lattices in Lie groups/arithmetic groups.  They share many algebraic properties, such as being finitely generated, residually finite, and virtually torsion free.  Furthermore, they both satisfy a Tits alternative (for the Tits alternative of a mapping class group see \cite{m}).  This comparison can be taken much further (see, e.g.\cite{j}).  This motivates the question of whether the mapping class group is linear, a question which still remains open.  While it is known that $\text{Aut}(F_n)$ is not linear, the analogy between $\text{Aut}(F_n)$/mapping class groups and lattices in Lie groups has been fruitful.
	
	The Grunewald Lubotzky representations can be used to prove other nice properties and extract useful information. Koberda showed that these same representations could detect the Nielsen-Thurston classification of automorphisms or mapping classes (\cite{k2}).  Following this, Hadari and Liu published papers in which they showed these representations can detect interesting dynamical properties of mapping classes and automorphisms of free groups (\cite{h}, \cite{h2},\cite{l}).  
	
	Mapping class groups are not the only groups that have similarities to $\text{Aut}(F_n)$.  
	The group $F_n$ is an example of a graph product.  A \underline{graph product} is a group constructed from an underlying graph by letting the vertices of the graph be generators of the group and the edges of the graph represent relations.  The group $F_n$ corresponds to the graph with $n$ vertices, but no edges.  The automorphism groups of graph products were studied under various conditions in \cite{cg}, \cite{crsv}, and \cite{gpr}.
	In particular, $F_n$ is a \underline{right-angled Artin group} (RAAG), where the generators are of infinite order and the edges represent commutator relations. Right-angled Artin groups interpolate between the infinite abelian group $\Z^n$ and the free group $F_n$.  A \underline{right-angled Coxeter group} (RACG) is like a RAAG, but the generators are each of order 2.  The similarities between free groups and RAAGs/RACGs are born out in their automorphism groups.  In particular, it is shown in \cite{cg} that both automorphism groups are generated by the same types of generators.
	
	While the automorphism groups of both RAAGS and RACGs have been studied to various ends (\cite{ac},\cite{c},\cite{cv},\cite{gs},\cite{kw},\cite{ss}), not much work has been done on the representations of their automorphism groups, apart from specific cases like $\Z^n$ and $F_n$.  Guirardel and Sale used the Grunewald Lubotzky construction to study automorphism groups of RAAGs (\cite{gs}).   However, no one has applied Grunewald and Lubotzky's ideas to construct representations of the automorphism group of a RACG.  We do so in this dissertation.
	
	Let $\Gamma$ be a finite graph, and let $W_\Gamma$ denote the right-angled Coxeter group associated to $\Gamma$.  Let $\pi: W_\Gamma\to G$ be an epimorphism onto some finite group $G$.
	Following Grunewald and Lubotzky \cite{gl}, we construct a virtual representation $\rho_{\Gamma,G,\pi}$ of the automorphism group $\text{Aut}(W_\Gamma)$.  That is to say, we consider the finite index subgroup $$\Gamma(G,\pi):=\{\varphi\in\text{Aut}(W_\Gamma)\;|\; \pi\circ\varphi = \pi\}$$ of $\text{Aut}(W_\Gamma)$ and construct a representation $\rho_{\Gamma,G,\pi}:\Gamma(G,\pi)\to\text{GL}_t(\Q)$ for some $t\in\Z$.  This representation then induces a representation of $\text{Aut}(W_\Gamma)$.  While the construction works for any choice of $G,\pi$, we focus on a standard choice of $G,\pi$ that depends on $|V(\Gamma)|$.  In particular, we choose $G=(W_\Gamma)^{ab}\cong (\Z/2\Z)^{|V(\Gamma)|}$, and we choose $\pi$ to be the abelianization map.  Because we always make this choice, we suppress the $G,\pi$ indices in our representation and write $\rho_{\Gamma}$.  The goal of this paper is to better understand $\text{Im}(\rho_\Gamma)$ as we vary $\Gamma$.  We approach this goal by computing the isomorphism class of $\text{Im}(\rho_\Gamma)$.  In so doing, we give block matrix descriptions of the matrices in $\text{Im}(\rho_{\Gamma})$ as well as descriptions of the linear dependencies within each block.  It turns out that $\text{Im}(\rho_{\Gamma})$ is the $2$-congruence subgroup of the integer matrices in the integer points of a linear algebraic group; hence $\text{Im}(\rho_{\Gamma})$ is arithmetic.
	
	Topologically, $\rho_{\Gamma}$ can be constructed as follows.  We first construct a certain $K(W_\Gamma,1)$ space $X$.  We take the cover $p: \hat X\to X$ corresponding to the finite-index subgroup $\ker(\pi)$.  The group $\Gamma(G,\pi)$ acts on the first rational homology $H_1(\hat{X};\Q)$ of $\hat{X}$.  This action is the representation $\rho_{\Gamma}$.  
	
	Using cellular homology, we can think of $H_1(\hat{X};\Q)$ as consisting of formal sums of edges of $\hat X$ (up to cellular 1-boundaries).  Since $G$ acts on $\hat X$ by deck transformations, we thus obtain an action of $\Q[G]$ on $H_1(\hat X;\Q)$. As a $\Q[G]$-module, we may decompose $H_1(\hat{X};\Q)$ as a direct sum of irreducible $\Q[G]$-submodules.  By grouping together the isomorphic irreducible submodules of $H_1(\hat{X};\Q)$, we obtain what is called the \underline{isotypic components} of $H_1(\hat{X};\Q)$.  We describe this in detail in Section \ref{sect isotypic components}.  Due to our choice of $G,\pi$, the isotypic components can be indexed by subsets $J\subseteq V(\Gamma)$ of the generating set of $W_{\Gamma}$.
	
	In Section \ref{sect constructing the representation} we show that $\Gamma(G,\pi)$ acts by $\Q[G]$-module automorphisms.  Thus we may consider the action of $\Gamma(G,\pi)$ on the isotypic components of $H_1(\hat{X};\Q)$.  This gives us a decomposition of $\rho_{\Gamma}$ into sub-representations $\rho_{\Gamma,J}$ which are much easier to compute.  It turns out that to compute $\rho_{\Gamma,J}$, we may restrict our attention to the subgraph $\Gamma_J$ of $\Gamma$ induced by $J$.  This is made precise in the following lemma which we prove in Section \ref{sect subgraph lemma}.  Here $1_J$ is an element of $G$ that acts by projecting onto the $J$-isotypic component and $\hat v$ is a cellular 1-chain (up to cellular 1-boundaries) that corresponds in a natural way to $v\in W_\Gamma$.  The precise meaning will be described later.
	
	\begin{lem}[The Subgraph Lemma]\label{subgraph} 
		Let $\Gamma$ be a finite graph, and let $J\subseteq V(\Gamma)$.  Then as $\Q[G]$-modules, $I_{J}(H_1(\hat{X}; \Q))=
		\langle 1_J(\hat{v}-\hat{w}) \hspace{1mm}|\hspace{1mm} v,w\in V(\Gamma_J)\rangle/
		\langle 1_J(\hat{v}-\hat{w}) \hspace{1mm}|\hspace{1mm} (v,w)\in E(\Gamma_{J})\rangle\cong I_{J}(\Q[G])^{k_{J}-1}$, where $k_J$ is the number of components of $\Gamma_J$.  
	\end{lem}
	
	In Chapter 3 we prove a number of decompositions that allow us to compute $\text{Im}(\rho_{\Gamma,J})$.  The idea for each of our decompositions is to write the matrices of $\text{Im}(\rho_{\Gamma,J})$ in an upper triangular block matrix form where each of the entries in each block either come from $\text{Im}(\rho_{\Gamma,A})$ for some $A\subsetneqq J$ or from a known group.  Thus repeated applications of the decompositions allow us to compute $\text{Im}(\rho_{\Gamma,J})$. 
	
	Section \ref{sect commutation relations} deals mainly with showing when elements of $\text{Im}(\rho_{\Gamma,J})$ commute.  In Section \ref{sect tml decomposition}, we decompose $\text{Im}(\rho_{\Gamma,J})$ by introducing a division of the set $J$.  A \underline{\textit{division}} of $J$ is a decomposition into subsets $J=\displaystyle\bigsqcup_{i=1}^m A_i$ such that each $A_i$ is a union of components of $\Gamma_J$ and contains a \underline{\textit{special point}} $a_i\in A_i$. The special point $a_i$ provides some level of control over which components of $\Gamma\setminus\text{st}(a_i)$ intersect $A_i$.  We will formally define special points in Section \ref{sect tml decomposition}.  After some preliminary work, we get the following decomposition theorem.
	
	\begin{thm}[TML Decomposition]\label{tml}
		Let $J=\displaystyle\bigsqcup_{i=1}^m A_i$ be a division of $J$ with special points $a_i\in A_I$.  Let $J_0:=\{v\;|\; v\in[a_i]_J\text{ for some } 1\leq i \leq m\}$.  Then $\text{Im}(\rho_{\Gamma,J})\cong\Big(\Big(\displaystyle\prod_{i=1}^m\text{Im}(\rho_{\Gamma,A_i})\Big)\times\text{Im}(\rho_{\Gamma,J_0})\Big)\ltimes\Z^r$ where $r:=|\{([v]_J,D)\;|\; v\in A_i\setminus[a_i]_J \text{ for some } 1\leq i\leq m, D\text{ is a component of }\Gamma\setminus\text{st}(v), D\cap A_i = \emptyset\}|$.
	\end{thm}
	
	In Section \ref{sect splitting points} we define the notion of a splitting point.  We then show that if a vertex set $J$ contains a splitting point, then $\text{Im}(\rho_{\Gamma,J})$ can be decomposed as a direct product (Proposition \ref{ss reduction2}).
	
	In Section \ref{sect separating sets} we introduce the notion of a separating set.  Given two vertices $v,w$ in distinct components of $\Gamma_J$, we define $$\text{sep}(v,w):=\{x\in J\;|\;\exists x^*\in[x]_J\text{ such that } D(x^*,v)\neq D(x^*,w)\}.$$  It is natural to explore this set because partial conjugations by these vertices act on $\hat{v}-\hat{w}$ as something other than an eigenvector.  This set also has the nice property that for every element $c$ outside of $\text{sep}(v,w)$, the whole set $\text{sep}(v,w)$ is in the same component of $\Gamma\setminus\text{st}(c)$ (Proposition \ref{supset2}).  We also prove another decomposition result (Lemma \ref{sep=22})
	
	We take a slight detour in Section \ref{sect compressible components} to deal with a difficult case not covered by the hypotheses of Lemma \ref{sep=22}.  To that end, we introduce the notion of a compressible component.  A \underline{\textit{compressible component}} is a component of $\Gamma_J$ which can be compressed to a single vertex without affecting $\text{Im}(\rho_{\Gamma,J})$.  This will be formally defined in Section \ref{sect compressible components}.  This gives us a new method of computing $\text{Im}(\rho_{\Gamma,J})$.  
	
	In Section \ref{sect computing}, we finally prove how to compute the isomorphism class of $\text{Im}(\rho_{\Gamma,J})$ for arbitrary $\Gamma,J$ (Theorem \ref{final thm}). 
	
	Up until this point in the dissertation we have focused on how to compute $\text{Im}(\rho_{\Gamma,J})$ for given $\Gamma,J$.  One can also ask the following question.  Which groups can be written as $\text{Im}(\rho_{\Gamma,J})$ for some $(\Gamma,J)$?  In Section \ref{sect star subgroups} we show that a certain family of groups can be obtained in this way.  To be more precise, consider the set $$G_I:=\{M\in\Gamma_n(2)\;|\;M_{i,j}=0\text{ for all }(i,j)\notin I \}$$ where $\Gamma_n(2)$ is the kernel of the map $\text{GL}_n(\Z)\to \text{GL}_n(\Z/2\Z)$ and $I\subseteq\{1,2,\dots, n\}^2$.  We give a condition on the index set $I$ that is equivalent to $G_I$ being a group.  Then every group $G_I$ can be written as $\text{Im}(\rho_{\Gamma,J})$ for some $(\Gamma,J)$.  In particular, we see that the intersection of the upper triangular matrix group with $\Gamma_n(2)$ can be written as the image of one of these representations.
	
	\section{Index of Notation}\label{sect index of notation}
	For the convenience of the reader, we collect here some of the more important notation that is
	consistent throughout the paper.
	\\
	
	\noindent $\Gamma$: a finite graph\\
	$W_{\Gamma}$: the right-angled Coxeter group associated to $\Gamma$\\
	$G$: a finite group, usually $W_\Gamma^{ab}\cong (\Z/2\Z)^n$ where $n=|V(\Gamma)|$\\
	$\pi$: an epimorphism from $W_\Gamma$ to $G$\\
	$\Gamma(G,\pi):=\{\varphi\in\text{Aut}(W_{\Gamma})\;|\; \pi\circ\varphi = \pi\}$\\
	$\rho_{\Gamma}$: the virtual representation of $\text{Aut}(W_{\Gamma})$ constructed in Section \ref{sect constructing the representation}\\
	$X$: the $K(W_\Gamma,1)$ space constructed in Section \ref{sect constructing the representation}\\
	$\hat{X}$: the cover of $X$ corresponding to $\ker(\pi)$\\
	$\text{Aut}^0(W_{\Gamma})$: the subgroup of $\text{Aut}(W_{\Gamma})$: generated by partial conjugations\\
	$\text{Out}^0(W_{\Gamma})$: $\text{Aut}^0(W_{\Gamma})/\text{Inn}(W_{\Gamma})$\\
	$\sigma_{D,v}$: the partial conjugation by $v$ on the component $D$ of $\Gamma\setminus\text{st}(v)$\\
	$J$: a subset of $V(\Gamma)$\\
	$\Gamma_J$: the subgraph of $\Gamma$ induced by $J$\\
	$k_J$: the number of components of $\Gamma_J$\\
	$[v]_J$: the component of $\Gamma_J$ containing $v$\\
	$\rho_{\Gamma,J}$: the projection of $\rho_{\Gamma}$ onto $\text{Aut}(I_J(H_1(\hat{X};\Q)))$\\
	$\mathcal{P}^{\Gamma,J}:=\{\sigma_{D,v}\;|\; v\in J, D\cap J\setminus [v]_J \neq \emptyset\}$\\ 
	$\bar{\sigma}$: the image of $\sigma$ in $\text{Out}(\Gamma)$\\
	$\hat\sigma:=\rho_{\Gamma,J}(\sigma)$\\
	$D(v,w)$: the component of $\Gamma\setminus\text{lk}(v)$ containing $w$\\
	$\Gamma_n(2):=\ker(\text{GL}_n(\Z)\to\text{GL}_n(\Z/2\Z))$\\
	$\text{sep}(v,w):=\{x\in J\;|\;\exists x^*\in[x]_J\text{ such that } D(x^*,v)\neq D(x^*,w)\}$\\ 
	$\text{Int}(v,x):=\displaystyle\bigcap_{w\in[v]_J}D(w,x)\cap J$\\
	$\mathfrak{D}(v):=\{D(w,y)\cap J\setminus[v]_J\;|\; w\in [v]_J, y\in J\setminus[v]_J\}.$
	
	\chapter{Setup}
	\section{Constructing the representation}\label{sect constructing the representation}
	Let $F_n$ be the free group with n generators.  For each surjection $\pi: F_n\to G$ onto a finite group $G$, Grunewald and Lubotzky constructed a representation $\rho_{G,\pi}$ of $\text{Aut}(F_n)$ \cite{gl}.
	We follow Grunewald and Lubotzky's construction to get a representation $\rho_{\Gamma,G,\pi}$ for $\text{Aut}(W_{\Gamma})$, where $W_{\Gamma}$ is the right angled Coxeter group associated to the graph $\Gamma$.  That is to say, $$W_{\Gamma}:=\langle v\in V(\Gamma)\; |\; v^2=1, [v,w]=1 \text{ for all }(v,w)\in E(\Gamma)\rangle.$$
	
	Let $\Gamma$ be a finite graph.  Let $G$ be a finite group and $\pi: W_{\Gamma}\to G$ be an epimorphism.  Let $R=\ker(\pi)$ and $\bar{R}:=R/[R,R]$.  The action of $W_{\Gamma}$ on $R$ by conjugation leads to an action of $G$ on $\bar{R}$.  
	Let $$\Gamma(G,\pi):=\{\varphi\in\text{Aut}(W_{\Gamma})\;|\; \pi\circ\varphi = \pi\}.$$  Then $\Gamma(G,\pi)$ is a finite index subgroup of $\text{Aut}(W_{\Gamma})$.  Furthermore, every $\varphi\in\Gamma(G,\pi)$ induces a G-equivariant linear automorphism $\bar\varphi$ of $\bar{R}$.  Indeed, given $w\in W_{\Gamma}$ and $\varphi\in\Gamma(G,\pi)$, we have that $\varphi(w)=wr_w$ for some $r_w\in R$.  Hence $\varphi(wrw^{-1})=wr_w\varphi(r)r_w^{-1}w^{-1}$ holds for every $r\in R$.  But this is equivalent to $w\varphi(r)w^{-1}$ modulo $[R,R]$, which implies that $\bar\varphi:\bar R \to \bar R$ is $G$-equivariant.  Tensoring the domain and codomain of $\varphi$ by $\Q$ induces a $G$-equivariant linear transformation $\hat\varphi:\bar R \otimes_\Z \Q\to \bar R \otimes_\Z\Q$.  This yields a representation $$\rho_{\Gamma, G, \pi}: \Gamma(G,\pi)\to\text{GL}_t(\Q),\quad\varphi\mapsto\hat{\varphi}$$  for some $t\in\Z$.  We wish to describe the image of this representation as we vary $\Gamma$ for a standard choice of $G,\pi$ that depends on $\Gamma$.
	
	Let $V(\Gamma)=\{v_1,\dots,v_n\}$.  Then our choice of finite group $G$ is the group $W_\Gamma^{ab}\cong (\Z/2\Z)^n$, and our epimorphism $\pi_\Gamma: W_\Gamma\to W_\Gamma^{ab}$ is the abelianization map.  We write $\pi$ when $\Gamma$ is understood.  Since we will always use this choice of $G,\pi$ unless we say otherwise, we compress the $G,\pi$ in our notation and write $\rho_{\Gamma}:=\rho_{\Gamma,G,\pi}$.
	
	We can translate the above description of our representation into a topological description as follows.
	Let $\Gamma$ be a finite graph, let $G$ be a finite group, and let and $\pi: W_{\Gamma}\to G$ be an epimorphism.  We construct a $K(W_{\Gamma},1)$ space $X$ as outlined in \cite{h3}, Section 4.2.  We start with the wedge sum $\displaystyle\bigvee_{v\in V(\Gamma)}S_v^1$.  For each $v\in V(\Gamma)$ we attach a 2-cell to the loop $vv$.  For each edge $(v,w)\in E(\Gamma)$, we attach a 2-cell to the loop $vw\bar{v}\bar{w}$.  We then attach some higher dimensional cells in such a way that the higher order homotopy groups become trivial.
	
	By covering space theory, there is a cover of $X$ corresponding to each subgroup of $\pi_1(X)=W_{\Gamma}$.
	Let $p:\hat{X}\to X$ be the cover of $X$ corresponding to the subgroup $R:=\ker(\pi)$.  Then $G$ is the deck transformation group of $\hat X$. Fix a lift $\hat{*}$ of the vertex $*$ in $X$ to the cover $\hat{X}$.  Given a deck transformation $g\in G$, we let $g\hat *$ be the image of $\hat *$ under the deck transformation $g$.  This gives us a labeling of the vertices of $\hat X$.
	
	For each loop $v$ in $X$, there is a lift $\hat v$ of $v$ based at $*$.  Note that the terminal vertex of $v$ is the vertex $\pi(v)\hat *$.  Given a deck transformation $g\in G$, we let $g\hat v$ denote the image of $\hat v$ under the deck transformation.  This is precisely the lift of $v$ based at the vertex $g\hat *$.   
	
	\begin{example}
		Let $\Gamma$ be the graph 
		\begin{center}
			\begin{tikzpicture}
			
			\filldraw[black] (0,0) circle (2pt) node[anchor=north east] {$v_1$};
			\filldraw[black] (1,0) circle (2pt) node[anchor=north west] {$v_2$};
			
			\end{tikzpicture}
		\end{center}
		
		Then the 1-skeleton of $X$ is the graph
		\begin{center}
			\begin{tikzpicture}
			\filldraw[black] (0,0) circle (2pt)
			node[anchor=west] {$*$};
			\draw[decoration={markings, mark=at position 0.5 with {\arrow{>}}},
			postaction={decorate}](-.5,0) circle (.5);
			\draw [decoration={markings, mark=at position 0 with {\arrow{>}}},
			postaction={decorate}](.5,0) circle (.5);
			\node at (-1.3, 0){$v_1$};
			\node at (1.3, 0){$v_2$};
			\end{tikzpicture}
		\end{center}
		
		and the 1-skeleton of $\hat{X}$ is the graph
		\begin{center}
			\begin{tikzpicture}
			\filldraw[black] (0,0) circle (2pt)
			node[anchor=north east] {$\hat *$};
			\filldraw[black] (4,0) circle (2pt)
			node[anchor=north west] {$e_1\hat *$};
			\filldraw[black] (0,4) circle (2pt)
			node[anchor=south east] {$e_2\hat *$};
			\filldraw[black] (4,4) circle (2pt)
			node[anchor=south west] {$(e_1+e_2)\hat *$};
			\draw[decoration={markings, mark=at position 0.5 with {\arrow{>}}},
			postaction={decorate}] (0,0) to[out=20,in=160] (4,0);
			\draw[decoration={markings, mark=at position 0.5 with {\arrow{>}}},
			postaction={decorate}] (4,0) to[out=110,in=-110] (4,4);
			\draw[decoration={markings, mark=at position 0.5 with {\arrow{>}}},
			postaction={decorate}] (0,0) to[out=110,in=-110] (0,4);
			\draw[decoration={markings, mark=at position 0.5 with {\arrow{>}}},
			postaction={decorate}] (0,4) to[out=20,in=160] (4,4);
			
			\draw[decoration={markings, mark=at position 0.5 with {\arrow{>}}},
			postaction={decorate}] (4,0) to[out=-160,in=-20] (0,0);
			\draw[decoration={markings, mark=at position 0.5 with {\arrow{>}}},
			postaction={decorate}] (4,4) to[out=-70,in=70] (4,0);
			\draw[decoration={markings, mark=at position 0.5 with {\arrow{>}}},
			postaction={decorate}] (0,4) to[out=-70,in=70] (0,0);
			\draw[decoration={markings, mark=at position 0.5 with {\arrow{>}}},
			postaction={decorate}] (4,4) to[out=-160,in=-20] (0,4);
			
			\node at (2,.5)[anchor=south] {$\hat v_1$};
			\node at (2,-.5)[anchor=north] {$e_1.\hat v_1$};
			\node at (2,4.5)[anchor=south] {$e_2.\hat v_1$};
			\node at (2,3.5)[anchor=north] {$(e_1+e_2).\hat v_1$};
			\node at (-.5,2)[anchor=east] {$\hat v_2$};
			\node at (.5,2)[anchor=west] {$e_2.\hat v_2$};
			\node at (3.5,2)[anchor=east] {$e_1.\hat v_2$};
			\node at (4.5,2)[anchor=west] {$(e_1+e_2).\hat v_2$};
			\end{tikzpicture}
		\end{center}
	\end{example}
	
	Let $\varphi\in\Gamma(G,\pi)$.  Since $X$ is a $K(W_{\Gamma},1)$ space, $\varphi$ can be considered as an automorphism $\pi_1(X)\to\pi_1(X)$.  This induces a unique  homotopy equivalence $\varphi_X: (X,*)\to (X,*)$ up to based homotopy (see, e.g. \cite{h3}).  
	
	The map $p\circ\varphi_X: \hat X \to X$ induces a map $(p\circ\varphi_X)_*: \pi_1(\hat X,\hat *)\to \pi_1(X,*)$.  By definition of $\Gamma(G,\pi)$, the map $\varphi$ stabilizes $\ker(\pi)=p_*(\pi_1(\hat X))$.  It follows that $(p\circ\varphi_X)_*(\pi_1(\hat X,\hat *))=p_*(\pi_1(\hat X,\hat *))$.  Therefore $(p\circ\varphi_X)$ lifts to a map $\hat\varphi: (\hat X,\hat *) \to (\hat X,\hat *)$.  It is not hard to see that $\hat\varphi$ is a lift of $\varphi_X$.  
	
	\begin{center} $ \begin{tikzcd}
		(\hat{X},\hat *) \arrow[dashrightarrow]{r}{\hat\varphi} \arrow[swap]{d}{p}\arrow[rightarrow]{rd}{}
		& (\hat{X},\hat *)\arrow[swap]{d}{p} \\
		(X,*) \arrow[rightarrow]{r}{\varphi_X} &
		(X,*)
		\end{tikzcd}$ \end{center}
	
	Since $\Gamma(G,\pi)$ acts trivially on $W_\Gamma/\ker(\pi)= G$, it follows that $\varphi$ fixes the $G$-orbit of $\hat *$.  But this is precisely $\hat X_0$.  Consequently, $\hat\varphi: (\hat X, \hat X^0)\to (\hat X, \hat X^0)$ is a map of pairs.  We thus get an induced homomorphism $\hat \varphi_*: H_1(\hat X,\hat X^0; \Q)\to H_1(\hat X,\hat X^0;\Q)$. 
	
	Let $C^{cw}_i(\hat X;\Q):=H_1(\hat X_i, \hat X_{i-1};\Q)$. By Section 2.2 of \cite{h3}, there is a relative cellular chain complex $$\dots\to H_2(\hat X^2, \hat X^1\cup \hat X^0;\Q)\to H_1(\hat X^1, \hat X^0\cup \hat X^0;\Q)\to H_1(\hat X^0, \emptyset\cup \hat X^0;\Q)$$  with homology groups isomorphic to $H_n(\hat X, \hat X^0;\Q)$.  But
	\begin{itemize}
		\item $H_2(\hat X^2, \hat X^1\cup\hat X^0;\Q) = H_2(\hat X^2, \hat X^1;\Q)=C^{cw}_2(\hat X;\Q)$.
		\item $H_1(\hat X^1, \hat X^0\cup \hat X^0;\Q)= H_1(\hat X^1, \hat X^0;\Q)=C^{cw}_1(\hat X; \Q)$.
		\item $H_0(\hat X^0, \emptyset\cup\hat X^0;\Q)=H_0(\hat X^0,\hat X^0;\Q)=0$.
	\end{itemize}
	
	This shows that $H_1(\hat X,\hat X^0;\Q)\cong C^{cw}_1(\hat X;\Q)/\delta_2(C^{cw}_2(\hat X;\Q))$.  Hence we have a well-defined action of $\Gamma(G,\pi)$ on cellular 1-chains mod cellular 1-boundaries.  From this point on we let $\hat v$ represent the cellular 1-chain corresponding to the vertex $v$ up to cellular 1-boundary.  
	
	\begin{example}
		Consider the inner automorphism $\sigma_v$ by the vertex $v$.  By definition, for any $w\in E(\Gamma)$, we have that $\sigma_v(w)=vwv$.  To compute the action of $\sigma_v$ on $\hat w$, we look at the lift of the path $vwv$.  Hence
		$\hat\sigma_v$ sends $\hat w$ to $\hat v + \pi(v)\hat w + \pi(vw)\hat v$.
		
		\begin{center}
			\begin{tikzpicture}
			\filldraw[black] (0,0) circle (2pt) node[anchor=north] {$\hat *$};
			\filldraw[black] (3,0) circle (2pt) node[anchor=north] {$\pi(v)\hat *$};
			\filldraw[black] (3,3) circle (2pt) node[anchor=south] {$\pi(vw)\hat *$};
			\filldraw[black] (0,3) circle (2pt) node[anchor=south] {$\pi(w)\hat *$};
			
			\draw[decoration={markings, mark=at position .5 with {\arrow{>}}},
			postaction={decorate}] (0,0) -- (3,0);
			\filldraw[black] (1.5,0) circle (0pt)
			node[anchor = north]{$\hat v$};
			\draw[decoration={markings, mark=at position .5 with {\arrow{>}}},
			postaction={decorate}] (3,0) -- (3,3);
			\filldraw[black] (3,1.5) circle (0pt)
			node[anchor = west]{$\pi(v)\hat w$};
			\draw[decoration={markings, mark=at position .5 with {\arrow{>}}},
			postaction={decorate}] (3,3) -- (0,3);
			\filldraw[black] (1.5,3) circle (0pt)
			node[anchor = south]{$\pi(vw)\hat v$};
			\draw[decoration={markings, mark=at position .5 with {\arrow{>}}},
			postaction={decorate}] (0,0) -- (0,3);
			\filldraw[black] (0,1.5) circle (0pt)
			node[anchor = east]{$\hat w$};
			
			\end{tikzpicture}
		\end{center}
	\end{example}
	
	
	\section{Reduction to $\text{Aut}^0(G,\pi)$}\label{sect reduction to aut}
	
	Let $\Gamma$ be a finite graph, and let $W_{\Gamma}$ be the right-angled Coxeter group associated to the graph $\Gamma$.  The goal of this section is to show that for the standard choice of finite group $G$ and epimorphism $\pi: W_{\Gamma}\to G$, the group $\Gamma(G,\pi)$ is just $\text{Aut}^0(W_{\Gamma})$, the group generated by partial conjugations of $W_\Gamma$.  We begin by describing $\text{Aut}(W_{\Gamma})$.  By \cite{cg}, $\text{Aut}(W_{\Gamma})$ is generated by three types of automorphisms:
	\begin{enumerate}
		\item Graph automorphisms
		\item Dominated transvections:  Choose distinct vertices $u,v$ such that $\text{st}(u)\subseteq \text{st}(v)$.  Then define $$\tau_{u,v}(w):=\begin{cases} uv\quad\text{if } w=u\\ w\quad\text{if } w\neq u\end{cases}$$
		\item Partial conjugations:  Choose a vertex $v$.  Let $D$ be a connected component of the graph $\Gamma\setminus \text{st}(v)$.  Then define $$\sigma_{D,v}(w):=\begin{cases} vwv^{-1}\quad\text{if }w\in D\\ w\quad\text{if } w\notin D\end{cases}$$
	\end{enumerate}
	
	We may then define the following two subgroups:
	\begin{itemize}
		\item $\text{Aut}^0(W_{\Gamma})$ is the group generated by partial conjugations
		\item $\text{Aut}^1(W_{\Gamma})$ is the group generated by graph automorphisms and dominated transvections.
	\end{itemize} 
	
	Given a right-angled coxeter group $W_{\Gamma}$, a \underline{word} is a finite sequence $w=(v_1^{\epsilon_1},v_2^{\epsilon_2},\dots, v_n^{\epsilon_n})$ where for each $1\leq i \leq n$ we have $v_i\in V(\Gamma)$ and $\epsilon_i\in\{\pm 1\}$.  We call $n$ the \underline{length} of the word $w$.  Each word represents an element of $W_\Gamma$, namely the element $v_1^{\epsilon_1}v_2^{\epsilon_2}\dots  v_n^{\epsilon_n}$ obtained by multiplying together the elements in the sequence.  Note that multiple different words may represent the same group element.  A word $w$ is said to be \underline{in reduced form} if there is no word representing the same group element with a smaller length than $w$.  Note that the reduced form of a group element need not be unique.  However, any two reduced words representing the same group element differ only by repeated swapping of the order of adjacent vertices (Lemma 2.3 of \cite{gpr}).
	
	The following lemma will help us to better understand $\text{Aut}^1(W_{\Gamma})$.
	\begin{lem}\label{aut1}
		Let $\Gamma$ be a finite graph.  If $\varphi\in\mathrm{Aut}^1(W_{\Gamma})$, then for all $v\in V(\Gamma)$ we have that $\displaystyle\varphi(v)=\prod_{i=1}^{k}v_{i}$ in reduced form for some $\{v_{i}\in V(\Gamma)\;|\; 1\leq i \leq k\}$ which form a complete subgraph of $\Gamma$.
	\end{lem}
	
	\begin{proof}
		We prove this by induction on the length of a word in $\text{Aut}^1(W_\Gamma)$.  The base case is trivial.
		
		Now assume that for all $\varphi\in\text{Aut}^1(W_{\Gamma})$ of length $\leq k$ that the induction hypothesis holds.  We can increase the length either by composing by a graph automorphism or a dominated transvection.  Since graph automorphisms take complete subgraphs to complete subgraphs, we need only consider composing by a dominated transvection.  Let $\varphi\in\text{Aut}^1(W_{\Gamma})$ such that $\displaystyle\varphi(v)=\prod_{i=1}^{k}v_{i}$ and let 
		$\tau_{u,w}$ be a dominated transvection.	If $u\notin\{v_i\;|\; 1\leq j \leq k\}$, then  $\displaystyle\tau_{u,w}\circ\varphi(v)=\prod_{i=1}^{k}v_{i}$ and we are done.  If, on the other hand, $u\in\{v_{i}\;|\; 1\leq i \leq k\}$, then for all $1\leq i \leq k$ we have $v_{i}\in \text{st}(u)\subseteq \text{st}(w)$ since $\{v_{i}\;|\; 1\leq i \leq k\}$ form a complete graph and $\tau_{u, w}$ is well-defined.  Therefore $\{v_{i}\;|\; 1\leq i \leq k\}\cup\{w\}$ form a complete subgraph, completing the induction.
	\end{proof}
	
	In order to prove that $\Gamma(G,\pi)=\text{Aut}^0(W_\Gamma)$, we first show the following lemma.
	
	\begin{lem}\label{no transvections}
		If $\varphi\in\mathrm{Aut}^1(W_{\Gamma})\setminus\{\mathrm{Id}\}$ then $\varphi\notin\Gamma(G,\pi)$.
	\end{lem}  
	\begin{proof}
		Let $\varphi\in\text{Aut}^1(W_{\Gamma})\setminus\{\text{Id}\}$.  If $\varphi$ is a graph isomorphism, then it is not in $\Gamma(G,\pi)$. By this and Lemma \ref{aut1}, we may assume for some $v\in V(\Gamma)$ that $\displaystyle\varphi(v)=\prod_{i=1}^k v_{i}\neq v$ in reduced form and the $v_i$ form a complete subgraph.  But then $\displaystyle\pi\Big(\prod_{i=1}^k v_{i}\Big)\neq\pi(v)$.
	\end{proof}
	
	We now have all the tools we need to prove the main result of this section.  
	\begin{lem}\label{gamma}
		$\Gamma(G,\pi)=\mathrm{Aut}^0(W_{\Gamma})$	
	\end{lem} 
	\begin{proof}
		By proposition 5.3 of \cite{cg}, $\text{Aut}^0(W_{\Gamma})$ is the group of conjugating automorphisms, i.e. automorphisms where the images of generators are conjugates.  Since $G$ is abelian, and $\pi$ is a homomorphism, for any $v\in V(\Gamma)$ the image of a conjugate of $v$ under $\pi$ is the same as for $v$.  Thus $\text{Aut}_0(W_{\Gamma})\leq\Gamma(G,\pi)$.
		
		To show the reverse inclusion, we will first show that $\text{Aut}^1(W_{\Gamma})$ normalizes $\text{Aut}^0(W_{\Gamma})$.  Let $\sigma=\sigma_{D, v}$ be partial conjugation by $v$ on the component $D$ of $\Gamma\setminus \text{st}(v)$ and $\gamma$ be a graph isomorphism.  Then for all $w\in V(\Gamma)$, we compute $$\gamma^{-1}\circ\sigma\circ\gamma(w)=\gamma^{-1}(v^{\epsilon}\gamma(w)v^{-\epsilon})=\gamma^{-1}(v)^{\epsilon}w\gamma^{-1}(v)^{-\epsilon}$$ where $\epsilon\in\{0,1\}$ depending on whether $\gamma(w)\in D$.  This is a conjugate of $w$, so $\gamma^{-1}\circ\sigma\circ\gamma\in\text{Aut}^0(W_{\Gamma})$.  Now let $\tau=\tau_{u,w}$ be a dominated transvection.  For any vertex $v\neq u$, let  $\sigma(v)=cvc^{-1}$;  then $\tau^{-1}\circ\sigma\circ\tau(v)=\tau^{-1}(c) v \tau^{-1}(c)^{-1}$, which is a conjugate of $v$.  Therefore to show that $\tau^{-1}\circ\sigma\circ\tau\in\text{Aut}^0(W_{\Gamma})$ it suffices to show that $\tau^{-1}\circ\sigma\circ\tau(u)$ is a conjugate of $u$.  
		If $w\in \text{st}(v)$, then $w$ and $v$ commute and $w\notin D$, so  $$\tau^{-1}\circ\sigma\circ\tau(u)=\tau^{-1}(v^{\epsilon}uv^{-\epsilon}w)=\tau^{-1}(v^{\epsilon}uwv^{-\epsilon})=\tau^{-1}(v)^{\epsilon}u\tau^{-1}(v)^{-\epsilon}$$ is a conjugate of $u$.
		If $u\in \text{st}(v)$, then $v\in\text{st}(u)\subseteq\text{st}(w)$.  Hence $w\in\text{st}(v)$ and we are in the case above.
		If $u,w\notin \text{st}(v)$, then $u$ and $w$ are in the same component of $\Gamma\setminus \text{st}(v)$ so $\tau^{-1}\circ\sigma\circ\tau(u)=\tau^{-1}(v^{\epsilon}uwv^{-\epsilon})=\tau^{-1}(v)^{\epsilon}u\tau^{-1}(v)^{-\epsilon}$ is a conjugate of $u$.  Since this covers all cases, $\tau^{-1}\circ\sigma\circ\tau\in\text{Aut}^0(W_{\Gamma})$.  Therefore $\text{Aut}^1(W_{\Gamma})$ normalizes $\text{Aut}^0(W_{\Gamma})$.  Since $\text{Aut}^0(W_\Gamma), \text{Aut}^1(W_\Gamma)$ generate $\text{Aut}(W_\Gamma)$. this implies that any $\phi\in\text{Aut}(W_{\Gamma})$ can be written in the form $\sigma\circ\varphi$ for some $\sigma\in\text{Aut}^0(W_{\Gamma})$, $\varphi\in\text{Aut}^1(W_{\Gamma})$.
		
		Now let $\phi\in\Gamma(G,\pi)$ and write $\phi=\sigma\circ\varphi$ as above.  Then since $\sigma\in\text{Aut}^0(W_{\Gamma})\leq \Gamma(G,\pi)$, it follows that $\varphi=\sigma^{-1}\circ\phi\in\Gamma(G,\pi)$.  By Lemma \ref{no transvections}, this implies $\varphi=\text{Id}$.  Therefore $\phi=\sigma\in\text{Aut}^0(C_{\Gamma})$.
	\end{proof}
	
	\section{Isotypic components}\label{sect isotypic components}
	
	The following information can be found in \cite{cr}.  Let $G$ be a finite group.  Then there are a finite number of irreducible left $\Q[G]$-modules.  Let $M_1, M_2,\dots M_s$ be the irreducible left $\Q[G]$-modules.  Every left $\Q[G]$-module $M$ may be written as a finite direct sum of irreducible left $\Q[G]$-modules $M=\displaystyle\bigoplus_{i=1}^s M_i^{r_i}$ for some $r_i\in\Z_{\geq 0}$. Given any $1\leq i \leq s$, the left $\Q[G]$-module $I_i(M):= M_i^{r_i}$ is called the $M_i$-isotypic component of the left $\Q[G]$-module $M$.  We may therefore write $M=\displaystyle\bigoplus_{i=1}^s I_i(M)$ as a direct sum of its isotypic components.
	
	Now $\Q[G]$ itself may be thought of as a left-$\Q[G]$ module, where the group action is multiplication on the left by elements of $\Q[G]$.  Thus we may write $\Q[G]=\displaystyle\bigoplus_{i=1}^s I_i(\Q[G])$.  Each $I_i(\Q[G])$ is not only a $\Q[G]$-module, it is also a ring with unity $1_i$.  From the theory of representations of finite groups, $Q[G]$ decomposes as a product of rings $B_1\times B_2\times\dots\times B_s$ where $B_i$ is the $M_i$-isotypic component.  Moreover, each $(0,0,\dots, 1_i,0,\dots, 0)$ acts on $M_i$ as identity and on $M_j$ as $0$ for each $j\neq i$.  Thus, if we view $B_i$ as a subring in this way, multiplying by $1_i$ is the same as projection to the $M_i$-isotypic component. 
	
	We now consider our standard finite group $G=W_\Gamma^{ab}\cong(\Z/2\Z)^n$.  Since $G$ is finite, abelian, and all elements are of order $2$, its irreducible rational representations are all one-dimensional over $\Q$.  In fact the irreducible rational representations of $G$ are precisely $\rho_J: G\to\Q^\times$ where $J\subseteq\pi(V(\Gamma))$ and $$\rho_J(\pi(v))=\begin{cases}-1\quad\text{if } \pi(v)\in J\\
	1 \quad\text{if } \pi(v)\notin J\end{cases}.$$  Let $M_J$ denote the irreducible module corresponding to the representation $\rho_J$.  Then for distinct $M_{J_1}$ and $M_{J_2}$ we have that $M_{J_1}\not\cong M_{J_2}$ as left $\Q[G]$-modules.  Therefore each $M_J$ is an isotypic component of $\Q[G]$.  We therefore index the isotypic components of $\Q[G]$-modules and their corresponding identity elements by subsets $J\subseteq \pi(V(\Gamma))$ rather than integers as above.
	
	Because the standard epimorphism $\pi:W_{\Gamma}\to G$ induces a bijection between $V(\Gamma)$ and $\pi(V(\Gamma))$, we can think of $J$ as indexing a subset of $V(\Gamma)$ rather than a subset of $\pi(V(\Gamma))$. We therefore identify $J$ with a subset of $V(\Gamma)$.  Hence the irreducible $\Q[G]$-modules are in bijective correspondence with $\mathcal{P}(V(\Gamma))$.  
	
	
	\section{Subgraph Lemma}\label{sect subgraph lemma}
	
	We now return to considering the representation $\rho_{\Gamma}$.  Recall this arose by considering the action of $\Gamma(G,\pi)$ on the $\Q[G]$-module $H_1(\hat{X};\Q)$.  It turns out to be much simpler to consider the action of $\Gamma(G,\pi)$ on the isotypic components of $H_1(\hat{X};\Q)$.  
	
	Let $J\subseteq V(\Gamma)$.  Let $\Gamma_{J}$ denote the subgraph of $\Gamma$ induced by $J$, let $k_{J}$ denote the number of connected components of $\Gamma_{J}$, and let $[v]_J$ denote the connected component of $\Gamma_J$ containing $v$.  It turns out that decomposing $H_1(\hat{X};\Q)$ into its isotypic components is simply a matter of understanding the subgraphs $\Gamma_J$  (the subgraph induced by the vertices in $J$).  To that end, we now prove the Subgraph Lemma (Lemma \ref{subgraph})
	
	\begin{customlem}{\ref{subgraph}}[The Subgraph Lemma]
		Let $\Gamma$ be a finite graph, and let $J\subseteq V(\Gamma)$.  Then as $\Q[G]$-modules, $I_{J}(H_1(\hat{X}; \Q))\cong
		\langle 1_J(\hat{v}-\hat{w}) \hspace{1mm}|\hspace{1mm} v,w\in V(\Gamma_J)\rangle/
		\langle 1_J(\hat{v}-\hat{w}) \hspace{1mm}|\hspace{1mm} (v,w)\in E(\Gamma_{J})\rangle\cong I_{J}(\Q[G])^{k_{J}-1}$.  
	\end{customlem}
	
	
	\begin{proof}[Proof of Subgraph Lemma]
		Let $d_i: C^{cw}_i(\hat{X}; \Q)\to C^{cw}_{i-1}(\hat{X}; \Q)$ be the boundary maps on the cellular chain complex and let $p_{i,J}: C^{cw}_i(\hat{X};\Q)\to  I_J(C^{cw}_i(\hat{X};\Q))$ denote projection onto the isotypic component corresponding to $J$.  Because $d_i(1_J\hat v)=1_Jd_i(\hat v)$, the $p_{i,J}$ induce a boundary map on the chain complex \\ $I_J(C^{cw}_i(\hat{X};\Q))$ such that the following diagram commutes.
		
		\begin{center} $ \begin{tikzcd}
			\dots \arrow[rightarrow]{r}{d_3} 
			& C^{cw}_2(\hat{X};\Q) \arrow{d}{}\arrow[rightarrow]{r}{d_2} \arrow[swap]{d}{p_{2,J}} 
			& C^{cw}_1(\hat{X};\Q) \arrow{d}{}\arrow[rightarrow]{r}{d_1} \arrow[swap]{d}{p_{1,J}}
			& C^{cw}_0(\hat{X};\Q) \arrow{d}{}\arrow[swap]{d}{p_{0,J}} \\
			\dots \arrow[rightarrow]{r}{\bar{d}_3} 
			&  I_J(C^{cw}_2(\hat{X};\Q)) \arrow[rightarrow]{r}{\bar{d}_2} 
			& I_J(C^{cw}_1(\hat{X};\Q)) \arrow[rightarrow]{r}{\bar{d}_1} 
			& I_J(C^{cw}_0(\hat{X};\Q))
			\end{tikzcd}$ \end{center}
		
		\noindent Furthermore $I_J(H^{cw}_1(\hat{X};\Q))\cong\ker(\bar d_1)/\text{Im}(\bar d_2)$ is just the first homology of the induced chain complex.  
		
		Let $\hat v$ be an element of $C^{cw}_1(\hat{X};\Q)$. Then $d_1(\hat v)=(\pi(v)-1)\hat{*}$.	
		Projecting onto $I_J(C^{cw}_i(\hat{X};\Q))$ is the same as multiplying by $1_J$.  Thus we compute
		$$\bar{d}_1(1_J\hat v)=
		1_J(\pi(v)-1)\hat{*} 
		=(\rho_J\circ\pi(v)-1)1_J\hat *
		=\begin{cases} -2\cdot 1_J\hat * \quad\text{if } v\in J\\
		0 \quad\text{if }v\notin J\end{cases}$$
		
		\noindent Therefore $\ker(\bar d_1)=\langle 1_J(\hat v - \hat w)\;|\; v,w\in V(\Gamma_J)\rangle\oplus \langle 1_J\hat v \;|\; v\notin V(\Gamma_J)\rangle$.
		
		We now wish to compute $\text{Im}(\bar d_2)$.  Let $f_v$ denote the 2-cell in $\hat{X}$ attached to the lift of $v^2$ based at $\hat *$.  Then $d_2(f_v)=(\pi(v)+1)\hat v$. Let $f_{v,w}$ denote the 2-cell in $\hat{X}$ attached to the lift of $[v,w]$ based at $\hat *$.  Then $d_2(f_{v,w})= (1-\pi(v))\hat{w}+(\pi(w)-1)\hat{v}.$ 
		
		\noindent Looking now at the induced boundary map, we see that
		\begin{center}
			$\bar d_2(1_Jf_v)=\displaystyle (\rho_J\circ\pi(v)+1)1_J\hat{v}=\begin{cases} 2\cdot 1_J\hat{v}\quad\text{if }v\notin J \\
			0 \quad\text{if } v\in J\end{cases}$
		\pagebreak
			
			$\bar d_2(1_Jf_{v,w})= (1-\rho_J\circ\pi(v))1_J\hat{w} + (\rho_J\circ\pi(w)-1)1_J\hat{v}=\begin{cases} 2\cdot 1_J(\hat w -\hat v)\quad\text{if } v,w\in J\\
			-2\cdot 1_J\hat w \quad \text{if } v\in J, w\notin J\\
			2\cdot 1_J\hat w \quad \text{if } v\notin J, w\in J\\
			0 \quad\text{if } v,w\notin J\end{cases}$
		\end{center}
		
		
		\noindent It follows that $\text{Im}(\bar d_2)=\langle 1_J(\hat v - \hat w) \;|\; (v,w)\in E(\Gamma_J)\rangle \oplus \langle 1_J\hat v \;|\; v\notin V(\Gamma_J)\rangle$. Therefore 
		$$I_J(H_1(\hat X; \Q))\cong I_J(H^{cw}_1(\hat X; \Q))=(\langle 1_J(\hat v - \hat w) \;|\; v,w\in V(\Gamma_J)\rangle\oplus \langle 1_J\hat v \;|\; v\notin V(\Gamma_J)\rangle)/$$
		$$(\langle1_J(\hat v - \hat w) \;|\; (v,w)\in E(\Gamma_J)\rangle\oplus \langle 1_J\hat v \;|\; v\notin V(\Gamma_J)\rangle).$$
		
		This gives us the first isomorphism.  To see the second one, let $\{v_1,\dots, v_{k_J}\}$ be a set of representatives of the components of $\Gamma_J$.  Then $\{1_J(\hat v_{i+1}-\hat v_i)\;|\; 1\leq i < k_J\}$ is a basis for $I_J(H^{cw}_1(\hat{X};\Q))$.  
	\end{proof}
	
	We now return to the automorphism group $\Gamma(G,\pi)$, which we showed is equal to $\text{Aut}^0(W_{\Gamma})$ (Lemma \ref{gamma}). Let $\rho_{\Gamma,J}$ denote the projection of $\rho_{\Gamma}$ onto $\text{Aut}(I_J(H_1(\hat{X};\Q)))$.  We will mostly be studying the images of these $\rho_{\Gamma,J}$.   When there is no room for ambiguity, given any $\sigma\in\text{Aut}^0(W_\Gamma)$, we let $\hat{\sigma}$ denote the map on $C_1^{cw}(\hat X;\Q)/\delta_2(C_2^{cw}(\hat X;\Q))$ induced by $\sigma$ described in Section \ref{sect constructing the representation}.  Since $\hat{\sigma}$ is linear, and elements of $I_J(H_1(\hat{X};\Q))$ are essentially linear combinations of elements of $C_1^{cw}(\hat X;\Q)/\delta_2(C_2^{cw}(\hat X;\Q))$, we also let $\hat{\sigma}$ denote the matrix $\rho_{\Gamma,J}(\sigma)$.  It will always be clear from context which definition of $\hat\sigma$ is being used.
	
	The following corollary tells us that we can ignore partial conjugations by elements not in $J$.  Recall that $[v]_J$ is the component of $\Gamma_J$ containing $v$.
	\begin{cor}\label{stay in J}
		For any $v\notin J$ and any partial conjugation $\sigma=\sigma_{D,v}$ of D by v, we have that $\hat{\sigma}=\text{Id}$.  Furthermore, for any partial conjugation $\sigma=\sigma_{D,v}$ such that $v\in J$ and $D\cap J\setminus[v]_J=\emptyset$, we have that $\hat{\sigma}=\text{Id}$.
	\end{cor}
	
	\begin{proof}
		For the first statement, if $w\notin D$, then $\sigma(w)=w$, hence $\hat\sigma(1_J\hat w) = 1_J\hat w$.  If $w\in D$, then $\sigma(w)=vwv^{-1}=vwv$.  In this case, the induced automorphism on $C_1(\hat{X};\Q)$ maps the 1-chain $\hat w$ to the 1-chain $\hat v + \pi(v)\hat w + \pi(vw)\hat v$.  Note that $v\notin J\implies 1_J\hat{v}=0$.  Projecting onto $I_J(H_1(\hat{X};\Q))$, we get that $\hat\sigma$ maps $1_J\hat w$ to $1_J(\hat v + \pi(v)\hat w + \pi(vw)\hat v) = 1_J\hat v  + \rho_J\circ\pi(v)1_J\hat w + \rho_J\circ\pi(vw) 1_J\hat v =  1_J\hat w$.  Therefore $\hat\sigma$ acts as identity on all the 1-chains mod boundaries in $I_J(C_1(\hat X; \Q))$, so $\hat\sigma=\Id$.
		
		For the second statement, let $1_J\hat w \in I_J(C_1(\hat X; \Q))$ be non-zero.  Then if $w\notin D$, we have that $\sigma(w)=w$.  If $w\in D$, then by hypothesis $w\in [v]_J$.  By the Subgraph Lemma (Lemma \ref{subgraph}), $1_J\hat v = 1_J \hat w$ inside of $I_J(H_1(\hat{X};\Q))$.  By the above calculation, $\hat\sigma(1_J\hat w ) =  1_J\hat v  + \rho_J\circ\pi(v)1_J\hat w + \rho_J\circ\pi(vw) 1_J\hat v =2(1_J\hat v)- 1_J \hat w$.  Inside of $I_J(H_1(\hat{X};\Q))$ this equals $1_J\hat w$.  Therefore $\hat\sigma=\Id$.  
	\end{proof} 
	
	To make use of the above corollary we establish the following notation.  Let $\Gamma$ be a graph, $J\subseteq V(\Gamma)$.  We define 
	$$\mathcal{P}^{\Gamma,J}:=\{\sigma_{D,v}\;|\; v\in J, D\cap J\setminus[v]_J\neq\emptyset\}.$$  
	We also define the following subgroup of $\text{Aut}^0(W_\Gamma)$.
	$$\text{Aut}^J(W_\Gamma):=\langle \mathcal{P}^{\Gamma,J}\rangle.$$
	
	Given $\sigma_{D,v}\in\mathcal{P}^{\Gamma,J}$, to compute $\hat\sigma_{D,v}$ it suffices to know how $\hat\sigma_{D,v}$ acts on vectors of the form $\hat b - \hat a$ for $a,b\in J$.  That computation is the content of the following lemma.
	\begin{lem}\label{computation}
		Let $\sigma_{D,v}\in\mathcal P^{\Gamma,J}$ and $a,b\in J$.
		\begin{enumerate}
			\item If $a,b\notin D$, then $\hat{\sigma}_{D,v}(1_J(\hat b - \hat a ))= 1_J(\hat b - \hat a)$.
			
			\item If $a,b\in D$, then $\hat\sigma_{D,v}(1_J(\hat b - \hat a ))= -1_J (\hat b - \hat a)$.
			
			\item If $b\in D, a\notin D$, then $\hat\sigma_{D,v}(1_J(\hat b - \hat a ))=2\cdot 1_J(\hat v - \hat b) + 1_J (\hat b - \hat a)$.
			
			\item If $a\in D, b\notin D$, then $\hat\sigma_{D,v}(1_J(\hat b - \hat a ))=2\cdot 1_J(\hat v - \hat a) - 1_J(\hat b - \hat a)$.
		\end{enumerate}
	\end{lem}
	
	\begin{proof}
		If $w\notin D$, then $\sigma_{D,v}(w)=w$, so $\hat\sigma_{D,v}(1_J\hat w) = 1_J\hat w$.  If $w\in D$, then $\sigma_{D,v}(w)=vwv$, so $\hat\sigma_{D,v}(1_J\hat w)= 1_J \hat v +\rho_J\circ \pi(v) 1_J \hat w + \rho_J\circ \pi(vw) 1_J \hat v = 2\cdot 1_J \hat v - 1_J\hat w$.  The result follows by linearity.
	\end{proof}
	
	Knowing how partial conjugations act on vectors of the form $\hat b - \hat a$ for $a,b\in J$, we see that writing $1_J$ all the time isn't really necessary.  Therefore we will abuse notation and will set $\hat v = 1_J \hat v$.  
	
	Corollary \ref{stay in J} allows us to restrict the domain of $\rho_{\Gamma,J}$ to $\text{Aut}^J(W_\Gamma)$ when computing its image. The following lemma will allow us to restrict the codomain of $\rho_{\Gamma,J}$ as well.
	
	\begin{lem}\label{congruence}
		For any choice of basis of $I_J(H_1(\hat{X};\Q))$ where every vector is of the form $\hat v - \hat w $ for some $v,w\in V(\Gamma)$, we have $\text{Im}(\rho_{\Gamma,J})\leq\Gamma_{k_J-1}(2)$ where $\Gamma_{n}(2):=\ker(\text{GL}_n(\Z)\to \text{GL}_n(\Z/2\Z))$.
	\end{lem}
	
	\begin{proof}
		Fix a basis $\{\hat{v}_{i}-\hat{v}_{i-1}\hspace{1mm}|\hspace{1mm} 1<i\leq k_{J}\}$ of $I_{J}(H_1(\hat{X};\Q))$.
		By the above commentary, we need only consider the images of elements of $\mathcal{P}^{\Gamma,J}$.  Let $\sigma_{D,v}\in\mathcal{P}^{\Gamma,J}$.  Then by Lemma \ref{computation}, we have $$\hat\sigma_{D,v}(\hat{v}_{i+1}-\hat{v}_i)=a(\hat{v}-\hat{v}_{i+1})+b(\hat{v}-\hat{v}_i)+c(\hat v_{i+1}-\hat v_i)$$ for some $a,b\in \{0,\pm 2\}$ and $c\in\{\pm 1\}$.  
		
		Now $\hat{v}-\hat{v}_i$ need not be a basis vector, but since $v,v_i\in J$ we can write $\hat{v}-\hat{v}_i$ as a linear combination of basis vectors with all coefficients equal to 1 or 0.  If  $v=v_j$ and $j<i$ then $\hat{v}-\hat{v}_i=\displaystyle\sum_{k=1}^{j-i} (\hat{v}_{j-k+1}-\hat{v}_{j-k})$ (similarly for $i<j$).  If $i=j$, then $\hat{v}-\hat{v}_i=0$.  Thus each basis vector maps to a linear combination of basis vectors where the coefficients are all $\pm2$s or 0s except the coefficient on the original basis vector, which is $\pm 1$.  This shows that $\hat{\sigma}\in\Gamma_n(2)$.
	\end{proof}
	
	By virtue of Lemma \ref{congruence} and Corollary \ref{stay in J}, we adopt the standing assumption that $\rho_{\Gamma,J}:\text{Aut}^J(W_\Gamma)\to \Gamma_{k_J-1}(2)$.  We now show that this is the most we can restrict the codomain in general.  However, we first introduce the following notation which will be used henceforth.  Given non-adjacent vertices, $v,w\in J$, let $D(v,w)$ denote the component of $\Gamma\setminus\text{lk}(v)$ containing $w$.  We use $\text{lk}(v)$ in this definition rather than $\text{st}(v)$ so that we have $D(v,v)=\{v\}$.  This will be useful later.
	
	
	\begin{thm}\label{discrete}
		Let $\emptyset\neq J\subseteq V(\Gamma)$.  Let $K$ be a set of representatives of the components of $\Gamma_J$. Assume that for each pair of distinct vertices $v,w\in K$ we have $D(v,w)\cap K= \{w\}$.  Then $\text{Im}(\rho_{\Gamma, J})=\Gamma_{k_J-1}(2)$.  In particular, if $\Gamma$ is discrete, then $\text{Im}(\rho_{\Gamma, J})=\Gamma_{|J|-1}(2)$ for every nonempty $J\subseteq V(\Gamma)$.
	\end{thm}
	
	\begin{proof}
		Let $K=\{v_1,v_2,\dots, v_{k_J}\}$.  It is known that for all $n\geq 1$ the group $\Gamma_n(2)$ is generated by $\{E_{i,j}, F_i\;|\; 1\leq i,j\leq n, i\neq j\}$ where $E_{i,j}$ is the matrix identical to the identity matrix except in the $(i,j)$-entry, which equals 2, and $F_i$ is the matrix identical to the identity matrix except in the $(i,i)$-entry, which equals -1 (see, for example, \cite{k}).  
		
		Fix the basis $\hat{v}_{k_J}-\hat{v}_1,\hat{v}_{k_J}-\hat{v}_2,\dots, \hat{v}_{k_J}-\hat{v}_{k_J-1}$ of $I_J(H_1(\hat{X};\Q))$.  By Lemma \ref{congruence}, we know that $\text{Im}(\rho_{\Gamma,J})\leq \Gamma_{k_J-1}(2)$.  Thus it suffices to show that each element of $\{E_{i,j}, F_i\;|\; 1\leq i,j\leq k_J-1, i\neq j\}$ is in $\text{Im}(\rho_{\Gamma,J})$.
		Let $1\leq i,j \leq k_J-1$ with $i\neq j$.  Then by direct calculation, $$\hat{\sigma}_{D(v_i,v_j),v_i}\cdot\hat{\sigma}_{D(v_{k_J},v_{j}),v_{k_J}}=E_{i,j}$$ $$\hat{\sigma}_{D(k_J,v_i),v_{k_J}}=F_i.$$ This completes the proof.
		
	\end{proof}
	
	We conclude this section with a couple of simple observations that will be used throughout
	\begin{lem}\label{simple}
		For any $J\subseteq V(\Gamma)$ the following hold.
		\begin{enumerate}
			\item For any partial conjugation $\sigma\in\mathcal{P}^{\Gamma,J}$, we have $\hat{\sigma}$ is of order 2.
			\item For any inner automorphism $\sigma$ relative to $J$ (i.e. $\sigma$ conjugates all of $J$ by some element $v$), we have $\hat{\sigma}=-\text{Id}$.
		\end{enumerate}
	\end{lem}
	\begin{proof}
		For the first statement, this follows immediately from the fact that $\sigma$ is of order 2, which follows from the fact that every generator $v\in V(\Gamma)$ of $W_{\Gamma}$ is of order 2.  For the second statement, this follows from the second case calculation in the proof of Lemma \ref{congruence}.
	\end{proof}
	
	
	\chapter{Computing $\text{Im}(\rho_{\Gamma,J})$}
	\section{Commutation relations}\label{sect commutation relations}
	In this section we prove a strong connection between commuting matrices in $\text{Im}(\rho_{\Gamma,J})$ and commuting outer automorphisms in $\text{Out}^0(W_{\Gamma}):=\text{Aut}^0(W_{\Gamma})/\text{Inn}(W_{\Gamma})$.
	To understand when two outer automorphisms commute, we must first recall the definition of a separating intersection of links (SIL).  We do so here. 
	
	\begin{definition}\label{SIL}
		Let $\Gamma$ be a graph.  A \underline{\textit{Separating Intersection of Links\textit{(SIL)}}} is a triple $(u,v\;|\;w)$ of vertices $u,v,w\in V(\Gamma)$ such that  $u$ and $v$ are non-adjacent and $w$ is not in the same component of $\Gamma\setminus(\text{lk}(u)\cap\text{lk}(v))$ as $u$ or $v$. One may also write $(u,v\;|\; D)$ where $D$ is the component of $\Gamma\setminus(\text{lk}(u)\cap\text{lk}(v))$ containing $w$. 
	\end{definition}
	
	One reason that SILs are useful is that if $(u,v\;|\;D)$ is a SIL, then $D$ is not only a component of $\Gamma\setminus(\text{lk}(u)\cup\text{lk}(v))$, it is also a component of $\Gamma\setminus\text{lk}(u)$ and  $\Gamma\setminus\text{lk}(v)$.  This will be the content of Lemma \ref{SIL comp eq}.  However, before we can prove that, we need
	the following lemma from \cite{gpr} (Lemma 4.3).
	
	\begin{lem}[GPR]\label{gpr}
		Let $\sigma_{C,v},\sigma_{D,w}$ be partial conjugations.  If $v,w$ are non-adjacent and $w\notin C$, then $C\cap D = \emptyset$ or $C\subseteq D$.
	\end{lem}
	
	For our purposes, the following form of the above lemma will be more useful.  Recall from Section \ref{sect subgraph lemma} the following notation.  Given non-adjacent vertices, $v,w\in J$, let $D(v,w)$ denote the component of $\Gamma\setminus\text{lk}(v)$ containing $w$.
	
	\begin{cor}\label{permissible}
		Let $u,v,w \in J$ be pairwise non-adjacent.  Then if $v\notin D(u,w)$ then $D(u,w)\subseteq D(v,w)$.
	\end{cor}
	
	\begin{proof}
		Consider the partial conjugations $\sigma_{D(u,w),u}$ and $\sigma_{D(v,w),v}$.  By Lemma \ref{gpr}, either $D(u,w)\cap D(v,w)=\emptyset$ or $D(u,w)\subseteq D(v,w)$.  But $w\in D(u,w)\cap D(v,w)$.  Therefore $D(u,w)\subseteq D(v,w)$.
	\end{proof}
	
	\begin{lem}\label{SIL comp eq}
		Let $v,w$ be non-adjacent vertices.  Then, $(v,w\;|\; D)$ is a SIL $\iff$ for all $d\in D$, we have $D(v,d)=D(w,d)=D$.
	\end{lem}
	\begin{proof}
		By the definition of a SIL, $D$ is a component of $\Gamma\setminus(\text{lk}(v)\cap\text{lk}(w))$.  Let $d\in D$.  Then by the definition of a SIL, we have that $w\notin D(v,d)$ and $v\notin D(w,d)$.  Since $D(v,d)\cap D(w,d)\neq\emptyset$, Corollary \ref{permissible} implies that $D(v,d)=D(w,d)$.  
		
		Let $u\in D(v,d)$.  Then there is a path $\alpha$ from $u$ to $d$ that does not pass through $\text{st}(v)$.  It follows that $\alpha$ does not pass through $\text{lk}(v)\cap\text{lk}(w)$, hence $u\in D$.  This shows that $D(v,d)\subseteq D$.  
		
		Now let $u\in D(v,d)^\complement$.  Let $\alpha$ be a path from $d$ to $u$.  We claim that $\alpha$ must pass through $\text{lk}(v)\cap\text{lk}(w)$.  Clearly it passes through $\text{lk}(v)$.  If $\alpha$ passed through $\text{lk}(w)$ prior to passing through $\text{lk}(v)$, then we would have a path $\beta$ from $d$ to $w$ that did not pass through $\text{lk}(v)$.  However $(v,w\;|\; d)$ is a SIL, so this is not the case.  Therefore $\alpha$ does not pass through $\text{lk}(w)$ prior to passing through $\text{lk}(v)$.  Similarly, $\alpha$ does not pass through $\text{lk}(v)$ prior to passing through $\text{lk}(w)$.  Since $\alpha$ passes through $\text{lk}(v)$, it follows that $\alpha$ passes through $\text{lk}(v)\cap\text{lk}(w)$.  
		This shows that $D(v,d)^\complement\subseteq D^\complement$ and hence $D(v,d)=D$.  By symmetry, $D(w,d)=D=D(v,d)$.  
		
		Next, assume that $D(v,d)=D(w,d)=D$ where $D$ is the component of $\Gamma\setminus(\text{lk}(v)\cap\text{lk}(w))$ containing $d$.  Then $v\notin D(v,d)=D$ and $w\notin D(w,d)=D$, so $(v,w\;|\; D)$ is a SIL.  
	\end{proof}  
	
	
	We now turn our attention to determining when matrices in $\rho_{\Gamma,J}(\mathcal{P}^{\Gamma,J})$ commute.  We establish the following notation.  Given an automorphism $\sigma\in\text{Aut}^0(W_{\Gamma})$, we let $\bar{\sigma}$ denote the image of $\sigma$ in $\text{Out}^0(W_{\Gamma})$.  The following lemma, though proved in \cite{gpr} is stated in our preferred form in \cite{ss} (Lemma 1.4).  It tells us when two outer automorphisms commute.
	
	\begin{lem}[SS]\label{conditions}
		Given two partial conjugations $\sigma_1:=\sigma_{C,u}$ and $\sigma_2:=\sigma_{D,v}$, we have that $\bar{\sigma}_1, \bar{\sigma}_2$ do not commute $\iff$ there exists some $w\in V(\Gamma)$ such that $(u,v\;|\;w)$ is a SIL and one of the following conditions is met:
		
		\begin{enumerate}
			\item $w\in C=D$
			\item $u\in D, v\in C$
			\item $v\in C, w\in D$
			\item $u\in D, w\in C$
		\end{enumerate}
	\end{lem}
	
	Recall that by Lemma \ref{simple}, every inner automorphism maps to $\pm\Id$.  Combining this with Lemma \ref{congruence}, we get the following commutative diagram.
	
	\begin{center} $ \begin{tikzcd}
		\text{Aut}^{J}(W_{\Gamma}) \arrow[rightarrow]{r}{} \arrow[swap]{d}{\rho_{\Gamma,J}} & \text{Out}^J(W_{\Gamma})  \arrow{d}{\bar{\rho}_{\Gamma,J}} \\
		\Gamma_{k_J-1}(2) \arrow[rightarrow]{r}{} & 	\Gamma_{k_J-1}(2)/\{\pm\Id\}
		\end{tikzcd}$ \end{center}
	
	\noindent We use this diagram to prove the following result. 
	
	\begin{prop}\label{if commute}
		Let $\sigma_{C,u},\sigma_{D,v}\in\mathcal{P}^{\Gamma,J}$.  If $\bar{\sigma}_{C,u}$  and $\bar{\sigma}_{D,v}$ commute then $\hat{\sigma}_{C,u}$ and $\hat\sigma_{D,v}$ commute.
	\end{prop}
	
	However before we prove this, we need the following lemma:
	
	\begin{lem}\label{nosqrt} There is no $A\in\Gamma_n(2)$ such that $A^2=-\text{Id}$
	\end{lem}
	\begin{proof}[Proof of Lemma \ref{nosqrt}]
		Let $A\in\Gamma_n(2)$.  Then either $A_{1,1}\equiv_4 1$ or $A_{1,1}\equiv_4 3$.  In either case, we see that $A_{1,1}^2\equiv_4 1$.  Furthermore, for all $2\leq i \leq n$, we have that $A_{1,i}A_{i,1}\equiv_4 0$.  Thus $(A^2)_{1,1}=A_{1,1}^2+\displaystyle\sum_{i=2}^n A_{1,i}A_{i,1}\equiv_4 1$.  But $(-\text{Id})_{1,1}\not\equiv_4 1$.  Therefore $A^2\neq -\Id$.  
	\end{proof}
	
	\begin{proof}[Proof of Proposition \ref{if commute}]
		If $\bar{\sigma}_1,\bar{\sigma}_2$ commute, then the above diagram implies that $[\hat{\sigma}_1,\hat{\sigma}_2]=\pm\text{Id}$. By the first statement of Lemma \ref{simple}, we have $([\sigma_1,\sigma_2])=(\hat{\sigma}_1\circ\hat{\sigma}_2)^2$.  Then by Lemma \ref{nosqrt}, we have $(\hat{\sigma}_1\circ\hat{\sigma}_2)^2\neq-\Id$.  Therefore $[\hat{\sigma}_1,\hat{\sigma}_2]=\text{Id}$, so $\hat{\sigma}_1$ and $\hat{\sigma}_2$ commute.
	\end{proof}
	
	The reverse direction to Proposition \ref{if commute} does not hold in general.  However it is easy to state when it holds. First we need the following definition.
	\begin{definition}	
		Let $\Gamma$ be a finite graph and let $J\subseteq V(\Gamma)$.  We say $(u,v\;|\;w)$ is a \underline{\textit{SIL relative to J}} if $(u,v\;|\; w)$ is a SIL, $u,v,w\in J$, and $[u]_J\neq[v]_J$.  We also say $(u,v\;|\;D)$ is a \underline{\textit{SIL relative to J}} if $(u,v\;|\; w)$ is a SIL relative to $J$ for some $w\in D$.  
	\end{definition}
	
	We require $[u]_J\neq [v]_J$ in the above definition because if $[u]_J=[v]_J$ then $\hat u = \hat v$.  Thus this requirement is in some sense analogous to the requirement for SILs that $u$ and $v$ be non-adjacent.  We now show that partial conjugations by elements $u,v\in J$ such that $[u]_J=[v]_J$ commute.  
	
	\begin{prop}\label{same comp commute}
		Let $\sigma_{C,u},\sigma_{D,v}\in\mathcal{P}^{\Gamma,J}$ be such that $[u]_J=[v]_J$.  Then $\hat\sigma_{C,u}$ and $\hat\sigma_{D,v}$ commute.
	\end{prop}
	
	\begin{proof}
		Fix a basis $\mathcal{B}$ of $I_J(H_1(\hat X;\Q))$ such that each element of $\mathcal{B}$ is of the form $\hat u - \hat w$ for some $w\notin[u]_J$.  Then for each $\hat u - \hat w \in \mathcal{B}$ we have that $\hat\sigma_{C,u}(\hat u - \hat w) = \pm (\hat u - \hat w)$.  Therefore $\hat\sigma_{C,u}$ is a diagonal matrix.  Similarly, since $\hat u = \hat v$, we get that $\hat\sigma_{D,v}$ is a diagonal matrix.  Therefore $\hat\sigma_{C,u}$ and $\hat\sigma_{D,v}$ commute.
	\end{proof}
	
	\begin{cor}\label{no SILs rel J}
		Given $u,v\in J$, if $u$ and $v$ do not form a SIL relative to $J$, then $\hat\sigma_{C,u},\hat\sigma_{D,v}$ commute for all $\sigma_{C,u},\sigma_{D,v}\in\mathcal{P}^{\Gamma,J}$.  
	\end{cor}
	
	\begin{proof}
		Let $\sigma_{C,u},\sigma_{D,v}\in\mathcal{P}^{\Gamma,J}$.  Assume that $u$ and $v$ do not form a SIL relative to $J$.  If $[u]_J=[v]_J$, then $\hat\sigma_{D,v}$ commutes with $\hat\sigma_{C,u}$ by Proposition \ref{same comp commute}.  If not, then by Proposition \ref{if commute}, $\hat\sigma_{D,v}$ commutes with $\hat\sigma_{C,u}$ unless $\bar\sigma_{D,v}$ does not commute with $\bar\sigma_{C,u}$.  By Lemma \ref{conditions}, this implies that $(u,v\;|\; w)$ is a SIL for some $w\in V(\Gamma)$ and one of the four conditions is met.  Since $u$ and $v$ do not form a SIL relative to $J$, it follows from Lemma \ref{SIL comp eq} that $w\notin C$ and $w\notin D$.    Thus we must have that $u\in D, v\in C$.
		
		We now show by contradiction that $(C\cup D)\cap J= J$.  Let $j\in J\setminus(C\cup D)$.  Then since $j\notin C = D(u,v)$, it follows that $v\notin D(u,j)$.  Corollary \ref{permissible} implies that $D(u,j)\subseteq D(v,j)$.  Similarly, since $j\notin D=D(v,u)$, we get that $D(v,j)\subseteq D(u,j)$.  This implies that $D(u,j)=D(v,j)$.  But then by Lemma \ref{SIL comp eq}, we have that $(u,v\;|\; D(u,j))$ is a SIL relative to $J$.  This is a contradiction.  Therefore $(C\cup D)\cap J = J$.  
		
		By direct computation, for any $w\in J$ we have  $[\sigma_{C,u},\sigma_{D,v}](w)= uvuvwvuvu$ so $[\hat\sigma_{C,u},\hat\sigma_{D,v}]=\text{Id}$.  This shows that $\hat\sigma_{C,u}$ and $\hat\sigma_{D,v}$ commute.
	\end{proof}
	
	We now have all the tools we need to say exactly when the images of two partial conjugations commute.
	
	\begin{prop}\label{commute}
		Let $\sigma_{C,u},\sigma_{D,v}\in\mathcal{P}^{\Gamma,J}$. Then $\hat \sigma_{C,u}$ and $\hat\sigma_{D,v}$ do not commute $\iff$ there exists some $w\in J$ such that $(u,v\;|\; w)$ is a SIL relative to $J$ and one of the following conditions is met:
		\begin{enumerate}
			\item $w\in C=D$
			\item $u\in D, v\in C$
			\item $v\in C, w\in D$
			\item $u\in D, w\in C$
			
		\end{enumerate}
	\end{prop}
	
	\begin{proof}
		Assume that $\hat\sigma_{C,u}$ and $\hat\sigma_{D,v}$ do not commute.  Then by the contrapositive of Proposition $\ref{if commute}$, we see that $\bar\sigma_{C,u}$ and $\bar\sigma_{D,v}$ do not commute.  By Lemma \ref{conditions}, this implies that there exists some $w\in J$ such that $(u,v\;|\; w)$ is a SIL and one of the four conditions is met.  Note that the contrapositive of Proposition \ref{same comp commute} implies that $[u]_J\neq [v]_J$. If $D(u,w)=C$, then $(u,v\;|\; C)$ is a SIL relative to $J$ and one of the four conditions is met.  Similarly, if $D(u,w)=C$, then $(u,v\;|\; C)$ is a SIL relative to $J$ and one of the four conditions is met.  Thus we can assume that $u\in D$ and $v\in C$.  In this case, it remains to show that $u$ and $v$ form a SIL relative to $J$.  This follows from the contrapositive to Corollary \ref{no SILs rel J}.
		
		Now assume that there exists some $w\in J$ such that $(u,v\;|\; w)$ is a SIL relative to $J$ and one of the four conditions holds.  We check case by case that $\hat\sigma_{C,u}$ and $\hat\sigma_{D,v}$ do not commute by finding a vector in $I_J(H_1(\hat X, \Q))$ which $[\hat\sigma_{C,u},\hat\sigma_{D,v}]$ does not fix.
		
		\begin{enumerate}
			\item $(w\in C=D)$:  Then $[\hat\sigma_{C,u},\hat\sigma_{D,v}](\hat{w}-\hat{u})=4\hat{v}-4\hat{u}+\hat{w}-\hat{u}$.
			\item $(u\in D, v\in C)$:  In this case, neither $(u,v\;|\; C)$ nor $(u,v\;|\; D)$ is a SIL.  Therefore $w\notin C\cup D$,  It follows that $[\hat\sigma_{C,u},\hat\sigma_{D,v}](\hat{w}-\hat{v})= 4\hat{v}-4\hat{u}+\hat{w}-\hat{v}$.
			\item $(v\in C, w\in D)$:  Since $w\in D$, we have that $(u,v\;|\; D)$ is a SIL.  Hence $u\notin D$.  It follows that $[\hat\sigma_{C,u},\hat\sigma_{D,v}](\hat{w}-\hat{u})= 4\hat{v}-4\hat{u}+\hat{w}-\hat{u}$.
			\item $(u\in D, w\in C)$:  Then
			$[\hat\sigma_{C,u},\hat\sigma_{D,v}](\hat{w}-\hat{v})= 4\hat{v}-4\hat{u}+\hat{w}-\hat{v}$.
		\end{enumerate}
		
		Since $[u]_J\neq [v]_J$, we have that $4\hat{v}-4\hat u\neq 0$ in $I_J(H_1(\hat{X};\Q))$.  Thus each of the above calculations show that $[\hat\sigma_{C,u},\hat\sigma_{D,v}]\neq\text{Id}$.  
	\end{proof}
	
	\section{TML Decomposition}\label{sect tml decomposition}
	The goal of this section is to prove a decomposition theorem that will allow us to reduce the size of our vertex set $J$ subject to certain conditions.  The basic idea of the argument is to partition $J$ into "components" $J=\displaystyle\bigsqcup_{i=1}^m A_i$ in a nice way.  We use this partition to break $\text{Im}(\rho_{\Gamma,J})$ into three different subgroups.  
	
	\begin{itemize}
		\item The subgroup $\mathcal{T}$ is generated by images of partial conjugations by elements of $A_i$ on subsets of $A_i$ for some $i$.  Thus, choosing the appropriate basis, you almost end up with block upper diagonal matrices with one block corresponding to each $A_i$.  Unfortunately, to make a complete basis, we need to include vectors which are a difference of vectors from distinct $A_i$.  We group these all into a block at the end.  After some slight modifications, you end up with matrices that look like this: $$\begin{bmatrix}M_1 & 0 & 0 & \dots & 0 \\
		0 & M_2 & 0 & \dots & 0 \\
		& & \ddots & &  \\
		0 & 0 & \dots & M_m & 0 \\
		0 & 0 & 0 & \dots & \text{Id}\end{bmatrix}.$$
		
		To clarify how $\mathcal{T}$ relates to our other groups, we compress the non-identity blocks and think of $\mathcal{T}$ as consisting of block 2x2 matrices
		
		$$\begin{bmatrix}M & 0 \\ 0 & \text{Id}\end{bmatrix}.$$
		
		The blocks for our other subgroups correspond to the blocks here.
		
		\item The subgroup $\mathcal{L}$ is generated by images of partial conjugations by elements of $A_i$ on things outside of $A_i$.  We choose our $A_i$ in such a way that such elements can only act as $\pm\text{Id}$ on the blocks corresponding to $A_i$.  This can be easily modified so as to act as $\text{Id}$ on these blocks.  In so doing, we end up with matrices of the following form:
		
		$$\begin{bmatrix}\Id & M \\
		0 & \text{Id}\end{bmatrix}.$$
		
		\item In making the modifications for the subgroup $\mathcal{L}$, some partial conjugations end up getting left out.  These get grouped together to form the subgroup $\mathcal{M}$.  Again modifications are made to ensure that $\mathcal{M}$ behaves nicely on the $A_i$ blocks, and we end up with matrices of the form $$\begin{bmatrix} \Id & 0 \\ 0 & M\end{bmatrix}.$$
	\end{itemize}
	
	When considering the block diagonal forms of these matrices, it seems natural to expect that $\text{Im}(\rho_{\Gamma,J})=(\mathcal{T}\times\mathcal{M})\ltimes \mathcal{L}$.  Furthermore, each of these subgroups is either isomorphic to $\text{Im}(\rho_{\Gamma,J'})$ for some $J'\subsetneqq J$ or to a known group.
	
	Before we define our partition condition, we must first make a different definition.
	
	\begin{definition}
		Let $A\subseteq J$ be a non-empty union of components of $\Gamma_J$.  Then a point $a\in A$ is a \underline{special point} if the following hold:
		\begin{enumerate}
			\item For each $v\in [a]_J$ and each component $D$ of $\Gamma\setminus\text{st}(v)$, either $D\cap J\setminus[v]_J\subseteq A$ or $D\cap J\setminus[v]_J\subseteq A^\complement$
			\item For each $v\in A^\complement$, we have that $A\subseteq D(v,a)$.  
		\end{enumerate}
	\end{definition}
	
	The reason a special point is special is that, for any $v\in J$, it provides some level of control over which components of $\Gamma\setminus\text{lk}(v)$ intersect $A$.  If $v\in A^\complement$, we see by the second condition that only one component of $\Gamma\setminus\text{lk}(v)$ intersects $A$.  If $v\in [A]_J$, the first condition ensures that any component that intersects $A$ (up to elements of $[v]_J$) is in fact a subset of $A$.  To see what happens to elements in $A\setminus[a]_J$, we prove the following lemma.
	
	\begin{lem}\label{special point}
		Let $A\subseteq J$ be a non-empty union of components of $\Gamma_J$,  Let $a\in A$ be such that for each $v\in [a]_J$ and each component $D$ of $\Gamma\setminus\text{st}(v)$ we have that either $D\cap J\setminus[v]_J\subseteq A$ or $D\cap J\setminus[v]_J\subseteq A^\complement$.  Then for all $b\in A\setminus[a]_J$ and every component $D$ of $\Gamma\setminus\text{st}(b)$ the following hold.
		
		\begin{enumerate}
			\item If $a\notin D$, then either $D\cap J\subseteq A $ or $D\cap J\subseteq A^\complement$.
			\item If $D\subseteq A^\complement$, we have that $D$ is a component of $\Gamma\setminus\text{st}(a)$.
		\end{enumerate}
	\end{lem}
	
	\begin{proof}
		\begin{enumerate}
			\item Let $d\in D$ so that $D=D(b,d)$.  Then by Corollary \ref{permissible}, we have that $D(b,d)\subseteq D(a,d)$.  But by hypothesis, we have that $D(a,d)\cap J\setminus[a]_J\subseteq A$ or $D(a,d)\cap J \setminus[a]_J\subseteq A^\complement$.  Since $[a]_J\cap D = \emptyset$, this shows that $D\cap J \subseteq A$ or $D\cap J \subseteq A^\complement$.
			
			\item Let $d\in D$ so that $D=D(b,d)$.  Since $a\notin D(b,d)$, Corollary \ref{permissible} implies that $D(b,d)\subseteq D(a,d)$.  Furthermore, since $d\in A^\complement$, by hypothesis we have that $D(a,d)\cap J \setminus[a]_J\subseteq A^\complement$. In particular, we see that $b\notin D(a,d)$.  Therefore Corollary \ref{permissible} implies that $D(a,d)\subseteq D(b,d)$.  This shows that $D=D(b,d)=D(a,d)$.
		\end{enumerate}
	\end{proof}
	
	We now define the partition condition necessary for the TML decomposition to hold.
	
	\begin{definition}
		Let $J\subseteq V(\Gamma)$.  We say  $J=\displaystyle\bigsqcup_{i=1}^m A_i$ is a \underline{division} of $J$ if the following hold.
		\begin{enumerate}
			\item Each $A_i$ is a union of components of $\Gamma_J$.
			\item Each $A_i$ has a special point $a_i$.
		\end{enumerate}
		
	\end{definition}
	
	To get a sense of this definition we state a few examples.
	\begin{example}
		Let $J=\displaystyle\bigsqcup_{i=1}^m[v_i]_J$.  Then define $A_i:=[v_i]_J$.  Then it is trivial that $J=\displaystyle\bigsqcup_{i=1}^m A_i$ is a division of $J$.  Thus every $J$ has a trivial division.  However, this division will not be useful to us.
	\end{example}
	\begin{example}
		Let $\Gamma$ have $m$ components $D_1,D_2,\dots, D_m$ that intersect $J$ for some $m>1$.  For each $1\leq i \leq m$, let $A_i:=D_i\cap J$.  Since each component of $\Gamma_J$ lies in a component of $\Gamma$, each $A_i$ is a union of components of $\Gamma_J$. For each $1\leq i \leq m$, fix some $a_i\in A_i$.  Then for any $1\leq i \leq m$, any $v\in [a_i]$, and any $w\in A_j$ with $j\neq i$, we have $D(v,a_j)\cap J=A_j$ and $D(a_j,a_i)\cap J=A_i$.  Thus each $a_i$ is a special point, and $J=\displaystyle\bigsqcup_{i=1}^m A_i$ is a division of $J$.
	\end{example}
	
	\begin{example}
		Let $\Gamma$ be a tree, and let $J$ be the set of leaves of $\Gamma$.  Given $v,w\in J$, we say $v\sim w$ if $\text{lk}(v)=\text{lk}(w)$.  Then $\sim$ is an equivalence relation.  Assume that there is more than one equivalence class under this relation, and let $A_1,A_2,\dots, A_m$ be the equivalence classes.  Since $J$ consists of pairwise non-adjacent vertices, each $A_i$ is a union of components of $\Gamma_J$.  For each $1\leq i \leq m$, fix some $a_i\in A_i$.  Then for any $v\sim w$ we have that $D(v,w)=\{w\}$.  Furthermore, if $v\nsim w$, then $\text{lk}(v)\cap\text{lk}(w)=\emptyset$.  Therefore for any $v\in A_i, w\in A_i^\complement$, we have that $D(w,v)\supseteq A_i$.  This shows that $J=\displaystyle\bigsqcup_{i=1}^m A_i$ is a division of $J$.
	\end{example}
	
	\begin{example}  In the previous examples, every point was a special point relative to $A_i$.  This need not be the case.  Consider, for example the following graph:
		
		\begin {center}
		\begin{tikzpicture}
		
		\draw (0,0) -- (1,2);
		\draw (0,0) -- (3,2);
		\draw (0,0) -- (2,0);
		\draw (0,0) -- (1,-1);
		\draw (0,0) -- (1,-2);
		\draw (2,1) -- (2,0);
		\draw (2,1) -- (1,2);
		\draw (2,1) -- (3,2);
		\draw (2,0) -- (2,-1);
		\draw (2,-1) -- (3,-1);
		\draw (2,-1) -- (3,-2);
		\draw (2,-1) -- (1,-1);
		\draw (2,-1) -- (1,-2);
		
		\draw (0,0) -- (-1,2);
		\draw (0,0) -- (-3,2);
		\draw (0,0) -- (-2,0);
		\draw (0,0) -- (-1,-1);
		\draw (0,0) -- (-1,-2);
		\draw (-2,1) -- (-2,0);
		\draw (-2,1) -- (-1,2);
		\draw (-2,1) -- (-3,2);
		\draw (-2,0) -- (-2,-1);
		\draw (-2,-1) -- (-3,-1);
		\draw (-2,-1) -- (-3,-2);
		\draw (-2,-1) -- (-1,-1);
		\draw (-2,-1) -- (-1,-2);
		
		
		\filldraw[black] (0,0) circle (2pt) node[anchor=north] {};
		
		\filldraw[black] (2,1) circle (2pt) node[anchor=north] {};
		\filldraw[black] (2,0) circle (2pt) node[anchor=west] {$v_3$};
		\filldraw[black] (2,-1) circle (2pt) node[anchor=west] {};
		\filldraw[black] (1,2) circle (2pt) node[anchor=south] {$v_1$};
		\filldraw[black] (3,2) circle (2pt) node[anchor=south] {$v_2$};
		\filldraw[black] (3,-1) circle (2pt) node[anchor=north] {$v_4$};
		\filldraw[black] (3,-2) circle (2pt) node[anchor=north] {$v_5$};
		\filldraw[black] (1,-1) circle (2pt) node[anchor=north] {$v_6$};
		\filldraw[black] (1,-2) circle (2pt) node[anchor=north] {$v_7$};
		
		\filldraw[black] (-2,1) circle (2pt) node[anchor=north] {};
		\filldraw[black] (-2,0) circle (2pt) node[anchor=east] {$v_{10}$};
		\filldraw[black] (-2,-1) circle (2pt) node[anchor=north] {};
		\filldraw[black] (-1,2) circle (2pt) node[anchor=south] {$v_8$};
		\filldraw[black] (-3,2) circle (2pt) node[anchor=south] {$v_9$};
		\filldraw[black] (-3,-1) circle (2pt) node[anchor=north] {$v_{11}$};
		\filldraw[black] (-3,-2) circle (2pt) node[anchor=north] {$v_{12}$};
		\filldraw[black] (-1,-1) circle (2pt) node[anchor=north] {$v_{13}$};
		\filldraw[black] (-1,-2) circle (2pt) node[anchor=north] {$v_{14}$};
		
		
		\end{tikzpicture}
	\end{center}
	
	Let $J=\{v_1,v_2,\dots, v_{14}\}$.  Let $A_1=\{v_1,v_2,\dots, v_7\}$, and let $A_2=\{v_8,v_9,\dots, v_{14}\}$.  Then $J=A_1\sqcup A_2$ is a division of $J$.  However, $v_4$ and $v_5$ are not special points with respect to $A_1$ and $v_{11}$ and $v_{12}$ are not special points with respect to $A_2$.
\end{example}

To simplify notation, we extend our partial conjugation notation as follows.  We write $\sigma_{D\cap J,v}:=\sigma_{D,v}$ whenever $\sigma_{D,v}$ is defined.  Furthermore, since conjugating any element of $[v]_J$ by $v$ acts as identity in the homology, we may add or remove elements of $[v]_J$ to the set being conjugated and still have the same automorphism.  Thus for an arbitrary subset $D\subseteq J$, we define $\sigma_{D, v}:=\sigma_{D',v}$ for any $D'$ such that $\sigma_{D',v}$ is defined and $D\setminus[v]_J = D'\setminus[v]_J$.  Thus, for example, we may write $\sigma_{J,v}$ to represent an inner automorphism relative to $J$.

Recall that the group $\mathcal{T}$ consists of block diagonal matrices that are non-identity on a number of distinct blocks.  It will be beneficial to us later on to look at these blocks individually.  Therefore, before working with the full subgroup $T$, we consider a subgroup $T_1$ consisting of a single block.  Recall that $\mathcal{P}^{\Gamma,J}:=\{\sigma_{D,v}\;|\; v\in J, D\cap J\setminus[v]_J\neq\emptyset\}$.  
\begin{lem}\label{T0}
	Let $J\subseteq V(\Gamma)$ be an arbitrary vertex set.  Let $A\subsetneqq J$ be a non-empty union of components of $\Gamma_J$.  Assume there exists an $a\in A$ such that for each $v\in [a]_J$ and each component $D$ of $\Gamma\setminus\text{st}(v)$, either $D\cap J\setminus[v]_J\subseteq A$ or $D\cap J\setminus[v]_J\subseteq A^\complement$.    We define the following subgroup of $\text{Im}(\rho_{\Gamma,J})$.
	
	$$\mathcal{T}_1:=\langle \hat\sigma_{D,v}\in\rho_{\Gamma,J}(\mathcal{P}^{\Gamma,A})\;|\; a\notin D\rangle.$$
	
	Then with respect to the appropriate bases, $$\mathcal{T}_1 = \Big\{\begin{bmatrix} M & 0 \\ 0 & \text{Id}\end{bmatrix}\;\Big|\; M\in\text{Im}(\rho_{\Gamma,A})\Big\}\cong \mathcal{T}_1\cong\text{Im}(\rho_{\Gamma,A}).$$
\end{lem}
\begin{proof}
	Let $\mathcal{B}_A$ be an ordered basis of $\langle \hat v - \hat w\;|\; v,w\in A\rangle\leq I_J(H_1(\hat X; \Q))$  consisting of vectors of the form $\hat v - \hat w$.  Let $S$ be an ordered list of representatives of the components of $\Gamma_{J\setminus A}$.  We extend $\mathcal{B}_A$ to an ordered basis $\mathcal{B}$ of $I_J(H_1(\hat X; \Q))$ so that $\mathcal{B} = \mathcal{B}_A\cup\{\hat s - \hat a\;|\; s\in S\}.$
	
	Let $G:=\{\sigma_{D,v}\in\mathcal{P}^{\Gamma,A}\;|\;a\notin D\setminus[v]_J\}.$  Let $\sigma_{D,v}\in G$.  By the definition of a special point, combined with Lemma \ref{special point}, we have that $D\cap J\setminus[v]_J\subseteq A$.  Therefore the matrix $\hat\sigma_{D,v}$ only differs from identity on the block corresponding to $\mathcal{B}_A$. 
	
	
	Let $\pi:\mathcal{T}_1\to \Gamma_{|\mathcal{B}_A|}(2)$ be the projection map onto the block corresponding to $\mathcal{B}_A$.  By the above argument $\pi$ is both well-defined and injective, hence an isomorphism onto its image.  
	
	Because $A\subseteq J$, we have a linear embedding $$T:I_A(H_1(\hat X; \Q))\to I_J(H_1(\hat X;\Q)),\quad T(1_A(\hat v - \hat w)) = 1_J(\hat v - \hat w).$$
	Fix the basis $T^{-1}(\mathcal{B}_A)$ of $I_A(H_1(\hat X; \Q))$.  Then for all $\sigma\in\rho_{\Gamma,J}^{-1}(\mathcal{T}_1)$ we have that $\rho_{\Gamma,A}(\sigma) = \pi\circ\rho_{\Gamma,J}(\sigma)$.  Thus it suffices to show that $\rho_{\Gamma,A}(\mathcal{P}^{\Gamma, A})\in\pi(\mathcal{T}_1)$.  By the hypothesis, for each $v\in[a]_J$ and each component $D$ of $\Gamma\setminus\text{st}(v)$, either $D\cap J \setminus[v]_J\subseteq A$ or $D\cap J \setminus[v]_J\subseteq A^\complement$.  Thus we need only show that $\rho_{\Gamma, A}(\sigma_{D(v,a),v})\in\pi(\mathcal{T}_1)$ for each $v\in A\setminus[a]_J$.  
	
	Note that $\sigma_{A,a}\in \langle G\rangle$, since $\sigma_{D,a}\in G$ for all components of $\Gamma\setminus\text{st}(a)$ which intersect $A\setminus[a]_J$.  Since $\sigma_{A,a}$ is inner relative to $A$, we have that  $\rho_{\Gamma,A}(\sigma_{A,a})=-\Id$ (Lemma \ref{simple}).  It follows that for any $v\in A\setminus[a]_J$ we have $$\rho_{\Gamma, A}(\sigma_{D(v,a),v})=\rho_{\Gamma, A}\Big(\displaystyle\sigma_{A, a}\circ\prod_{ a\notin D}\sigma_{D,v}\Big)\in\pi(\mathcal{T}_1).$$
	
\end{proof} 

The following corollary will be useful in proving other similar decomposition theorems later.

\begin{cor}\label{remove}
	Let $J\subseteq V(\Gamma)$ be an arbitrary vertex set.  Fix some $v\in J$.  Then $\{\hat\sigma_{D,w}\;|\; v\notin D\}$ generates $\text{Im}(\rho_{\Gamma,J})$.
\end{cor}

\begin{proof}
	This follows immediately from Lemma \ref{T0} by letting $A=J$.
\end{proof}

With Lemma \ref{T0} in hand, understanding the subgroup $\mathcal{T}$ is fairly straightforward.

\begin{cor}\label{T}
	Let $J=\displaystyle\bigsqcup_{i=1}^m A_i$ be a division of $J$ with special points $a_i\in A_I$.  We define the following subgroup of $\text{Im}(\rho_{\Gamma,J})$.
	$$\mathcal{T}:=\langle\hat\sigma_{D,v}\;|\; v\in A_i, 1\leq i \leq m, D\cap A_i\setminus[v]_J\neq\emptyset, a_i\notin D\rangle.$$  Then $\mathcal{T}\cong\displaystyle\prod_{i=1}^m\text{Im}(\rho_{\Gamma,A_i})$.
\end{cor}

\begin{proof}
	For each $1\leq i \leq m$, let $\mathcal{B}_{A_i}$ be an ordered basis of $\langle \hat v - \hat w\;|\; v,w\in A_i\rangle\leq I_J(H_1(\hat X; \Q))$  consisting of vectors of the form $\hat v - \hat w$.  
	Fix the following ordered basis of $I_J(H_1(\hat{X};\Q))$.\\
	
	$\mathcal{B}_1,\mathcal{B}_2,\dots, \mathcal{B}_m,\hat{a}_2-\hat{a}_1, \hat{a}_3-\hat{a}_{1},\dots, \hat{a}_{m}-\hat{a}_{1}$.\\
	
	Consider the set of automorphisms
	
	\begin{center}
		$G_i:=\{\sigma_{D,v}\;|\; v\in A_i, D\cap A_i\setminus [v]_J\neq\emptyset, a_i\notin D\}$.  
	\end{center}
	Then $\displaystyle\bigsqcup_{i=1}^m \rho_{\Gamma,J}(G_i)$ is a generating set for $\mathcal{T}$.  Furthermore, by Lemma \ref{special point}, each $\hat\sigma_{D,v}\in\rho_{\Gamma,J}(G_i)$ only acts non-trivially on the block corresponding to $\mathcal{B}_i$.  Note that our choice of basis corresponds to the choice of basis in Lemma \ref{T0}.  Therefore by Lemma \ref{T0}, we have that
	
	$$\mathcal{T}=\Big\{\begin{bmatrix}
	M_1 & & & &\\
	& M_2 & & &\\
	& & \ddots & & &\\
	& & & M_m &\\
	& & & & \text{Id}_{m-1}
	\end{bmatrix}\;\Big|\; M_i\in \text{Im}(\rho_{\Gamma,A_i})\Big\}\cong \displaystyle\prod_{i=1}^{m}\text{Im}(\rho_{\Gamma,A_i}).$$
	
\end{proof}

Next, we turn our attention to the subgroup $\mathcal{M}$.  

\begin{lem}\label{M}
	Let $J=\displaystyle\bigsqcup_{i=1}^m A_i$ be a division of $J$ with special points $a_i\in A_i$.  Let $$\mathcal{M}:=\Big\langle
	\Big(\displaystyle\prod_{A_i\subseteq D}(\hat\sigma_{A_i, a_i})\Big)\hat\sigma_{D,v}\;|\; v\in [a_j]_J\text{ for some } 1\leq j \leq m, D\cap J\setminus[v]_J\subseteq A_j^\complement\Big\rangle.$$  Let $J_0:=\{v\;|\; v\in[a_i]_J\text{ for some } 1\leq i \leq m\}$.  Then with respect to the appropriate bases 	$$\mathcal{M}=\Big\{\begin{bmatrix} \text{Id}_{|J|-m} & 0 \\ 0 & M\end{bmatrix}\;\Big|\; M\in \text{Im}(\rho_{\Gamma,J_0})\Big\}\cong\text{Im}(\rho_{\Gamma,J_0}).$$  
\end{lem}

\begin{proof}
	For each $1\leq i \leq m$, let $\mathcal{B}_i$ be an ordered basis of $\langle \hat v - \hat w\;|\; v,w\in A_i\rangle\leq I_{J}(H_1(\hat X; \Q))$ consisting of vectors of the form $\hat v - \hat w$.  
	Fix the following ordered basis of $I_J(H_1(\hat{X};\Q))$.\\
	
	$\mathcal{B}_1,\mathcal{B}_2,\dots, \mathcal{B}_m,\hat{a}_2-\hat{a}_1, \hat{a}_3-\hat{a}_{1},\dots, \hat{a}_{m}-\hat{a}_{1}$.\\ 
	
	Let $\Big(\displaystyle\prod_{A_i\subseteq D}(\hat\sigma_{A_i, a_i})\Big)\cdot\hat\sigma_{D,v}$ be a generator of $\mathcal{M}$. Then for each $A_i$ such that $i\neq j$ and $A_i\cap D\neq \emptyset$, by definition of special point we have that $A_i\subseteq D$.  It follows that $D\cap J\setminus[v]_J=\displaystyle\bigcup_{A_i\subseteq D} A_i$.  Therefore for each $d\in D\cap J \setminus[v]_J$, there exists an $i\neq j$ such that $d\in A_i\subseteq D$.  It follows that	$\Big(\displaystyle\prod_{A_i\subseteq D}(\sigma_{A_i, a_i})\Big)\circ\sigma_{D,v}(d)=va_ida_iv$.  Hence $	\Big(\displaystyle\prod_{A_i\subseteq D}(\hat\sigma_{A_i, a_i})\Big)\hat\sigma_{D,v}$ fixes $\mathcal B_i$ for each $1\leq i \leq m$.  
	
	Let $\hat V_{end}$ be the vector space spanned by	$\hat a_2-\hat a_1, \hat a_3-\hat a_1, \dots, \hat a_m-\hat a_1$. Let $\pi:\mathcal{M}\to\Gamma_{m-1}(2)$ be the projection onto $\hat{V}_{end}$.  By the above argument, $\pi$ is both well-defined and injective.  Since $J_0\subsetneqq J$, we have a linear embedding 
	$$T: I_{J_0}(H_1(\hat X, \Q))\to I_J(H_1(\hat X,\Q)),\quad T(1_{J_0}(\hat v - \hat w))=1_J(\hat v - \hat w).$$  Fix the basis $T^{-1}(\{1_J(\hat a_i-\hat a_1)\;|\; 2\leq i \leq m\})$ of $I_{J_0}(H_1(\hat X,\Q))$. Then for each $\sigma\in\rho_{\Gamma,J}^{-1}(\mathcal M)$, we have that $\rho_{\Gamma,J_0}(\sigma)=\pi\circ\rho_{\Gamma,J}(\sigma)$.	
	
	Note that $\rho_{\Gamma,J_0}(\sigma_{A_i,a_i})=\text{Id}$.  Thus for each generator $\Big(\displaystyle\prod_{A_i\subseteq D}(\hat\sigma_{A_i, a_i})\Big)\hat\sigma_{D,v}$ of $\mathcal{M}$, we have that $\pi(\Big(\displaystyle\prod_{A_i\subseteq D}(\hat\sigma_{A_i, a_i})\Big)\hat\sigma_{D,v})=\rho_{\Gamma,J_0}(\hat\sigma_{D,v})$.  
	
	We now show that the condition $D\cap J \setminus[v]_J\subseteq A_j^\complement$ in the definition of $\mathcal{M}$ can be replaced by the condition $D\cap J_0\setminus[v]_J\neq \emptyset$.  From this it follows that $\mathcal{M}\cong\pi(\mathcal{M})=\langle\rho_{\Gamma,J_0}(\mathcal{P}^{\Gamma,J_0})\rangle= \text{Im}(\rho_{\Gamma,J_0})$, which proves the lemma.
	
	Let $\Big(\displaystyle\prod_{A_i\subseteq D}(\hat\sigma_{A_i, a_i})\Big)\circ\hat\sigma_{D,v}$ be a generator of $\mathcal{M}$.  If $v\in A_j$ and $D\cap J\setminus[v]_J\subseteq A_j^\complement$, then for some $i\neq j$ we must have that $D\cap A_i\neq \emptyset$.  Since $a_i$ is a special point and $v\notin A_i$, we have that $D\supseteq A_i$.  In particular, $a_i\in D$.  This shows that $D\cap J_0\setminus[v]_J\neq\emptyset$.
	
	Now assume that $D\cap J_0\setminus[v]_J\neq\emptyset$.  Then for some $i\neq j$, we have that $a_i\in D$.  In particular, this shows that $D\cap A_i^\complement\neq\emptyset$.  But since $v\in[a_j]_J$, the definition of a special point implies that $D\cap J \setminus[v]_J\subseteq A_j^\complement$.  This completes the proof.
\end{proof}

Just as $\mathcal{T}$ can be divided into distinct subgroups with similar properties to $\mathcal{T}$, the same can be said of the group $\mathcal{L}$.  It will be beneficial to us later to work with these smaller subgroups, so we prove their properties first.  The same results about $\mathcal{L}$ will follow from them.

\begin{lem}\label{Li}
	Let $J\subseteq V(\Gamma)$ be an arbitrary vertex set.  Let $A\subsetneqq J$ be a non-empty union of components of $\Gamma_J$ containing a special point $a$.  We define the following subgroup of $\text{Im}(\rho_{\Gamma,J})$.
	$$\mathcal{L}_1:=\langle\hat\sigma_{D,v}\cdot\hat\sigma_{D,a}\;|\; v\in A\setminus[a]_J, D\text{ is a component of }\Gamma\setminus\text{lk}(v), D\cap J\subseteq A^\complement\rangle.$$  Then for the appropriate basis $$\Z^r\cong \mathcal{L}_1\leq\Big\{\begin{bmatrix}\text{Id}_{k_J-|S|} & M\\ 0 &  \text{Id}_{|S|-1}\end{bmatrix}\;\Big|\; M\in M_{k_J-|S|,|S|-1}(2\Z)\Big\}$$ where $r:=|\{([v]_J,D)\;|\; v\in A\setminus[a]_J, D\text{ is a component of }\Gamma\setminus\text{st}(v), D\cap J\subseteq A^\complement\}|$.  \\ Furthermore $\mathcal{L}_1$ is normal in $\text{Im}(\rho_{\Gamma,J})$.  
\end{lem}

\begin{proof}
	Let $\mathcal{B}_A$ be an ordered basis of $\langle \hat v - \hat w\;|\; v,w\in A_i\rangle\leq I_J(H_1(\hat X; \Q))$ consisting of vectors of the form $\hat v - \hat w$.  Let $S$ be an ordered list of representatives of the components of $\Gamma_{J\setminus A}$.  We extend $\mathcal{B}_A$ to an ordered basis $\mathcal{B}$ of $I_J(H_1(\hat X; \Q))$ so that $\mathcal{B} = \mathcal{B}_A\cup\{\hat s - \hat a\;|\; s\in S\}.$
	
	
	Fix a generator $\hat\sigma_{D,v}\cdot\hat\sigma_{D,a}$ of $\mathcal{L}_1$.  Let $d\in D$.  First note that by Lemma \ref{special point} this generator is well-defined.  This also implies that $D$ is independent of the choice of representative of $[v]_J$.  By direct computation, the only vectors that $\hat\sigma_{D, v}\cdot\hat\sigma_{D,a}$ does not fix are vectors of the form $\hat w - \hat a$ where $w\in D$.  It maps these vectors to $-2(\hat{v}-\hat{a})+(\hat{w}-\hat{a})$. Therefore 
	$$\mathcal{L}_1\leq\Big\{\begin{bmatrix}\text{Id}_{k_J-|S|} & M\\ 0 &  \text{Id}_{|S|-1}\end{bmatrix}\;\Big|\; M\in M_{k_J-|S|,|S|-1}(2\Z)\Big\}.$$  
	
	It follows that each generator $\hat\sigma_{D,v}\cdot\hat\sigma_{D,a}$ of $\mathcal{L}_1$ is of infinite order and commutes with every other generator.  Furthermore, since the entries where distinct generators differ from identity are distinct to those generators, $\mathcal{L}_1\cong \Z^r $.
	
	We now show that $\mathcal{L}_1$ is normal in $\text{Im}(\rho_{\Gamma,J})$.  Let $L:=\hat\sigma_{C,v}\cdot\hat\sigma_{C,a}\in\mathcal{L}_1$.  Let $M:=\hat\sigma_{D,w}\in\rho_{\Gamma,J}(\mathcal{P}^{\Gamma,J})$.  We break into a number of cases:
	
	\begin{itemize}
		\item $(w\in A, D\cap J\subseteq A)$:  In this case, $D\cap C\cap J=\emptyset$.  In particular, $D\neq C$.  Furthermore, $w\notin C$.  By Proposition \ref{commute}, $M$ and $L$ commute unless $(v,w\;|\; C)$ is a SIL relative to $J$ (this is equivalent to $(a,w\;|\;  C)$ being a SIL relative to $J$) and either $v\in D$ or $a\in D$.  By Lemma \ref{SIL comp eq}, this implies that $C$ is a component of $\Gamma\setminus\text{st}(w)$, so $\sigma_{C,w}$ is well-defined.
		\begin{itemize}
			\item If both $v,a\in D$, then 	$$\sigma_{D,w}\circ\sigma_{ C,v}\circ\sigma_{ C,a}\circ\sigma_{D,w}(u)=\begin{cases}
			wavwuwvaw\quad\text{if }u\in C\\
			u\quad\text{else}
			\end{cases}$$
			so that
			$$MLM^{-1}=L^{-1}\in\mathcal{L}_1$$
			
			\item
			If $v\in D$, $a\notin D$, then 
			$$\sigma_{D,w}\circ\sigma_{ C,v}\circ\sigma_{ C,a}\circ\sigma_{D,w}(u)=\begin{cases}
			awvwuwvwa\quad\text{if }u\in  C\\
			u\quad\text{else}
			\end{cases}$$
			so that
			$$MLM^{-1}=(\hat\sigma_{ C,w}\cdot\hat\sigma_{ C,a})\cdot(\hat\sigma_{ C,v}\cdot\hat\sigma_{ C,a})^{-1}\cdot(\hat\sigma_{ C,w}\cdot\hat\sigma_{ C,a})\in\mathcal{L}_1$$ 
			
			\item
			If $v\notin D$, $a\in D$, then 
			$$\sigma_{D,w}\circ\sigma_{ C,v}\circ\sigma_{ C,a}\circ\sigma_{D,w}(u)=\begin{cases}
			wawvuvwaw\quad\text{if }u\in  C\\
			u\quad\text{else}
			\end{cases}$$
			so that
			$$MLM^{-1}=(\hat\sigma_{ C,v}\cdot\hat\sigma_{C,a})\cdot
			(\hat\sigma_{C,w}\cdot\hat\sigma_{ C,a})^{-2}\in\mathcal{L}_1$$ 
		\end{itemize}
		
		\item $(w\in A, D\cap J\nsubset A, a\notin D)$: If $D= C$, then by direct computation $MLM^{-1}=L^{-1}\in\mathcal{L}_1$.  If not, let $d\in D\cap J\setminus A$.  Then Corollary \ref{permissible} implies that $D=D(w,d)\subseteq D(a,d)$.  The definition of special point implies that $D(a,d)\setminus[a]_J \subseteq A^\complement$.  This shows that $v\notin D$. Since $w\notin C$, Proposition \ref{commute} implies that $M$ and $L$ commute.
		
		\item $(w\in A, D\cap J\nsubset A, a\in D)$:  By Lemma \ref{simple}, every inner automorphism relative to $J$ maps to $-\Id$.  Since the image of every partial conjugation is of order 2 (Lemma \ref{simple}) and every pair of partial conjugations by $v$ commute, $M=\hat\sigma_{J,w}\cdot\displaystyle\prod_{D'\neq D}\hat\sigma_{D',w}=-\displaystyle\prod_{D'\neq D}\hat\sigma_{D',w}$.  This reduces this case to the previous two cases.
		
		\item $(w\notin A, D\cap A = \emptyset)$: If $D= C$, then by direct calculation $MLM^{-1}=L^{-1}\in\mathcal{L}_1$.  If not, then since $v,a\notin D$, Proposition \ref{commute} implies that $M$ and $L$ commute unless $w\in C$.  In addition, we must have that $(v,w\;|\; D)$ is a SIL relative to $J$, or $(a,w\;|\; D)$ is a SIL relative to $J$.  We show that the existence of one of these SILs implies the existence of the other.
		
		Assume that $(v,w\;|\; D)$ is a SIL relative to $J$.  Let $d\in D\cap J$.  By Lemma \ref{special point}, we have that $ D(a,d)= D(v,d)=D$.  Then Lemma \ref{SIL comp eq} implies that  $(a,w\;|\;D )$ is a SIL relative to $J$.
		
		
		Now assume that $(a,w\;|\; D)$ is a SIL relative to $J$.  Let $d\in D\cap J$.  Since $D\cap A = \emptyset$, we have that $v\notin D=D(a,d)$.  Therefore by Corollary \ref{permissible}, we have that $D(a,d)\subseteq D(v,d)$.  Since $w\in C$, we have that $C=D(v,w)$.  
		Since $a\notin C$, it follows that $w\notin D(v,a)$.  Applying Corollary \ref{permissible}, we see that $D(v,a)\subseteq D(w,a)$.  Then since $a\notin D=D(w,d)$, we have that $d\notin D(w,a)$.  It follows that $d\notin D(v,a)$, which implies that $a\notin D(v,d)$.  Applying Corollary \ref{permissible}, we get that $D(v,d)\subseteq D(a,d)$.  Thus $D=D(a,d)=D(v,d)$.  Finally, Lemma \ref{SIL comp eq} implies that $(v,w\;|\; D)$ is a SIL relative to $J$. 
		
		We can now assume that $w\in C$ and that both $(v,w\;|\; D)$ and $(a,w\;|\; D)$ are SILs relative to $J$.  Since $D(v,d)=D\neq C = D(v,w)$, we have that $D\cap C = \emptyset$.  By direct computation, $$\sigma_{D,w}\circ\sigma_{C,v}\circ\sigma_{ C,a}\circ\sigma_{D,w}(u)=\begin{cases}
		avuva\quad\text{if }u\in C\\
		vawavwuwvawaw\quad\text{if } u\in D\\
		u\quad\text{else}
		\end{cases}$$
		so that
		$$MLM^{-1}=(\hat\sigma_{ C,v}\cdot\hat\sigma_{C,a})\cdot(\hat\sigma_{ D,v}\cdot\hat\sigma_{D,a})^{-2}\in\mathcal{L}_1$$ 
		
		\item $(w\notin A, D\cap A \neq \emptyset)$:  
		Let $d\in D\cap A$.  By definition of a special point, we have that $A\subseteq D(w,a)=D(w,d)=D$. Then $M=-\displaystyle\prod_{D'\neq D}\hat\sigma_{D',w}$ and this reduces to the previous case.
	\end{itemize} 
	
	This shows that $\mathcal{L}_1$ is normal in $\text{Im}(\rho_{\Gamma,J})$. 
\end{proof}


\begin{cor}\label{L}
	Let $J=\displaystyle\bigsqcup_{i=1}^m A_i$ be a division of $J$ with special points $a_i\in A_I$.  Let $$\mathcal{L}:=\langle\hat\sigma_{D,v}\cdot\hat\sigma_{D,a_i}\;|\; 1\leq i \leq m, v\in A_i\setminus[a_i]_J, D\text{ is a component of }\Gamma\setminus\text{st}(v), D\cap J\subseteq A_i^\complement\rangle.$$  Then for the appropriate basis, $$\Z^r\cong \mathcal{L}\leq\Big\{\begin{bmatrix}\text{Id}_{|J|-m} & M\\ 0 &  \text{Id}_{m-1}\end{bmatrix}\;\Big|\; M\in M_{|J|-m,m-1}(2\Z)\Big\}$$ where $r:=|\{([v]_J,D)\;|\; v\in A_i\setminus[a_i]_J \text{ for some } 1\leq i\leq m, D\text{ is a component of }\Gamma\setminus\text{st}(v), \\ D\cap J\subseteq A_i^\complement\}|$.  Furthermore $\mathcal{L}$ is normal in $\text{Im}(\rho_{\Gamma,J})$.  
\end{cor}

\begin{proof}
	For each $1\leq i \leq m$, let $\mathcal{B}_i$ be an ordered basis of $\langle \hat v - \hat w\;|\; v,w\in A_i\rangle\leq I_{J}(H_1(\hat X; \Q))$ consisting of vectors of the form $\hat v - \hat w$.  
	Fix the following ordered basis of $I_J(H_1(\hat{X};\Q))$.\\
	
	$\mathcal{B}_1,\mathcal{B}_2,\dots, \mathcal{B}_m,\hat{a}_2-\hat{a}_1, \hat{a}_3-\hat{a}_{1},\dots, \hat{a}_{m}-\hat{a}_{1}$.\\
	
	By direct computation,
	
	$$\mathcal{L}\leq\Big\{\begin{bmatrix}\text{Id}_{|J|-m} & M\\ 0 &  \text{Id}_{m-1}\end{bmatrix}\;\Big|\; M\in M_{|J|-m,m-1}(2\Z)\Big\}.$$  
	Let $$\mathcal{L}_i:=\langle\hat\sigma_{D,v}\cdot\hat\sigma_{D,a_i}\;|\; v\in A_i\setminus[a_i]_J, D\cap A_i\setminus[v]_J = \emptyset, a_i\notin D\rangle.$$  Note that our choice of basis corresponds to the choice of basis made in Lemma \ref{Li}. Then by Lemma \ref{Li}, for each $1\leq i \leq m$ we have that $\mathcal{L}_i\cong\Z^{r_i}$ where $r_i:=|\{([v]_J,D)\;|\; v\in A_i\setminus[a_i]_J, D\text{ is a component of }\Gamma\setminus\text{st}(v), D\cap J\subseteq A_i^\complement\}|$.  Since each $\mathcal{L}_i$ only differs from identity on rows corresponding to vectors in $\mathcal{B}_i$, it follows that $\mathcal{L}=\displaystyle\prod_{i=1}^m \mathcal{L}_i\cong\Z^r$.  Furthermore, since each $\mathcal{L}_i$ is normal in $\text{Im}(\rho_{\Gamma,J})$, we have that $\mathcal{L}$ is normal in $\text{Im}(\rho_{\Gamma,J})$.
\end{proof}
We are now ready to prove the TML decomposition (Theorem \ref{tml}).


\begin{customthm}{\ref{tml}}[TML Decomposition]
	Let $J=\displaystyle\bigsqcup_{i=1}^m A_i$ be a division of $J$ with special points $a_i\in A_I$.  Let $J_0:=\{v\;|\; v\in[a_i]_J\text{ for some } 1\leq i \leq m\}$.  Then $\text{Im}(\rho_{\Gamma,J})\cong\Big(\Big(\displaystyle\prod_{i=1}^m\text{Im}(\rho_{\Gamma,A_i})\Big)\times\text{Im}(\rho_{\Gamma,J_0})\Big)\ltimes\Z^r$ where $r:=|\{([v]_J,D)\;|\; v\in A_i\setminus[a_i]_J \text{ for some } 1\leq i\leq m, D\text{ is a component of }\Gamma\setminus\text{st}(v), D\cap J\subseteq A_i^\complement\}|$.
\end{customthm}

\begin{proof}
	For each $1\leq i \leq m$, let $\mathcal{B}_i$ be an ordered basis of $\langle \hat v - \hat w\;|\; v,w\in A_i\rangle\leq I_{J}(H_1(\hat X; \Q))$ consisting of vectors of the form $\hat v - \hat w$.  
	Fix the following ordered basis of $I_J(H_1(\hat{X};\Q))$.\\
	
	$\mathcal{B}_1,\mathcal{B}_2,\dots, \mathcal{B}_m,\hat{a}_2-\hat{a}_1, \hat{a}_3-\hat{a}_{1},\dots, \hat{a}_{m}-\hat{a}_{1}$.\\
	
	Let $\hat{V}_{end}$ be the vector space spanned by $\hat{a}_2-\hat{a}_1, \hat{a}_3-\hat{a}_{1},\dots, \hat{a}_{m}-\hat{a}_{1}$.  Consider the following subgroups of $\text{Im}(\rho_{\delta})$
	\begin{itemize}
		\item
		$\mathcal{T}:=\langle\hat\sigma_{D,v}\;|\; v\in A_i, 1\leq i\leq m, D\cap A_i\setminus[a_i]_J\neq \emptyset\rangle$.
		\item
		$\mathcal{M}:=\Big\langle
		\Big(\displaystyle\prod_{A_i\subseteq D}(\hat\sigma_{A_i, a_i})\Big)\hat\sigma_{D,v}\;|\; v\in [a_j]_J\text{ for some } 1\leq j \leq m, D\cap J\setminus[v]_J\subseteq A_j^\complement\Big\rangle.$
		\item $\mathcal{L}:=\langle\hat\sigma_{D,v}\cdot\hat\sigma_{D, a_i}\;|\; 1\leq i \leq m, v\in A_i\setminus[a_i]_J, D\text{ is a component of }\Gamma\setminus\text{lk}(v), D\cap J\subseteq A_i^\complement\rangle$.
	\end{itemize}
	Our choice of basis corresponds to the choices of bases made in Corollary \ref{T}, Lemma \ref{M} and Corollary \ref{L}.  Thus by Corollary \ref{T}, Lemma \ref{M}, and Corollary \ref{L}, we have that  $$\prod_{i=1}^m \text{Im}(\rho_{\Gamma,A_i})\cong\mathcal T\leq \Big\{\begin{bmatrix} M & 0 \\ 0 & \text{Id}_{m-1}\end{bmatrix}\Big\}.$$
	
	$$\mathcal{M}=\Big\{\begin{bmatrix} \text{Id}_{|J|-m} & 0 \\ 0 & M\end{bmatrix}\;\Big|\; M\in \text{Im}(\rho_{\Gamma,J_0})\Big\}\cong \text{Im}(\rho_{\Gamma,J_0}).$$ 
	
	$$ \Z^r\cong\mathcal{L}\leq\Big\{\begin{bmatrix}\text{Id}_{|J|-m} & M\\ 0 &  \text{Id}_{m-1}\end{bmatrix}\;\Big|\; M\in M_{|J|-m,m-1}(2\Z)\Big\}.$$ 
	
	We also know from Corollary \ref{L} that $\mathcal{L}$ is normal in $\text{Im}(\rho_{\Gamma,J})$.  From this and the above block matrix descriptions of $\mathcal{T},\mathcal{M}$, and $\mathcal{L}$, we see that $\mathcal{TML}=(\mathcal{T}\times\mathcal{M})\ltimes\mathcal{L}$.  It remains to show that $\mathcal{TML}=\text{Im}(\rho_{\Gamma,J})$.  It suffices to show that $\rho_{\Gamma,J}(\mathcal{P}^{\Gamma,J})\in\mathcal{TML}$.
	
	First, consider $\sigma_{D,v}\in\mathcal{P}^{\Gamma,J}$ where $v\in[a_i]_J$ for some $1\leq i \leq m$. By the definition of divisibility, either $D\cap J \setminus[v]_J\subseteq A_i$ or $D\cap J \setminus[v]_J\subseteq A_i^\complement$.  In the first case, $\hat\sigma_{D,v}\in\mathcal{T}$.  
	In the second case, since $\hat\sigma_{A_k,a_k}\in\mathcal{T}$, we see that $$\hat\sigma_{D,v}=
	\Big(\displaystyle\prod_{A_k\subseteq D}\hat\sigma_{A_k,a_k}\Big)\cdot\Big(\Big(\displaystyle\prod_{A_k\subseteq D}(\hat\sigma_{A_k, a_k})\Big)\hat\sigma_{D,v}\Big)\in\mathcal{TM}.$$  In particular, we see that $\hat\sigma_{J,v}=-\text{Id}\in\mathcal{TML}$.  
	
	Next, consider $\sigma_{D,v}\in\mathcal{P}^{\Gamma,J}$ for $v\notin J_0$.  Let $v\in A_i$.  If $D\cap J \setminus[v]_J\subseteq A_i^\complement$, then  $\hat\sigma_{D,v}=(\hat\sigma_{D,v}\cdot\hat\sigma_{D,a_i})\cdot\hat\sigma_{D,a_i}\in\mathcal{TML}$.  If $D\cap A_i\setminus[a_i]_J\neq \emptyset$, then $\hat\sigma_{D,v}\in\mathcal{T}$.  Finally, if $a_i\in D$, then $\hat\sigma_{D,v}=-\displaystyle\prod_{C\neq D}\hat\sigma_{D,v}\in\mathcal{TML}$ by the previous arguments.  This completes the proof.
\end{proof}

\section{Splitting Points}\label{sect splitting points}
In this section we give a condition that will allow us to decompose $\text{Im}(\rho_{\Gamma,J})$ as a direct sum of the images of representations with restricted domains.

\begin{definition}\label{delta ss2}
	We say that $v\in J$ is a \underline{splitting point} if there exist non-empty disjoint subsets $A,B\subseteq J$,  such that $J=A\sqcup B \sqcup [v]_J$ and $v$ is a special point of both $A\sqcup[v]_J$ and $B\sqcup[v]_J$.  
\end{definition}

\begin{prop}\label{ss reduction2}
	Let $\Gamma$ be a graph, $J$ a set of pairwise non-adjacent vertices, and $v\in J$ a splitting point.  Let $A,B$ be as in the definition of splitting set.  Then we have $$\text{Im}(\rho_{\Gamma,J})=\text{Im}(\rho_{\Gamma,A\cup[v]_J})\times\text{Im}(\rho_{\Gamma,B\cup[v]_J}).$$
\end{prop}

\begin{proof} Let $\mathcal{B}_A$ be a basis of $\langle \hat v - \hat w\;|\; v,w\in A\rangle\leq I_J(H_1(\hat X; \Q))$, and let $\mathcal{B}_B$ be a basis of $\langle \hat v - \hat w\;|\; v,w\in B\rangle\leq I_J(H_1(\hat X; \Q))$.  Fix some $a\in A, b\in B$.  Then we use the ordered basis
	
	$\mathcal{B}=\mathcal{B}_A, \hat v-\hat a, \hat v-\hat b, \mathcal{B}_B$.
	
	We define the following subgroups of $\text{Im}(\rho_{\Gamma,J})$:
	\begin{itemize}
		
		\item $\mathcal{A}:=\langle\hat\sigma_{D,w}\in\mathcal{P}^{\Gamma,A\sqcup[v]_J}\;|\; v\notin D\setminus[w]_J \rangle$
		\item $\mathcal{B}:=\langle\hat\sigma_{D,w}\in\mathcal{P}^{\Gamma,B\sqcup[v]_J}\;|\; v\notin D\setminus[w]_J\rangle$
	\end{itemize}
	
	Then by the definition of a splitting point, for any generator $\hat\sigma_{D,w}$ of $\mathcal{A}$, we must have that $D\cap J\subseteq A$.
	Similarly, for any generator $\hat\sigma_{D,w}$ of $\mathcal{B}$, we must have that $D\cap J\subseteq B$.  By the definitions of splitting point, $v$ is a special point with respect to the subsets $A\sqcup[v]_J$ and $B\sqcup[v]_J$. Note that our choice of basis corresponds to the choice of basis in Lemma \ref{T0}.  Hence by Lemma \ref{T0} we have
	
	$$\mathcal{A}=\Big\{\begin{bmatrix}M & 0 \\ 0 & \text{Id}_{|\mathcal{B}_B|+1}\end{bmatrix}\;\Big|\; M\in\text{Im}(\rho_{\Gamma,A\sqcup [v]_J})\Big\}\cong\text{Im}(\rho_{\Gamma,A\sqcup [v]_J})$$ 
	
	$$\mathcal{B}=\Big\{\begin{bmatrix}\text{Id}_{|\mathcal{B}_A|+1} & 0 \\ 0 & M\end{bmatrix}\;\Big|\; M\in\text{Im}(\rho_{\Gamma,B\sqcup [v]_J})\Big\}\cong\text{Im}(\rho_{\Gamma,B\sqcup [v]_J})$$ 
	
	By the block diagonal descriptions above, it follows that $\mathcal{AB}=\mathcal{A}\times\mathcal{B}\cong \text{Im}(\rho_{\Gamma,A\sqcup[v]_J})\times\text{Im}(\rho_{\Gamma,B\sqcup[v]_J})$.  Thus it remains to show that $\text{Im}(\rho_{\Gamma,J})=\mathcal{AB}$.  By Corollary \ref{remove}, it suffices to show that $\hat\sigma_{D,w}\in \mathcal{AB}$ for all $\hat\sigma_{D,w}$ such that $v\notin D\setminus[w]_J$.  But this is immediate from our definitions of $\mathcal{A}$ and $\mathcal{B}$.  
\end{proof}

\section{Separating Sets}\label{sect separating sets}
Consider a vector of the form $\hat{v}-\hat{w}$ in $I_J(H_1(\hat X; \Q))$ where $[v]_J\neq [w]_J$.  For any partial conjugation $\sigma_{D,u}$, the action of $\hat\sigma_{D,u}$ on $\hat v - \hat w$ depends upon whether or not $v$ or $w$ are in $D$.  If neither $v$ nor $w$ are in $D$, then $\hat\sigma_{D,u}$ acts as identity on $\hat v- \hat w$.  If both $v$ and $w$ are in $D$, then $\hat\sigma_{D,u}$ acts as $-\Id$ on $\hat v-\hat w$.  However, if $D$ separates $v$ and $w$, i.e. exactly one of these two vertices is in $D$, then $\hat v-\hat w$ is not an eigenvector (unless $u\in[v]_J\cup [w]_J$).  This motivates us to make the following definition.

\begin{definition}
	Let $v,w\in J$ be such that $[v]_J\neq[w]_J$.  Then $$\text{sep}(v,w):=\{x\in J\;|\;\exists x^*\in[x]_J\text{ such that } D(x^*,v)\neq D(x^*,w)\}.$$
\end{definition}

Right away from the definition, we see that $[v]_J\cup [w]_J\subseteq \text{sep}(v,w)$.  Besides determining which partial conjugations induce homomorphisms for which $\hat v-\hat w$ is not an eigenvector, $\text{sep}(v,w)$ has a couple of nice properties that we outline below.

\begin{prop}\label{supset2}
	If $c\notin\text{sep}(v,w)$ then $ D(c,v)\supseteq \text{sep}(v,w)$.
\end{prop}

\begin{proof}
	Let $x\in\text{sep}(v,w)$.  Then there exists a $x^*\in [x]_J$ such that $D(x^*,v)\neq D(x^*,w)$.  If $x^*\notin D(c,v)$, then by Corollary \ref{permissible}, we have $D(c,v)\subseteq D(x^*,v)$. But $w\in D(c,v)$ and $w\notin D(x^*,v)$.  This is a contradiction, therefore $x^*\in D(c,v)$.  It follows that $x\in D(c,v)$.  This proves the proposition.
\end{proof}

\begin{prop}\label{inclusion2}
	If $a,b\in \text{sep}(v,w)$ then $\text{sep}(a,b)\subseteq\text{sep}(v,w)$
\end{prop}
\begin{proof}
	Let $a,b\in\text{sep}(v,w)$.  Then by definition of $\text{sep}(v,w)$, there exist $a^*\in[a]_J$ and $b^*\in [b]_J$ such that $D(a^*,v)\neq D(a^*,w)$ and $D(b^*,v)\neq D(b^*,w)$.  Next, let $c\in\text{sep}(a,b)$.  By definition, of $\text{sep}(a,b)$, there exists $c^*\in[c]_J$ such that $D(c^*,a^*)=D(c^*,a)\neq D(c^*,b)=D(c^*,b^*)$.  Corollary \ref{permissible} implies that $D(c^*,a^*)\subseteq D(b^*,a^*)$.  Since $D(b^*,v)\neq D(b^*,w)$, it follows that either $D(b^*,a^*)\neq D(b^*,v)$ or $D(b^*,a^*)\neq D(b^*,w)$.  Assume without loss of generality that $D(b^*,a^*)\neq D(b^*,v)$.  Then since $D(c^*,a^*)\subseteq D(b^*,a^*)$, we have that $D(c^*,a^*)\neq D(c^*,v)$.  It follows from Corollary \ref{permissible} that $D(c^*,v)\subseteq D(a^*,v)$.  But $w\notin D(a^*,v)$.  Therefore $w\notin D(c^*,v)$, so $D(c^*,v)\neq D(c^*,w)$.  This shows that $c^*\in\text{sep}(v,w)$, hence $c\in\text{sep}(v,w)$. The statement follows.
\end{proof}



Earlier we observed that $[v]_J\cup [w]_J\subseteq\text{sep}(v,w)$.  If we have that $\text{sep}(v,w)=[v]_J\cup [w]_J$, then $\hat v - \hat w$ is always an eigenvector.  This motivates us to explore what happens in this case.  We prove the following decomposition result.

\begin{lem}\label{sep=22}
	If $\text{sep}(v,w)=\{v\}\cup [w]_J$, then $\text{Im}(\rho_{\Gamma,J})\cong(\text{Im}(\rho_{\Gamma,A\cup [v]_J})\times\text{Im}(\rho_{\Gamma,A^\complement}))\ltimes\Z^r $ where $A:= D(v,w)$ and
	$r=|\{ ([a]_J,D)\;|\; a\in A, D\text{ is a component} \\\text{ of }\Gamma\setminus\text{st}(a), D\cap J \subseteq (A\sqcup[v]_J)^\complement\}|$.
\end{lem}

\begin{proof}
	We define the following subgroups:
	
	\begin{itemize}
		\item $\mathcal{A}:=\langle\hat\sigma_{D,a}\in \mathcal{P}^{\Gamma,A\cup [v]_J}\;|\; v\notin D\rangle$
		\item $\mathcal{B}:=\langle\hat\sigma_{D,b}\in\mathcal{P}^{\Gamma, A^\complement}\;|\; v\notin D\rangle$. 
		\item $\mathcal{L}:=\langle \hat\sigma_{D,a}\cdot\hat\sigma_{D,v}\;|\; a\in A, D\text{ is a component of }\Gamma\setminus\text{st}(v), D\cap J \subseteq (A\sqcup[v]_J)^\complement\rangle$.
	\end{itemize}
	
	Let $\{a_1,\dots, a_s\}$ be a set of representatives of the components of $\Gamma_A$ and let $\{b_1,\dots, b_t\}$ be a set of representatives of the components of $\Gamma_{A^\complement\setminus[v]_J}$.  Consider the basis $$\hat v-\hat a_1,\hat v-\hat a_2,\dots, \hat v-\hat a_s,\hat v-\hat b_1,\hat v-\hat b_2,\dots, \hat v-\hat b_t.$$
	
	We show that $v$ is a special point with respect to $A\sqcup[v]_J$.  First note that by hypothesis, $[v]_J=\{v\}$.  Thus if $a\in A=D(v,w)$, then $D(v,a)\cap J\setminus[v]_J=D(v,w)\cap J\setminus[v]_J=A\subseteq A\sqcup [v]_J$, and if $b\in A^\complement$, then $D(v,b)\cap D(v,w)=\emptyset$, so $D(v,b)\cap J\setminus[v]_J\subseteq (A\cup[v]_J)^\complement$.  Furthermore, if $b\in (A\cup[v]_J)^\complement$, then $b\notin D(v,w)\cap J=A$.  By Corollary \ref{permissible}, this implies that $D(v,w)\subseteq D(b,w)$.  But since $b\notin\text{sep}(v,w)$, we have that $D(b,w)=D(b,v)$.  This shows that $v$ is a special point with respect to $A\sqcup[v]_J$.  
	
	The above argument also shows that for all $j\in J\setminus[v]_J$ we have that either $D(v,j)\cap J \setminus[v]_J\subseteq A^\complement$ or $D(v,j)\cap J \setminus[v]_J\subseteq A$.
	Note that our choice of basis corresponds to the choice of basis in Lemma \ref{T0}.  Applying Lemma \ref{T0}, we have that $$\mathcal{A}=\Big\{\begin{bmatrix} A & 0 \\ 0 & 1\end{bmatrix}\;\Big|\; A\in \text{Im}(\rho_{\Gamma, A\cup [v]_J})\Big\}\cong \text{Im}(\rho_{\Gamma, A\cup [v]_J})$$
	
	$$\mathcal{B}=\Big\{\begin{bmatrix} 1 & 0 \\ 0 & B\end{bmatrix}\;\Big|\; B\in\text{Im}(\rho_{\Gamma,A^\complement})\Big\}\cong \text{Im}(\rho_{\Gamma,A^\complement})$$
	
	We now turn to the subgroup $\mathcal{L}$.  Note that our choice of basis corresponds to the choice of basis in Lemma \ref{Li}. Since $v$ is a special point with respect to $A\sqcup[v]_J$, Lemma \ref{Li} implies that
	
	$$\mathcal{L}\cong \Z^r\leq\Big\{\begin{bmatrix} 1 & L \\ 0 & 1\end{bmatrix}\;|\; L\in M_{s,t}(2\Z)\Big\}$$ and $\mathcal{L}$ is normal in $\text{Im}(\rho_{\Gamma,J})$.  Therefore $\mathcal{ABL}=(\mathcal{A}\times\mathcal{B})\ltimes\mathcal{L}$.  By Corollary \ref{remove}, it suffices to show that $\mathcal{ABL}$ contains $\{\hat\sigma_{D,u}\;|\; v\notin D\}$.  
	
	First note that $\hat\sigma_{D,v}\in\mathcal{A}\mathcal{B}$ for any choice of $D$.  In particular, this shows that $-\text{Id}\in \mathcal{ABL}$.  
	
	Let $a\in A$ and let $x\in J$ be such that $x\notin [a]_J$ and $v\notin D(a,x)$ .  By Corollary \ref{permissible}, we have that $D(a,x)\subseteq D(v,x)$.  If $x\in A$, this implies that $ D(a,x)\subseteq A$, so that $\hat\sigma_{D(a,x),a}\in \mathcal{A}$.  If $x\in A^\complement$, then $D(a,x)\subseteq A^\complement$.  Thus $\hat\sigma_{D(a,x),v}\cdot\hat\sigma_{D(a,x),a}\in \mathcal{L}$.  Since $\hat\sigma_{D(a,x),v}\in\mathcal{B}$, we have that $\hat\sigma_{D(a,x),a}\in \mathcal{BL}$.  Thus $\mathcal{ABL}$ includes $\hat\sigma_{D,a}$ for all $a\in A$ and all $D$ such that $v\notin D$.
	
	Finally, let $b\in A^\complement\setminus[v]_J$. In particular, we have that $b\notin[v]_J\cup [w]_J = \text{sep}(v,w)$.  Since $b\notin A=D(v,w)$, Corollary \ref{permissible} implies that $A =D(v,w)\subseteq D(b,w)$.  But $D(b,w)= D(b,v)$.  Therefore $D(b,v)\supseteq A\sqcup [v]_J$.  Now let $x\in J$ be such that $x\notin [b]_J$ and $v\notin D(b,x)$.  By the above statement, it follows that $D(b,x)\subseteq A^\complement\setminus[v]_J$.  Therefore $\hat\sigma_{D(b,x),b}\in\mathcal{B}$.  Thus $\mathcal{ABL}$ includes $\hat\sigma_{D,b}$ for all $D$ such that $v\notin D$.  This shows that $\mathcal{ABL}$ contains a complete set of generators of $\text{Im}(\rho_{\Gamma,J})$.  
\end{proof}

\section{Compressible Components}\label{sect compressible components}
We now take a slight detour to deal with the hypotheses of Lemma \ref{sep=22}.  Ideally, we would like a similar statement to Lemma \ref{sep=22} but without requiring that $[v]_J=\{v\}$.  To see that the statement of Lemma \ref{sep=22} does not hold in this case, we provide a counter-example.

\begin{example}
	Let $\Gamma$ be the graph 
	
	\begin {center}
	\begin{tikzpicture}
	
	
	\filldraw[black] (-1,0) circle (2pt) node[anchor=east] {$v_1$};
	\filldraw[black] (-1,1) circle (2pt) node[anchor=east] {$v_2$};
	\filldraw[black] (0,2) circle (2pt) node[anchor=east] {};
	\filldraw[black] (-1,3) circle (2pt) node[anchor=south] {$v_3$};
	\filldraw[black] (1,3) circle (2pt) node[anchor=south] {$v_4$};
	\filldraw[black] (1,1) circle (2pt) node[anchor=west] {$v_5$};
	\filldraw[black] (1,0) circle (2pt) node[anchor=west] {$v_6$};
	\filldraw[black] (0,-1) circle (2pt) node[anchor=east] {};
	\filldraw[black] (1,-2) circle (2pt) node[anchor=north] {$v_7$};
	\filldraw[black] (-1,-2) circle (2pt) node[anchor=north] {$v_8$};
	
	\draw (-1,0) -- (-1,1);
	\draw (-1,1) -- (0,2);
	\draw (0,2) -- (-1,3);
	\draw (0,2) -- (1,3);
	\draw (0,2) -- (1,1);
	\draw (1,1) -- (1,0);
	\draw (1,0) -- (0,-1);
	\draw (0,-1) -- (1,-2);
	\draw (0,-1) -- (-1, -2);
	\draw (0,-1) -- (-1,0);
	
	\end{tikzpicture}
\end{center}
and let $J$ be the vertex set $\{v_1,v_2,\dots, v_8\}$.  Then $\Gamma$ has a unique minimal separating set, namely $\text{sep}(v_1,v_5)=\{v_1,v_2,v_5,v_6\}=[v_1]_J\sqcup[v_5]_J$.  However $D(v_2,v_5)\cap J\setminus[v_2]_J \nsubset D(v_1,v_5)\cap J \setminus[v_1]_J$ and $D(v_6,v_1)\cap J\setminus[v_6]_J\nsubset D(v_5,v_1)\cap J \setminus[v_5]_J$.  Thus if $A:=D(v_1,v_5)\cap J$, then $A\sqcup[v_1]_J$ does not have a special point.  Similarly, if $A:=(D(v_5,v_1)\cap J)\sqcup[v_5]_J$, then $A$ does not have a special point.  This was a crucial point in the proof of Lemma \ref{sep=22} because it allowed us to apply Lemma \ref{Li}. 

It is possible that the conclusion of Lemma \ref{sep=22} could still hold.  If this were the case, letting $A=D(v_1,v_5)\cap J$, then applying Theorem \ref{discrete}, we would see that $$\text{Im}(\rho_{\Gamma,J})\cong(\text{Im}(\rho_{\Gamma,\{v_1,v_2,v_3,v_4,v_5,v_6\}})\times\text{Im}(\rho_{\Gamma,\{v_7,v_8\}}))\ltimes\Z^2\cong (\Gamma_3(2)\times\Z/2\Z)\ltimes \Z^2.$$  

There is, however, an alternate way to solve this problem.  Consider the slightly modified graph $\Gamma'$
\begin {center}
\begin{tikzpicture}


\filldraw[black] (-1,0) circle (2pt) node[anchor=east] {$v_{1,2}$};
\filldraw[black] (0,1) circle (2pt) node[anchor=east] {};
\filldraw[black] (-1,2) circle (2pt) node[anchor=south] {$v_3$};
\filldraw[black] (1,2) circle (2pt) node[anchor=south] {$v_4$};
\filldraw[black] (1,0) circle (2pt) node[anchor=west] {$v_{5,6}$};
\filldraw[black] (0,-1) circle (2pt) node[anchor=east] {};
\filldraw[black] (1,-2) circle (2pt) node[anchor=north] {$v_7$};
\filldraw[black] (-1,-2) circle (2pt) node[anchor=north] {$v_8$};

\draw (-1,0) -- (0,1);
\draw (0,1) -- (-1,2);
\draw (0,1) -- (1,2);
\draw (0,1) -- (1,0);
\draw (1,0) -- (0,-1);
\draw (0,-1) -- (1,-2);
\draw (0,-1) -- (-1, -2);
\draw (0,-1) -- (-1,0);

\end{tikzpicture}
\end{center}
with vertex set $J':=\{v_{1,2},v_3,v_4,v_{5,6},v_7,v_8\}$ obtained by compressing the edges $(v_1,v_2)$ and $(v_5,v_6)$ to a point.  This graph is useful because $\text{Im}(\rho_{\Gamma,J})\cong\text{Im}(\rho_{\Gamma',J'})$.  We give a sketch of why this is true.  We will give a more formal argument later.

Any possible issue lies with partial conjugations by $v_{1,2}$ (or symmetrically $v_{5,6}$).  The partial conjugations $\sigma_{\{v_3\},v_{1,2}}, \sigma_{\{v_4\},v_{1,2}},\sigma_{\{v_7\},v_{1,2}}$, and $\sigma_{\{v_8\},v_{1,2}}$ have clear analogues in $\mathcal{P}^{\Gamma,J}$.  However, the partial conjugation $\sigma_{\{v_{5,6}\},v_{1,2}}$ does not.  By considering instead the product of partial conjugations $\sigma:=\sigma_{\{v_3,v_4,v_5,v_6\},v_1}\circ\sigma_{\{v_3\},v_2}\circ\sigma_{\{v_4\},v_2}$, we are able to remedy this problem.  The corresponding matrix $\hat\sigma$ acts on the the symmetric difference of the sets being conjugated $\{v_3,v_4,v_5,v_6\}\triangle \{v_3\}\triangle\{v_4\}=\{v_5,v_6\}$ by the 1-chain $\hat v_1 = \hat v_2$.  

Now $\text{sep}(v_{1,2},v_{5,6})=\{v_{1,2},v_{5,6}\}$ is a minimal separating set.  Applying Lemma \ref{sep=22}, we get that $$\text{Im}(\rho_{\Gamma,J})\cong(\text{Im}(\rho_{\Gamma,\{v_{1,2},v_{5,6}\}})\times\text{Im}(\rho_{\Gamma,J\setminus\{v_{1,2}\}}))\ltimes\Z^4\cong(\Z/2\Z\times\text{Im}(\rho_{\Gamma,J\setminus\{v_{1,2}\}}))\ltimes\Z^4.$$  We see that $v_{1,2}$ is a splitting point in $\Gamma\setminus\{v_{5,6}\}$, Proposition \ref{ss reduction2} and Theorem \ref{discrete} imply that $$\text{Im}(\rho_{\Gamma,J})\cong(\Z/2\Z\times\text{Im}(\rho_{\Gamma,\{v_{1,2}, v_3,v_4\}})\times\text{Im}(\rho_{\Gamma,\{v_{1,2}, v_7,v_8\}}))\ltimes\Z^4\cong (\Z/2\Z\times\Gamma_2(2)^2)\ltimes\Z^4.$$

As $(\Z/2\Z\times\Gamma_2(2)^2)\ltimes\Z^4\ncong (\Gamma_3(2)\times\Z/2\Z)\ltimes \Z^2$, we see that the conclusion of Lemma \ref{sep=22} does not hold for this example.
\end{example}


We now formalize the above method of compressing components of $\Gamma_J$.

\begin{definition}
Let $\Gamma$ be a finite graph and let $J\subseteq V(\Gamma)$ be a vertex set.  Given $v,x\in J$ such that $[v]_J\neq [x]_J$, we say that $x$ is a \underline{$v$-\textit{expander}} if the set $$\text{Int}(v,x):=\displaystyle\bigcap_{w\in[v]_J}D(w,x)\cap J$$ cannot be written as a symmetric difference of elements of $$\mathfrak{D}(v):=\{D(w,y)\cap J\setminus[v]_J\;|\; w\in [v]_J, y\in J\setminus[v]_J\}.$$  We say $[v]_J$ is a \underline{\textit{compressible component}} relative to $J$ if no $x\in J\setminus[v]_J$ is a $v$-expander.  A component that is not compressible is \underline{\textit{incompressible}}.
\end{definition}  

\begin{prop}\label{compressible}
If $J$ contains a compressible component $[v]_J$, then there exists a graph $\Gamma'$ with vertex set $J'=(J\setminus[v]_J\sqcup\{v'\})$ such that $\text{Im}(\rho_{\Gamma,J})\cong\text{Im}(\rho_{\Gamma',J'})$.  
\end{prop}

\begin{proof}
Let $\Gamma'$ be the graph where the subgraph $\Gamma_{[v]_J}$ induced by $[v]_J$ is replaced by a single point $v'$.  Then the edges incident to $v'$ in $\Gamma'$ are precisely the edges incident to only one vertex in $[v]_J$ in $\Gamma$.  Fix a basis $\mathcal{B}$ of $I_J(H_1(\hat X; \Q))$.  Let $\mathcal{B}'$ be the basis of $I_J(H_1(\hat X; \Q))$ where for every $w\in[v]_J$, every occurrence of $\hat w$ is replaced by $\hat v'$.  

Consider the map $$f: J\to J',\quad f(j)=\begin{cases} v'\quad\text{if } j\in[v]_J\\ j\quad\text{else}\end{cases}.$$ Note that $f$ induces a bijection between the components of $\Gamma_J$ and the components of $\Gamma'_{J'}$. We show that for each $a,b\in J$ such that $[a]_J\neq [b]_J$ we have that $D(f(a),f(b))\cap J' \subseteq f(D(a,b)\cap J)$, with equality when $a\notin [v]_J$.

If $a\in[v]_J$, then $f(a)=v'$.  But $\Gamma'\setminus\text{st}(v')$ is a subgraph of $\Gamma\setminus\text{st}(a)$.  Since $f|_{J\setminus[v]_J}=\text{Id}$, the result follows.  If $a\notin [v]_J$, then $f(a)=a$.  Let $x\in D(f(a),f(b))\cap J'$ and let $\alpha_x$ be a path in $\Gamma'\setminus\text{st}(f(b))$ from $f(b)$ to $x$.  If $\alpha_x$ does not pass through $v'$, then $\alpha_x$ only passes through points in $J\setminus[v]_J$.  It follows that $\alpha_x$ induces a path $\beta_x$ in $\Gamma\setminus[v]_J$ from $b$ to $x$ such that $f(\beta_x)=\alpha_x$.  The result follows.  Assume that $\alpha_x$ passes through $v'$.  Without loss of generality, we may assume that $\alpha_x$ does not contain a loop.  Then we may write $\alpha_x = (\alpha_1, v',\alpha_2)$ for some (possibly empty) paths $\alpha_1,\alpha_2$ that do not pass through $v'$.  Then $\alpha_1, \alpha_2$ induce paths $\beta_1,\beta_2$ in $\Gamma\setminus[v]_J$ such that $f(\beta_1)=\alpha_1$ and $f(\beta_2)=\alpha_2$.  Furthermore, there exist $u,w\in[v]_J$ such that the terminal vertex of $\beta_1$ is adjacent to $u$ and the initial vertex of $\beta_2$ is adjacent to $w$.  By definition of $[v]_J$, there is a path $\beta$ in $\Gamma$ from $u$ to $w$ that only passes through vertices in $[v]_J$.  Then $f((\beta_1,\beta,\beta_2))=\alpha_x$.  The result follows. 

Next, we show that $\text{Im}(\rho_{\Gamma',J'})$ contains the images of all partial conjugations.  Let $\sigma_{D,a}\in\mathcal{P}^{\Gamma,J}$.  Let $b\in D$ so that $D=D(a,b)$.  By the previous claim, we have that $D(f(a),f(b))\cap J' \subseteq f(D(a,b)\cap J)$.  
We may therefore write $$f(D\cap J)=\displaystyle\bigsqcup_{w\in I}D(f(a),f(w))\cap J'$$ for some index set $I$.  It follows that
$$\rho_{\Gamma,J}(\sigma_{D,a})=\displaystyle\prod_{w\in I}\rho_{\Gamma',J'}(\sigma_{D(f(a),f(w)),f(a)})\in\text{Im}(\rho_{\Gamma',J'}).$$

Finally, we show that $\rho_{\Gamma',J'}(\mathcal{P}^{\Gamma',J'})\subseteq \text{Im}(\rho_{\Gamma,J})$.  First, let $\sigma_{D,v'}\in\mathcal{P}^{\Gamma',J'}$.  Then $D$ is a component of $\Gamma'\setminus\text{st}(v')$.  By construction, we have that $$\Gamma'\setminus\text{st}(v')=\Gamma\setminus\Big(\displaystyle\bigcup_{w\in[v]_J}\text{st}(w)\Big)=\bigcap_{w\in[v]_J}\Gamma\setminus\text{st}(w).$$  Let $x\in D$, so that $D=D(v',x)$.  Then by the above observation, $$D\cap J=D(v',x)\cap J=\text{Int}(v,x).$$  By the definition of compressible, we may write $D\cap J=D_1\triangle D_2\triangle \dots \triangle D_m$ where for each $1\leq i \leq m$ we have $D_i\in\mathfrak{D}(v)$.  Let $D_i=D(w_i,y_i)$.  Then $$\rho_{\Gamma',J'}(\sigma_{D,v'})=\displaystyle\prod_{i=1}^m\rho_{\Gamma,J}(\sigma_{D_i,w_i})\in\text{Im}(\rho_{\Gamma,J}).$$

Now let $\sigma_{D,a}\in\mathcal{P}^{\Gamma',J'}$ for some $a\neq v'$.  It follows that $f^{-1}(a)=\{a\}$.  Let $b\in J$ be such that $f(b)\in D$.  Then $D=D(f(a),f(b))\cap J'\subseteq f(D(a,b)\cap J)$.  Let $x\in D(a,b)\cap J \setminus [b]_J$, and let $\alpha_x$ be a path in $\Gamma\setminus\text{st}(a)$ from $b$ to $x$.  If $\alpha_x$ does not intersect $[v]_J$, then $\alpha_x$ induces a path $\beta_x$ in $\Gamma'\setminus\text{st}(a)$ from $f(b)=b$ to $x$.  If $\alpha_x$ intersects $[v]_J$, we may write $\alpha_x = (\alpha_1,\alpha_2,\dots, \alpha_m)$ where each of the even-indexed $\alpha_i$ lies entirely in $[v]_J$ and each of the odd-indexed $\alpha_i$ lies entirely outside $[v]_J$ ($\alpha_1$ might be empty).  Then each of the odd-indexed $\alpha_i$ induces a path $\beta_i$ in $\Gamma'\setminus\text{st}(a)$ and $f(\alpha)=(\beta_1, v', \beta_3, v', \dots)$ is a path in $\Gamma'\setminus\text{st}(a)$ from $f(b)$ to $x$.  This shows that $D=f(D(a,b)\cap J)$.  Therefore $\rho_{\Gamma',J'}(\sigma_{D,a})=\rho_{\Gamma,J}(\sigma_{D,a})\in\text{Im}(\rho_{\Gamma,J})$.  This completes the proof.

\end{proof}

Unfortunately, there exist vertex sets $J$ with incompressible components of $\Gamma_J$, as the following example shows.

\begin{example}
Let $\Gamma$ be the graph 

\begin {center}
\begin{tikzpicture}

\filldraw [black] (-1,0) circle (2pt) node[anchor=north]{$v_9$};
\filldraw [black] (0,0) circle (2pt) node[anchor=north]{$v_{10}$};
\filldraw [black] (-1,1) circle (2pt) node[anchor=east]{};
\filldraw [black] (0,1) circle (2pt) node[anchor=east]{};
\filldraw [black] (1,1) circle (2pt) node[anchor=east]{};
\filldraw [black] (2,1) circle (2pt) node[anchor=west]{};
\filldraw [black] (-1,2) circle (2pt) node[anchor=south]{$v_2$};
\filldraw [black] (0,2) circle (2pt) node[anchor=south]{$v_3$};
\filldraw [black] (1,2) circle (2pt) node[anchor=south]{$v_4$};
\filldraw [black] (2,2) circle (2pt) node[anchor=south]{$v_5$};

\draw (-1,0) -- (0,0);
\draw (-1,0) -- (-1,1);
\draw (-1,0) -- (1,1);
\draw (0,0) -- (0,1);
\draw (0,0) -- (2,1);
\draw (-1,1) -- (-1,2);
\draw (-1,1) -- (0,2);
\draw (0,1) -- (0,2);
\draw (0,1) -- (1,2);
\draw (1,1) -- (1,2);
\draw (1,1) -- (2,2);
\draw (2,1) -- (-1,2);
\draw (2,1) -- (2,2);

\filldraw [black] (-2,2) circle (2pt) node[anchor=south]{$v_1$};
\filldraw [black] (-2,1) circle (2pt) node[anchor=north]{};
\filldraw [black] (-3,1) circle (2pt) node[anchor=east]{};
\filldraw [black] (-4,1) circle (2pt) node[anchor=east]{};
\filldraw [black] (-2,0) circle (2pt) node[anchor=north]{$v_8$};
\filldraw [black] (-3,0) circle (2pt) node[anchor=north]{$v_7$};
\filldraw [black] (-4,0) circle (2pt) node[anchor=north]{$v_6$};

\draw (-1,2) -- (-2,2);
\draw (-1,2) -- (-3,1);
\draw (-2,2) -- (-4,1);
\draw (-2,2) -- (-2,1);
\draw (-2,1) -- (-2,0);
\draw (-3,1) -- (-3,0);
\draw (-4,1) -- (-4,0);
\draw (-2,1) -- (-3,0);
\draw (-3,1) -- (-4,0);
\draw (-4,1) -- (-1,0);
\draw (-1,1) -- (-2,0);
\end{tikzpicture}
\end{center}
and let $J$ be the vertex set $\{v_1,v_2,\dots, v_{10}\}$.  Then $[v_1]_J$ and $[v_9]_J$ are both incompressible.  Furthermore, the separating set $\text{sep}(v_1,v_9)=[v_1]_J\sqcup[v_9]_J$ fails the hypothesis of Lemma \ref{sep=22}.  As this is not the only separating set in this graph, we can get around this fact.  However, it is not inconceivable that every separating set could fail in this way.
\end{example}

To avoid the issue presented in the previous example, we will show that for every finite graph $\Gamma$ and $J\subseteq V(\Gamma)$ such that $\Gamma_J$ has at least 2 components, there must be a minimal separating set $\text{sep}(v,w)$ such that either $[v]_J$ or $[w]_J$ is compressible.  We first prove a couple of lemmas.  The first lemma gives us a nice consequence of a vertex being a $v$-expander.  The second lemma tells us that every incompressible component $[v]_J$ has at least two distinct vertices $x$ and $y$, both of which are $v$-expanders, and that are separated by an element of $[v]_J$.

\begin{lem}\label{incompSIL}
Let $v,x\in J$ be such that $[v]_J\neq [x]_J$.  If $x$ is a $v$-expander,  then for all $y\in J\setminus([v]_J\cup [x]_J)$ we have that $D(v,x)\neq D(y,x)$.
\end{lem}
\begin{proof}
We prove the contrapositive.  Let $y\in J\setminus([v]_J\cup[x]_J)$ be such that $D(v,x) = D(y,x)$.  Then for all $w\in [v]_J$, we have $D(y,x)\neq D(y,v)=D(y,w)$.  Hence by Corollary \ref{permissible}, we have that $D(y,x)\subseteq D(w,x)$ for all $w\in [v]_J$.  It follows that $D(v,x)\cap J=D(y,x)\cap J\subseteq\displaystyle\text{Int}(v,x)$.  But $\text{Int}(v,x)\subseteq D(v,x)$.  Hence $D(v,x)=\text{Int}(v,x)$ which shows that $x$ is not a $v$-expander.
\end{proof}

\begin{lem}\label{atLeast2}
Let $v\in J$ be such that $[v]_J$ is incompressible.  Then there exist $x,y\in J \setminus[v]_J$ such that $\text{Int}(v,x)\neq\text{Int}(v,y)$ and both $x$ and $y$ are $v$-expanders.
\end{lem}

\begin{proof}
First note that for any $x,y\in J\setminus[v]_J$, if $y\notin\displaystyle\text{Int}(v,x)$, then there exists a $w\in[v]_J$ such that $D(w,x)\neq D(w,y)$.  Hence $x\notin\text{Int}(v,y)$.  It follows that $\{\text{Int}(v,x)\; x\in J\setminus[v]_J\}$ forms a partition of $J\setminus[v]_J$.  

By definition, there exists some $x\in J\setminus[v]_J$ such that $x$ is a  $v$-expander.  Assume that $y$ is a $v$-expander only if $\text{Int}(v,x)=\text{Int}(v,y)$.  Since $x$ is a $v$-expander, it follows that $\text{Int}(v,x)\neq J\setminus[v]_J$.  

Let $c\in\text{Int}(v,x)^\complement\setminus[v]_J$.  Then by assumption $\text{Int}(v,c)$ can be written as a symmetric difference of elements of $\mathfrak{D}(v)$.   Since $\{\text{Int}(v,j)\;|\; j\in J\setminus[v]_J\}$ forms a partition of $J\setminus[v]_J$, it follows that $
\displaystyle\bigcup_{c\;\in(\text{Int}(v,x))^\complement\setminus[v]_J}\text{Int}(v,c)$ can be written as a symmetric difference of elements of $\mathfrak{D}(v)$.  Let $\{v_1,v_2,\dots, v_n\}\subseteq J\setminus[v]_J$ be such that  $\displaystyle\bigsqcup_{i=1}^n D(v,v_i)\cap J\setminus[v]_J = J\setminus[v]_J$.  Then\\

$\text{Int}(v,x) =(D(v,v_1)\cap J\setminus[v]_J)\triangle (D(v,v_2)\cap J\setminus[v]_J)\triangle\dots \triangle(D(v,v_n)\cap J\setminus[v]_J)\triangle$

$\displaystyle\bigcup_{c\;\in(\text{Int}(v,x))^\complement\setminus[v]_J}\text{Int}(v,c).$\\

This contradicts the fact that $x$ is a $v$-expander.  Therefore there exists $y\in J\setminus[v]_J$ and $w\in [v]_J$ such that $D(w,x)\neq D(w,y)$ and $y$ is a  $v$-expander.
\end{proof}

We are now ready to prove the Proposition.

\begin{prop}\label{minSepComp}
There is no finite graph $\Gamma$ and $J\subseteq V(\Gamma)$ such that $\Gamma_J$ contains at least 2 components and every minimal separating set consists solely of incompressible components.
\end{prop}

\begin{proof}
If $J$ contains less than 2 incompressible components then this is trivial.  Assume that $J$ contains at least two incompressible component.  
\setcounter{clm}{0}
\begin{clm}
There exist $v,w\in J$ such that $[v]_J$ is an incompressible component, $w$ is not a  $v$-expander, and $D(v,w)\cap J$ consists solely of compressible components. 
\end{clm}

Let $v_1,v_2\in J$ be such that $[v_1]_J\neq[v_2]_J$ and $[v_1]_J, [v_2]_J$ are both incompressible components.  Then by Lemma \ref{atLeast2}, there exists a $w_3\in J\setminus([v_1]_J\cup[v_2]_J)$ such that $\text{Int}(v_2,v_1)\neq \text{Int}(v_2,w_3)$ and $w_3$ is a $v_2$-expander.  Without loss of generality, we may assume that $D(v_2,v_1)\neq D(v_2,w_3)$.  If $D(v_2,w_3)$ consists solely of compressible components, then we are done.  If not, there exists a $v_3\in D(v_2,w_3)$ such that $[v_3]_J$ is incompressible.  By Corollary \ref{permissible}, we have that $D(v_2,v_3)=D(v_2,w_3)\subseteq D(v_1,w_3)$.  It follows that $D(v_1,w_3)=D(v_1,v_3)$ and $D(v_2,v_3)\subseteq D(v_1,v_3)$.  By Lemma \ref{incompSIL}, we have that $D(v_1,v_3)=D(v_1,w_3)\neq D(v_2,w_3)=D(v_2,v_3)$.  Therefore $D(v_1,v_3)\supsetneqq D(v_2,v_3)$.  

Assume that we have constructed $v_1,\dots, v_n$ such that $[v_1]_J$, $[v_2]_J$, $\dots, [v_n]_J$ are incompressible and $D(v_1,v_n)\supsetneqq D(v_2,v_n)\supsetneqq \dots\supsetneqq D(v_{n-1},v_n)$.  Then by Lemma \ref{atLeast2}, there exists a $w_{n+1}\in J\setminus[v_n]_J$ such that $\text{Int}(v_n, v_{n-1})\neq \text{Int}(v_n,w_{n+1})$ and $w_{n+1}$ is a  $v_n$-expander.  Without loss of generality, we may assume that $D(v_n,v_{n-1})\neq D(v_n, w_{n+1})$.  If $D(v_n,w_{n+1})$ consists solely of compressible components, then we are done.  If not, there exists a $v_{n+1}\in D(v_n,w_{n+1})$ such that $[v_{n+1}]_J$ is incompressible.  By Corollary \ref{permissible}, we have that $D(v_n,v_{n+1})=D(v_n,w_{n+1})\subseteq D(v_{n-1},w_{n+1})$.  It follows that $D(v_{n-1},w_{n+1})=D(v_{n-1},v_{n+1})$ and $D(v_n,v_{n+1})\subseteq D(v_{n-1},v_{n+1})$.  By Lemma \ref{incompSIL}, we have that $D(v_{n-1},v_{n+1})=D(v_{n-1},w_{n+1})\neq D(v_n,w_{n+1})=D(v_n,v_{n+1})$.  Therefore $D(v_{n-1},v_{n+1})\supsetneqq D(v_{n},v_{n+1})$.  Furthermore, by the contrapositive of Corollary \ref{permissible}, it follows that $D(v_{n-1},v_{n+1})=D(v_{n-1},v_n)$.  Hence $D(v_1,v_{n+1})\supsetneqq D(v_2,v_{n+1})\supsetneqq\dots\supsetneqq D(v_{n-1},v_{n+1})\supsetneqq D(v_n,v_{n+1})$. Since $\Gamma$ is a finite graph, this process must eventually terminate in a vertex $w_N$ such that $w_N$ is a $v_{N-1}$-expander and $D(v_{N-1},w_N)$ consists solely of compressible components.  This proves Claim 1.

\begin{clm} Let $v,w\in J$ be such that $[v]_J$ is incompressible, $w$ is a $v$-expander, and $D(v,w)\cap J$ consists solely of compressible components.  Then $\text{sep}(v,w)\setminus([v]_J\cup[w]_J)$ contains only compressible components.  
\end{clm}
Suppose on the contrary that $x\in\text{sep}(v,w)\setminus([v]_J\cup[w]_J)$ is such that $[x]_J$ is incompressible.  Then there exists an $x^*\in[x]_J$ such that $D(x^*,v)\neq D(x^*,w)$.  Then Corollary \ref{permissible} implies that $D(x^*,w)\subseteq D(v,w)$ consists solely of compressible components.  Since $w$ is a $v$-expander, Lemma \ref{incompSIL} implies that $D(x^*,w)\neq D(v,w)$.  The contrapositive of Corollary \ref{permissible} thus implies that $D(v,x^*)=D(v,w)$.  But this contradicts the fact that $D(v,w)$ consists solely of compressible components.  Therefore no such $[x]_J$ exists.  This proves Claim 2.

By Claim 1, there exists $v,w\in J$ such that $[v]_J$ is incompressible, $w$ is a $v$-expander, and $D(v,w)$ consists solely of compressible components.  Claim 2 implies that $[v]_J$ is the only incompressible component in $\text{sep}(v,w)$.  It follows that any minimal separating set in $\text{sep}(v,w)$ has at most one incompressible component.  This proves the statement.
\end{proof}

\section{Computing $\text{Im}(\rho_{\Gamma,J})$}\label{sect computing}
If a separating set $\text{sep}(v,w)$ is minimal, then by Proposition \ref{inclusion2}, we see that $\text{sep}(v,w)=\text{sep}(a,b)$ for all $a,b\in\text{sep}(v,w)$.  This makes minimal separating sets particularly nice to work with.  The following lemma gives another nice property of minimal separating sets.

\begin{lem}\label{min sep}
	Let $\text{sep}(v,w)$ be minimal with respect to inclusion.  Then for each $a\in\text{sep}(v,w)$ there exists an $a^*\in[a]_J$ such that for all $b\in \text{sep}(v,w)\setminus[a]_J$ we have that $D(a^*,b)\cap\text{sep}(v,w)\setminus[a]_J=[b]_J$.
\end{lem}

\begin{proof} 
	First, let $a\in[v]_J\cup[w]_J$.  Without loss of generality, assume that $a\in[v]_J$.  If $\text{sep}(v,w)=[v]_J\cup[w]_J$ then the statement is trivial.  If not, let $c\in\text{sep}(v,w)\setminus([v]_J\cup[w]_J)$.  Then by Proposition \ref{inclusion2} combined with minimality, we have that $\text{sep}(v,w)=\text{sep}(c,w)$.  Thus we may assume that $a\notin [v]_J\cup [w]_J$.
	
	Let $a^*\in[a]_J$ be such that $D(a^*,v)\neq D(a^*,w)$. Then for any $c\in \text{sep}(v,w)$ such that $c\notin[a]_J\cup[v]_J$, we show that  $D(a^*,v)\neq D(a^*,c)$.  This implies that $D(a^*,v)\cap\text{sep}(v,w)\setminus[a]_J=[v]_J$. 
	
	To begin, Proposition \ref{inclusion2} implies that $\text{sep}(v,c)\subseteq\text{sep}(v,w)$.  Then by minimality $\text{sep}(v,c)=\text{sep}(v,w)$.  Hence there exists a $w^*\in[w]_J$ such that $D(w^*,v)\neq D(w^*,c)$.  Since $D(a^*,v)\neq D(a^*,w)=D(a^*,w^*)$, Corollary \ref{permissible} implies that $D(a^*,v)\subseteq D(w^*,v)$.  But $c\notin D(w^*,v)$.  Therefore $c\notin D(a^*,v)$, as claimed.
	
	Next, let $b\in\text{sep}(v,w)\setminus([a]_J\cup [v]_J)$.  Then Proposition \ref{inclusion2} combined with minimality tells us that $\text{sep}(v,w)=\text{sep}(b,v)$. We showed above that $D(a^*,b)\neq D(a^*,v)$.  But this is the only assumption we needed to run the above argument.  Therefore $D(a^*,b)\cap\text{sep}(b,v)\setminus[a]_J = [b]_J$.
	
\end{proof}

Using Lemma \ref{min sep}, we make the following definition.

\begin{definition}
	Let $\text{sep}(v,w)$ be minimal with respect to inclusion. For each $x\in\text{sep}(v,w)\setminus([v]_J\cup[w]_J)$, fix an element $x^*\in [x]_J$ via Lemma \ref{min sep}.  Then for the given choices for all $[x]_J\subseteq \text{sep}(v,w)$, we define $$\text{sep}^*(v,w):=\{x^*\;|\; x\in\text{sep}(v,w)\}.$$  
\end{definition}

Note that this definition depends on the choice of $x^*$.  However, no matter what choice is made, it contains a unique representative of each component $[x]_J$ of $\Gamma_J$ that intersects $\text{sep}(v,w)$.
Furthermore, Lemma \ref{min sep} says that for each $a^*,b^*\in\text{sep}^*(v,w)$ we have that $D(a^*,b^*)\cap\text{sep}^*(v,w)=\{b^*\}$.  

With this definition in hand, we prove the following corollary.

\begin{cor}\label{cor sep*}
	Let $\text{sep}(v,w)$ be minimal with respect to inclusion.  Then for all distinct $a^*,b^*,c^*\in \text{sep}^*(v,w)$, we have $D(a^*,c^*)=D(b^*,c^*)$.  
\end{cor}

\begin{proof}
	By Lemma \ref{min sep}, we have that $c^*\notin D(b^*,a^*)$ and $b^*\notin D(a^*,c^*)$.  By a double application of Corollary \ref{permissible}, this implies that  $D(b^*,c^*)=D(a^*,c^*)$. This completes the proof.	
\end{proof}

We now consider a set that is slightly bigger than $\text{sep}(v,w)$.  We define this set now.

\begin{definition}
Let $\text{sep}(v,w)$ be minimal with respect to inclusion.  Fix a set $\text{sep}^*(v,w)$.  Then we define $$C(v,w):=\{c\in J\;|\; \exists a^*,b^*\in \text{sep}^*(v,w)\text{ such that } D(a^*,b^*)=D(a^*,c)\}.$$ 
\end{definition}

Using properties of $\text{sep}(v,w)$, we can show that $C(v,w)$ contains a special point.  This will allow us to construct a division and apply the TML-decomposition.  However, if $C(v,w)=J$, then this application will be trivial. We therefore deal with this case separately.

Note that if $\text{sep}(v,w)=J$ is minimal, then $C(v,w)=J$.  This case is easily dealt with.  Indeed the minimality of $\text{sep}(v,w)$, combined with Proposition \ref{inclusion2}, implies that $\text{sep}(a,b)=J$ for all $a,b\in J$ such that $[a]_J\neq [b]_J$.  Then by Theorem \ref{discrete}, we have that $\text{Im}(\rho_{\Gamma,J})\cong\Gamma_{k_J-1}(2)$.  Thus we may safely ignore this case.

\begin{prop}\label{Cij=J2}
Let $\text{sep}(v,w)\subsetneqq J$ be minimal.  Let $\text{sep}(v,w)\neq [v]_J\cup [w]_J$.  Then if $C(v,w)=J$, then $\delta$ has a splitting point.

\end{prop}

\begin{proof}
Pick some $c\notin \text{sep}(v,w)$ and some $s^*,t^*\in \text{sep}^*(v,w)$ such that $D(t^*,s^*)=D(t^*,c)$.  Then define  $A:=D(t^*,s^*)\setminus [s^*]_J$ and let $B:=D(t^*,s^*)^\complement\setminus[s^*]_J$.  We will show that $s^*$ is a splitting point with splitting $A\sqcup [s^*]_J\sqcup B$.

First, we have that given any $a\in A$ and any $r^*\in \text{sep}^*(v,w)\setminus([s^*]_J\cup [t^*]_J)$, Corollary \ref{cor sep*} implies that $D(t^*,a)=D(t^*,s^*)=D(r^*,s^*)$.  By Lemma \ref{min sep}, we have that $D(r^*,s^*)\neq D(r^*,t^*)$.  This in turn implies that $a\notin D(r^*,t^*)$.  By Corollary \ref{cor sep*}, we have $D(r^*,t^*)=D(s^*,t^*)$.  Thus $a\notin D(s^*,t^*)$.  Equivalently, $t^*\notin D(s^*,a)$.  Then Corollary \ref{permissible} implies that $D(s^*,a)\subseteq D(t^*,a)=D(t^*,s^*)$.  Finally, we conclude that $D(s^*,a)\cap J\setminus[s^*]_J\subseteq A$.  This also implies that for all $b\in B$ we have $D(s^*,b)\cap J\setminus[s^*]_J\subseteq B$.  


Next, note that given any $b\in B$ we have $b\notin D(t^*,s^*)$.  By Corollary \ref{permissible}, we have that $D(t^*,s^*)\subseteq D(b,s^*)$.  This implies that $B^\complement\subseteq D(b,s^*)\setminus[b]_J$.  

It remains to show that for any $a\in A$ we have $A^\complement\subseteq D(a,s^*)\setminus[a]_J$.  Note that since $a\in \text{sep}(v,w)^\complement$ and $s\in\text{sep}(v,w)$, Proposition \ref{supset2} implies that $\text{sep}(v,w)\subseteq D(a,s^*)$.  Thus it suffices to show that $D(a,b)=D(a,s^*)$ for any $b\in B\setminus\text{sep}(v,w)$.  Assume that $B\nsubseteq \text{sep}(v,w)$ and let $b\in B\setminus \text{sep}(v,w)$.  By our hypothesis, there exists $x^*,y^*\in \text{sep}^*(v,w)$ such that $D(y^*,x^*)=D(y^*,b)$.  By Corollary \ref{cor sep*}, we have $D(y^*,b)=D(y^*,x^*)=D(t^*,x^*)=D(t^*,b)\neq D(t^*,a)$.  This shows that $a\notin D(y^*,b)$.  Corollary \ref{permissible} then implies that $D(y^*,b)\subseteq D(a,b)$.  It follows that $x^*\in D(a,b)$.  Proposition \ref{supset2} implies that $\text{sep}(v,w)\subseteq D(a,b)$.  Therefore $D(a,b)=D(a,x^*)=D(a,s^*)$, which was to be shown. 

\end{proof}  

We now present an algorithm that can compute the isomorphism class of $\text{Im}(\rho_{\Gamma,J})$ for general $(\Gamma,J)$.  

\begin{thm}\label{final thm}
Given an arbitrary finite graph $\Gamma$ and vertex set $J\subseteq V(\Gamma)$, the group $\text{Im}(\rho_{\Gamma,J})$ can be computed up to isomorphism.
\end{thm}

\begin{proof}
We induct on $|J|$.

First, by using repeated applications of Proposition \ref{compressible}, we may assume that for all $v\in J$ either $[v]_J=\{v\}$ or $[v]_J$ is incompressible.  By Proposition \ref{minSepComp}, there exists some minimal separating set $\text{sep}(v,w)$ such that $[v]_J=\{v\}$.  
If $\text{sep}(v,w)=J$, then by minimality and Proposition \ref{inclusion2}, we have that $\text{sep}(a,b)=J$ for all distinct $a,b\in J$.  Hence   $\text{Im}(\rho_{\Gamma,J})=\Gamma_{k_J-1}(2)$ by Theorem \ref{discrete}.  Thus we may assume that $\text{sep}(v,w)\neq J$. Furthermore, by Lemma \ref{sep=22}, we may assume that $\text{sep}(v,w)\neq[v]_J\cup[w]_J$.  
We prove a couple of nice facts about $C(v,w)$:
\setcounter{clm}{0}
\begin{clm}
	 For any $a^*,b^*\in \text{sep}^*(v,w)$ and any $j\in C(v,w)^\complement$, we have that $D(a^*,j)=D(b^*,j)$ .
\end{clm}   

Let $a^*,b^*\in \text{sep}^*(v,w)$ be such that $a^*\neq b^*$ and let $j\in C(v,w)^\complement$. Then by definition of $C(v,w)$, we have that $D(a^*,j)\neq D(a^*,b^*)$ and $D(b^*,j)\neq D(b^*,a^*)$.  Therefore by a double application of Corollary \ref{permissible}, we have that $D(a^*,j)=D(b^*,j)$.  If $a^*=b^*$ then $D(a^*,j)=D(b^*,j)$.  This proves the claim.

\begin{clm}
	For any $u^*\in \text{sep}^*(v,w)$ and any $j\in C(v,w)^\complement$, we have that $D(u^*,j)\cap C(v,w)\setminus[u^*]_J=\emptyset$.
\end{clm}

	Let $u^*\in\text{sep}^*(v,w)$, $j\in C(v,w)^\complement$.  Let $c\in C(v,w)$ such that $[u^*]_J\neq [c]_J$.  
	By definition of $C(v,w)$, there exists $x^*,y^*\in\text{sep}^*(v,w)$ such that $D(x^*,c)=D(x^*,y^*)$. 
	By Claim 1, $D(u^*,j)=D(x^*,j)$.  But by definition of $C(v,w)$, we have that $D(x^*,j)\neq D(x^*,y^*)=D(x^*,c)$.  Therefore $c\notin D(u^*,j)$.  This proves the claim.
	
	
	\begin{clm}
		 For any $j\in C(v,w)^\complement$, we have that $D(j,v)\supseteq C(v,w)$
	\end{clm}  Let $j\in C(v,w)^\complement$.   Given any $c\in C(v,w)$, there exists $x^*,y^*\in \text{sep}^*(v,w)$ such that $D(x^*,c)=D(x^*,y^*)$. By Proposition \ref{supset2}, we have that $D(j,v)\supseteq\text{sep}(v,w)$.  This implies that $D(j,v)=D(j,y^*)$.  By definition of $C(v,w)$, we have $D(x^*,y^*)\neq D(x^*,j)$.  Therefore by Corollary \ref{permissible}, we have $D(x^*,c)=D(x^*,y^*)\subseteq D(j,y^*)=D(j,v)$. Thus $c\in D(j,v)$.  This proves the claim.

If $C(v,w)=J$, then since $\text{sep}(v,w)\neq J$ and $\text{sep}(v,w)\neq[v]_J\cup [w]_J$, we are done by Proposition \ref{Cij=J2} combined with Proposition \ref{ss reduction2}.  If not,
then we may write $J=\displaystyle\bigsqcup_{i=1}^m A_i$ where $A_1 = C(v,w)$ and for all $1<i\leq m$, we have $A_i=[a_i]_J$ for some $a_i\in J$.  We show that this is a non-trivial division of $J$.  Hence we are done by the TML decomposition (Theorem \ref{tml}).

It is trivial to verify that for any $a\in J$, we have that $a$ is a special point in $[a]_J$.  Hence we need only show that $A_1$ contains a special point.  We show that $v$ is a special point in $A_1$.  We assumed earlier that $[v]_J=\{v\}$.  Hence by default $v\in\text{sep}^*(v,w)$.  Then Claim 2 implies that for all $j\in J$ either $D(v,j)\cap J\subseteq A_1$ or $D(v,j)\cap J\subseteq A_1^\complement$, and Claim 3 implies that for all $j\in A_1^\complement$ we have that $A_1\subseteq D(j,v)$.    Therefore $v$ is a special point in $A_1$.

  
\end{proof}

\chapter{Additional Results}\label{chapter additional results}

\section{Star groups}\label{sect star subgroups}
In this section we show that every element in a certain family of subgroups of $\Gamma_n(2)$ can be written as $\text{Im}(\rho_{\Gamma,J})$ for some $(\Gamma,J)$. We will use the following notation from the proof of Theorem \ref{discrete}.  Given $1\leq i,j \leq n$, $i\neq j$, let $E_{i,j}$ denote the matrix identical to the identity matrix except in the $(i,j)$-entry, which equals $2$ and let $F_i$ denote the matrix identical to the identity matrix except in the $(i,i)$-entry, which equals $-1$.  

\begin{definition}
Let $I\subseteq \{1,2,\dots, n\}^2$ be a subset with the following properties. For all $i\in\{1,2,\dots, n\}$ we have $(i,i)\in I$, and for each $(i,j),(j,k)\in I$ we have $(i,k)\in I$.  Then we call $I$ a \underline{star set of width n}. Given any star set of width n, the \underline{star group associated to $I$} is the group $$G_I:=\{A\in \Gamma_n(2)\;|\; A_{i,j}=0 \text{ for all }(i,j)\notin I\}.$$
\end{definition}

It is not immediately clear that $G_I$ is a group, in particular that it is closed under inverses.  This will follow from Proposition \ref{gen set} in which we compute a generating set for $G_I$.  First, however, we need the following lemma.


\begin{lem}\label{ij to ji}
Let $I$ be a star set of width n, and let $S_i:=\{(i,j)\;|\; 1\leq j \leq n, (i,j)\in I\}$. Assume that $|S_m|$ is maximal.  If $(i,m)\in I$, then $(m,i)\in I$
\end{lem}


\begin{proof}
Assume that $(i,m)\in I$.  It follows that for all $1\leq j \leq n$ such that $(m,j)\in I$ we have that $(i,j)\in I$.  Hence $|S_i|\geq |S_m|$.  But $|S_m|$ is maximal.  Hence $|S_i|=|S_m|$.  But since $(m,j)\in I \implies (i,j)\in I$, it follows that $i\in \{j\;|\; (i,j)\in I\}=\{j\;|\; (m,j)\in I\}$.  Therefore $(m,i)\in I$.  
\end{proof}

We now compute a generating set for each $G_I$.
\begin{prop}\label{gen set}
Let $I$ be a star set.  Then $G_I$ is the subgroup of $\Gamma_n(2)$ generated by $\{F_i \;|\; 1\leq i \leq n\}\cup\{E_{i,j}\;|\; (i,j)\in I, i\neq j\}$
\end{prop}

\begin{proof}
Let $H_I$ be the group generated by  $\{F_i \;|\; 1\leq i \leq n\}\cup\{E_{i,j}\;|\; (i,j)\notin I, i\neq j\}$.  It is clear that $H_I\leq G_I$.  Thus it suffices to show that $G_I\leq H_I$.  We proceed by induction on the width of $I$. The base case is trivial.

Assume for induction that for every star set $I$ of width less than $n$, we have $G_I=H_I$.  Let $I$ be a subgroup of width $n$, and let $A\in G_I$.  To show that $G_I\leq H_I$, we will take an arbitrary element of $G_I$ and multiply it by elements of $H_I$ to reduce it to the identity matrix.

Since $A\in \Gamma_n(2)$, we must have $A_{1,1}\equiv_2 1$. Assume that $|A_{1,1}|>1$.  We show that we can multiply $A$ by elements of $H_I$ to obtain a matrix $A'\in G_I$ such that $|A'_{1,1}|<|A_{1,1}|$. By induction, this will allow us to assume that $|A_{1,1}|=1$.  If $A_{1,1}$ is negative, then $(F_1A)_{1,1}$ is positive.  Since $F_1\in H_I$, we may assume that $A_{1,1}$ is positive.  If $A_{i,1}$ is a multiple of $A_{1,1}$ for all $i$ then $A$ is not invertible.  Therefore there exists an $i$ such that $A_{i,1}$ is not a multiple of $A_{1,1}$.  Multiplying by $F_i$ on the left if necessary, we may assume that $A_{i,1}$ is positive.

Assume that $A_{i,1}>A_{1,1}$.  Then $(i,1)\in I$, so $E_{i,1}\in H_I$.  By the division algorithm, there exists $q,r\in \Z$ such that $A_{i,1}=qA_{1,1}+r$ and $0\leq r < A_{1,1}$.  If $q=2k$ for some $k\in\Z$, then $(E_{i,1}^{-k}A)_{i,1}=r<A_{1,1}$ and $(E_{i,1}^{-k}A)_{j,1}=A_{j,1}$ for all $j\neq i$.  If $q=2k-1$ for some $k\in Z$, then  $-A_{1,1}<r-A_{1,1}=A_{i,1}-2kA_{1,1}<0$.  Therefore  $0<(F_{i}E_{i,1}^{-k}A)_{i,1}<A_{1,1}$ and $(F_iE_{i,1}^{-k}A)_{j,1}=A_{j,1}$ for all $j\neq i$.  We may therefore assume that $0<A_{i,1}<A_{1,1}$.

Let $S_i$ be as in Lemma \ref{ij to ji}.  Up to reindexing, we may assume that $|S_1|$ is maximal.  Since $(i,1)\in I$, Lemma \ref{ij to ji} implies that $(1,i)\in I$.  Therefore $E_{1,i}\in H_I$.  Since $0<A_{i,1}<A_{1,1}$, it follows that $|A_{1,1}-2A_{i,1}|<A_{1,1}$.  Hence $|E_{1,i}^{-1}A|_{1,1}<A_{1,1}$ and $(E_{1,i}^{-1}A)_{j,1}=A_{j,1}$ for all $j\neq 1$.  Thus we can reduce to the case where $|A_{1,1}|=1$.

If $A_{1,1}=-1$, then $F_{1}\in H_I$ and $(F_{1}A)_{1,1}=1$.  Thus we may assume that $A_{1,1}=1$.  For all $i\neq 1$ we have that $A_{i,1}$ is even.  Assume that $A_{i,1}=2k\neq 0$.  Then $(i,1)\in I$, hence $E_{i,1}\in G_I$.  It follows that $E_{i,1}^{-k}A\in G_I$.  But $(E_{i,1}^{-k}A)_{i,1}=0$ and $(E_{i,1}^{-k})_{j,1}=A_{j,1}$ for all $j\neq i$.  Repeating this process for each $i$ such that $A_{i,1}\neq 0$, we reduce to a matrix of the form $\begin{bmatrix} 1 & * \\ 0 & *\end{bmatrix}$. 

For each $(1,i)\in I$, we have that $E_{1,i}\in H$.  If $A_{1,i}=2k\neq 0$, then $(AE_{1,i}^{-k})_{1,i}=0$ and $(AE_{1,i}^{-k})_{1,j}=A_{1,j}$ for all $j\neq i$. Repeating this process for each $i$ such that $A_{1,i}\neq 0$, we reduce to a matrix of the form $\begin{bmatrix} 1 & 0 \\ 0 & B\end{bmatrix}$.  

Let $I':=\{(i,j)\in I \;|\; i\neq 1, j\neq 1\}$.  Then $I'$ is a star set of rank $n-1$. By induction, $B\in H_{I'}$.  But there is a natural embedding $$\iota: H_{I'}\to H_I,\quad B\mapsto \begin{bmatrix}1 & 0 \\ 0 & B\end{bmatrix}.$$ Therefore $A\in H_I$, which completes the proof.  \end{proof}

Now that we have a generating set for the star groups, we will show that every star group is the image of some representation $\rho_{\Gamma,J}$.  To that end, we first give a condition on $J$ which results in $\text{Im}(\rho_{\Gamma,J})$ being a star group.

\begin{thm}\label{singleton image}
Let $\{v_1,\dots, v_n\}$ be a set of representatives of the components of $\Gamma_J$.  Assume that $D(v_n,v_i)\cap \{v_1,\dots, v_n\}=\{v_i\}$ for all $v_i\in \{v_1,\dots, v_{n-1}\}$.  Let $I:=\{(i,j)\;|\; 1\leq i,j\leq n-1, D(v_i,v_j)\cap \{v_1,\dots, v_n\}=\{v_j\}\}$.  Then $I$ is a star set of width $n-1$ and $\text{Im}(\rho_{\Gamma,J})\cong G_I$.
\end{thm}

\begin{proof}
First assume that $D(v_i,v_j)\cap\{v_1,\dots, v_n\}=\{v_j\}$ and $D(v_j,v_k)\cap\{v_1,\dots, v_n\}=\{v_k\}$.  Then since $v_j\notin D(v_i,v_k)$, Corollary \ref{permissible} implies that $D(v_i,v_k)\subseteq D(v_j,v_k)$, so that $D(v_i,v_k)\cap \{v_1,\dots, v_n\}=\{v_k\}$.  This shows that $I$ is a star set.

Fix the following basis of $I_J(H_1(\hat X; \Q))$. 
\begin{center}
$\hat{v}_n-\hat{v}_1, \hat{v}_n-\hat{v}_2,\dots,\hat{v}_{n}-\hat{v}_{n-1}$.
\end{center}
\noindent By Corollary \ref{remove}, $\{\hat\sigma_{D,v}\;|\; v_n\notin D\}$ is a generating set.  Fix some $\hat\sigma_{D(v_i,v_j),v_i}$ such that $v_n\notin D(v_i,v_j)$.  Then by Corollary \ref{permissible}, we have that $D(v_i,v_j)\cap \{v_1,\dots, v_n\}\subseteq D(v_n,v_j)\cap \{v_1,\dots, v_n\}=\{v_j\}$ so $D(v_i,v_j)\cap \{v_1,\dots, v_n\}=\{v_j\}$. If $i=n$, then by direct calculation $\hat\sigma_{D(v_i,v_j),v_i}=F_j$.  If $i\neq n$, then by direct calculation $\hat\sigma_{D(v_i,v_j),v_i}=F_jE_{i,j}$.  Therefore by Proposition \ref{gen set} $$\text{Im}(\rho_{\Gamma,J})\cong\langle \{F_i\;|\; 1\leq i \leq n-1\}\cup \{E_{i,j}\;|\; 1\leq i,j \leq n-1, (i,j)\in I\}\rangle\cong G_I.$$
\end{proof}

Now that we have a theorem that recognizes when the image of a given representation is a star group, we can show that all such groups can be written as $\text{Im}(\rho_{\Gamma,J})$ for some $(\Gamma,J)$.

\begin{thm}
Let $I$ be a star set of width $n$.  Then there exists a graph $\Gamma$ and a vertex set $J$ such that $\text{Im}(\rho_{\Gamma,J})\cong G_I$.
\end{thm}

\begin{proof}
Let $J=\{v_1,v_2,\dots, v_n,v_{n+1}\}$ and let $W:=\{w_{i,j}\;|\; 1\leq i< j \leq n+1\}$.  Let $$V(\Gamma):=J\sqcup W$$ and let $$E(\Gamma):=\{(v_i,w_{j,k})\;|\; (i,j)\in I\text{ or }(i,k)\in I\text{ or }i=n+1\}$$

By construction, $\Gamma$ is a bipartite graph, and $J$ consists of pairwise non-adjacent vertices.  Since $(v_{n+1},w_{i,j})\in E(\Gamma)$ for all $1\leq i<j\leq n+1$, we have that $D(v_{n+1},v_i)\cap J = \{v_i\}$ for all $1\leq i \leq n+1$.  

Assume that $(i,j)\in I$.  It follows that $j\neq n+1$.  Let $(v_j, w_{b,c})\in E(\Gamma)$.  Then either $(j,b)\in I$ or  $(j,c)\in I$.  Since $I$ is a star set, it follows that either $(i,b)\in I$ or $(i,c)\in I$.  Hence $(v_i,w_{b,c})\in E(\Gamma)$.  This shows that $\text{lk}(v_j)\subseteq \text{lk}(v_i)$, which implies that $D(v_i,v_j)\cap J=\{v_j\}$.   

Now let $1\leq i,j\leq n$ be such that $D(v_i,v_j)\cap J = \{v_j\}$.  Then since $(v_j, w_{j,n+1},v_{n+1})$ is a path in $\Gamma$, we must have that $w_{j,n+1}\in\text{lk}(v_i)$.  This implies that $(i,j)\in I$.  

We have shown that $D(v_{n+1},v_i)\cap J = \{v_i\}$ for all $1\leq i \leq n$ and that $I=\{(i,j)\;|\; 1\leq i,j \leq n, D(v_i,v_j)\cap J = \{v_j\}\}$.  Therefore by Theorem \ref{singleton image}, we have that $\text{Im}(\rho_{\Gamma,J})\cong G_I$.
\end{proof}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% BIBLIOGRAPHY %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%{
%\singlespacing

%\bibliographystyle{plain}
%\bibliography{Xbib}
%}
\singlespacing

\begin{thebibliography}{CRSV}
\bibitem[AC]{ac} J. Aramayona, C. Martinez-Perez, On the first cohomology of automorphism groups of graph groups, J. Algebra 452, 2016
\bibitem[C]{c} C. Cunningham, On the automorphism groups of universal right-angled Coxeter groups, Thesis (Ph.D.)-Tufts University, 2015
\bibitem[CG]{cg} L.J. Corredor, M.A. Gutierrez, A generating set for the automorphism group of a graph product of abelian groups, Internat. J. Algebra Comput. 22(1), 2012
\bibitem[CR]{cr} C.W. Curtis, I. Reiner, Representation theory of finite groups and associative algebras, Interscience Publishers, 1962
\bibitem[CRSV]{crsv} R. Charney, K. Ruane, N. Stambaugh, A. Vuayan, The automorphism group of a graph product with no SIL, Illinois J. Math 54(1), 2010
\bibitem[CV]{cv} R. Charney, K. Vogtmann, Subgroups and quotients of automorphism groups of RAAGs, Proc. Sympos. Pure Math 82, 2011
\bibitem[F]{f} N. Fullarton, A generating set for the palindromic Torelli group,\\ Algebr. Geom. Topol. 15(6), 2015
\bibitem[GL]{gl} F. Grunewald, A. Lubotzky, Linear representations of the automorphism group of a free group, Geom. Funct. Anal. 18(5), 2009
\bibitem[GLLM]{gllm} F. Grunewald, M. Larsen, A. Lubotzky, J. Malestein, Arithmetic quotients of the mapping class group, Geom. Funct. Anal. 25(5), 2015
\bibitem[GPR]{gpr} M. Gutierrez, A. Piggott, K. Ruane, On the automorphisms of a graph product	of abelian groups, Groups Geom. Dyn. 6(1), 2012
\bibitem[GS]{gs} V. Guirardel, A. Sale, Vastness properties of automorphism groups of RAAGs, 2017
\bibitem[H]{h} A. Hadari, Every infinite order mapping class has an infinite order action on the homology of some finite cover, arXiv:1508.01555, 2015
\bibitem[H2]{h2} A. Hadari, Homological eigenvalues of lifts of pseudo-Anosov mapping classes to finite covers, arXiv:1712.01416, 2017

\bibitem[H3]{h3} A. Hatcher, Algebraic topology, Cambridge University Press, 2002
\bibitem[J]{j} L. Ji, A tale of two groups: arithmetic groups and
mapping class groups, 2010
\bibitem[K]{k} R. Kobayashi, A finite presentation of the level 2 principal congruence subgroup of $\text{GL}(n;\Z)$, 	Kodai Math. J. 38(3), 2015
\bibitem[K2]{k2} T. Koberda, Asymptotic linearity of the mapping class group and a homological version of the Nielsen-Thurston classification, Geom. Dedicata 156, 2012
\bibitem[KW]{kw} A. Kaul, M. White, Centralizers of Coxeter elements and inner
automorphisms of right-angled Coxeter groups, Int. J. Algebra	 3(10), 2009 
\bibitem[L]{l} Y. Liu, Virtual homological spectral radii for automorphisms of surfaces, arXiv:1710.05039, 2017
\bibitem [M]{m} J. McCarthy, A "Tits-alternative" for subgroups of surface mapping class groups, 1985
\bibitem[S]{s} J. Serre, Linear representations of finite groups, Springer, 1977
\bibitem[SS]{ss} A. Sale, T. Susse, Outer automorphism groups of right-angled Coxeter
groups are either large or virtually abelian, arXiv:1706.07873, 2017
\end{thebibliography}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% DEDICATION PAGE  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%\newpage
\newpage

{\singlespacing
\begin{center}
{
\par
\vspace{1.2in}
%\large
DEDICATION
\par
\vspace{0.17in}
to
\par
\vspace{1.2in}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%5
My wife
\par
\vspace{0.17in}
Lauren Elizabeth Fehr, and
\par
\vspace{0.17in}
My daughter
\par
\vspace{0.17in}
Marie Rhetta Morgan
\par
\vspace{1.2in}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
For
\par
\vspace{0.17in}
Loving me so well
\par
\vfill
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
}
\end{center}
}


\end{document}
