Actions of cofinite groups on cofinite graphs

Abstract

We defined group actions on cofinite graphs to characterize a unique way of uniformly topologizing an abstract group with profinite topology, induced by the cofinite graphs, so that the aforesaid action becomes uniformly equicontinuous.

Keywords:

• Profinite graph
• Cofinite graph
• Profinite group
• Cofinite group
• Equicontinuous group action

1 Introduction

1.1 Topological graphs

Definition 1.1

Topological Graphs A topological graph [1] is a topological space $\Gamma$ that is partitioned into two closed subsets $V\left(\Gamma \right)$ and $E\left(\Gamma \right)$ together with two continuous functions $s,t:E\left(\Gamma \right)\to V\left(\Gamma \right)$ and a continuous function $\overline{\phantom{e}}:E\left(\Gamma \right)\to E\left(\Gamma \right)$ satisfying the following properties: for every $e\in E\left(\Gamma \right)$,

(1)	$\overline{e}\ne e$ and $\overline{\overline{e}}=e$;
(2)	$t\left(\overline{e}\right)=s\left(e\right)$ and $s\left(\overline{e}\right)=t\left(e\right)$.

The elements of $V\left(\Gamma \right)$ are called vertices. An element $e\in E\left(\Gamma \right)$ is called a (directed) edge with source $s\left(e\right)$ and target $t\left(e\right)$; the edge $\overline{e}$ is called the reverse or inverse of $e$.

A map of graphs $f:\Gamma \to \Delta$ is a function that maps vertices to vertices, edges to edges, and preserves sources, targets, and inverses of edges. Analogously, we will call a map of graphs a graph isomorphism if and only if it is a bijection.

An orientation of a topological graph $\Gamma$ is a closed subset $E^{+}\left(\Gamma \right)$ consisting of exactly one edge in each pair $\{e,\overline{e}\}$. In this situation, setting $E^{-}\left(\Gamma \right)=\{e\in E\left(\Gamma \right)\mid \overline{e}\in E^{+}\left(\Gamma \right)\}$ we see that $E\left(\Gamma \right)$ is a disjoint union of the two closed (hence also open) subsets $E^{+}\left(\Gamma \right)$, $E^{-}\left(\Gamma \right)$.

Note 1.2

Let $\Gamma$ be a topological graph. The following are equivalent:

(1)	$\Gamma$ admits an orientation;
(2)	there exists a continuous map of graphs from $\Gamma$ to the discrete graph with a single vertex and a single edge and its inverse;
(3)	there exists a continuous map of graphs $f:\Gamma \to \Delta$ for some discrete graph $\Delta$.

Conceivably there are topological graphs that do not admit closed orientations. However such graphs will not concern us. Therefore, unless otherwise stated, by a topological graph we will henceforth mean a topological graph that admits an orientation.

We will be interested in equivalence relations on graphs that are compatible with the graph structure:

Definition 1.3

Compatible Equivalence Relation An equivalence relation $R$ on a graph $\Gamma$ is compatible if the following properties hold:

(1)	$R=R_V\cup R_E$ where $R_V$, $R_E$ are equivalence relations on $V\left(\Gamma \right)$, $E\left(\Gamma \right)$, precisely the restriction of $R$;
(2)	if $(e_1,e_2)\in R$, then $(s\left(e_1\right),s\left(e_2\right))\in R$, $(t\left(e_1\right),t\left(e_2\right))\in R$, and $({\overline{e}}_1,{\overline{e}}_2)\in R$;
(3)	for all $e\in E\left(\Gamma \right)$, $(e,\overline{e})\notin R$;

Note 1.4

If $K$ is a compatible equivalence relation on $\Gamma$, then there is a unique way to make $\Gamma ∕K$ into a graph such that the canonical map $\Gamma \to \Gamma ∕K$ is a map of graphs. It is defined by setting $s\left(K\left[e\right]\right)=K\left[s\left(e\right)\right]$,$t\left(K\left[e\right]\right)=K\left[t\left(e\right)\right]$, and$\overline{K\left[e\right]}=K\left[\overline{e}\right]$.

Conversely, if $\Delta$ is a graph and $f:\Gamma \to \Delta$ is a surjective map of graphs, then $K=f^{-1}f=\{(a,b)\in \Gamma \times \Gamma \mid f\left(a\right)=f\left(b\right)\}$ is a compatible equivalence relation on $\Gamma$ and$f$ induces an isomorphism of graphs such that $\Gamma ∕K\cong \Delta$.

Note 1.5

If $R_1$ and$R_2$ are compatible equivalences on $\Gamma$, then so is $R_1\cap R_2$.

Theorem 1.6

Let $R$ be any cofinite equivalence relation on a topological graph $\Gamma$. Then there exists a compatible cofinite equivalence [2] relation$S$ on$\Gamma$ such that $S\subseteq R$.

Proof

Extend the source and target maps $s,t:E\left(\Gamma \right)\to V\left(\Gamma \right)$ to all of $\Gamma$ so that they are both the identity map on $V\left(\Gamma \right)$. Then $s,t:\Gamma \to \Gamma$ are continuous maps satisfying the following properties:

•	$s^2=s$, $t^2=t$, $st=t$, and $ts=s$;
•	$s\left(x\right)=x⟺t\left(x\right)=x⟺x\in V\left(\Gamma \right)$.

Similarly, extend the edge inversion map $\overline{\phantom{e}}:E\left(\Gamma \right)\to E\left(\Gamma \right)$ to all of $\Gamma$ by also letting it be the identity map on $V\left(\Gamma \right)$. Then $\overline{\phantom{e}}:\Gamma \to \Gamma$ is a continuous map satisfying the following conditions for all $x\in \Gamma$:

•	$\overline{\overline{x}}=x$;
•	$\overline{x}=x⟺x\in V\left(\Gamma \right)$;
•	$s\left(\overline{x}\right)=t\left(x\right)$ and $t\left(\overline{x}\right)=s\left(x\right)$.

Now define $S_1=\{(x,y)\in \Gamma \times \Gamma \mid (s\left(x\right),s\left(y\right))\in R\}={(s\times s)}^{-1}\left[R\right]$, $S_2=\{(x,y)\in \Gamma \times \Gamma \mid (t\left(x\right),t\left(y\right))\in R\}={(t\times t)}^{-1}\left[R\right]$, and $S_3=\{(x,y)\in \Gamma \times \Gamma \mid (\overline{x},\overline{y})\in R)={(\overline{\phantom{e}}\times \overline{\phantom{e}})}^{-1}\left[R\right]$. Then, by the Correspondence Theorem [2], $S_1$, $S_2$, $S_3$ are cofinite equivalence relations on $\Gamma$. Let $S_4=R\cap S_1\cap S_2\cap S_3$ and observe that

(i)	$S_4$ is a cofinite equivalence relation on $\Gamma$;
(ii)	if $(e_1,e_2)\in S_4$, then $(s\left(e_1\right),s\left(e_2\right))\in S_4$, $(t\left(e_1\right),t\left(e_2\right))\in S_4$, and $({\overline{e}}_1,{\overline{e}}_2)\in S_4$.

Finally, choose a closed orientation $E^{+}\left(\Gamma \right)$ of $\Gamma$ and form the restrictions $S_V=S_4\cap [V\left(\Gamma \right)\times V\left(\Gamma \right)]$, $S_{E^{+}}=S_4\cap [E^{+}\left(\Gamma \right)\times E^{+}\left(\Gamma \right)]$, and $S_{E^{-}}=S_4\cap [E^{-}\left(\Gamma \right)\times E^{-}\left(\Gamma \right)]$. Then it is easy to check that $S=S_V\cup S_{E^{+}}\cup S_{E^{-}}$ is a compatible cofinite equivalence relation on $\Gamma$ and $S\subseteq R$, as required.□

The previous proof actually shows a little more, which is worth noting. Given a closed orientation $E^{+}\left(\Gamma \right)$ for $\Gamma$, we say that a compatible equivalence relation $R$ on $\Gamma$ is orientation preserving if whenever $(e,e^{\prime })\in R$ and $e\in E^{+}\left(\Gamma \right)$, then also $e^{\prime }\in E^{+}\left(\Gamma \right)$. Since the equivalence relation $S$ that we constructed in the proof of Theorem 1.6 is also orientation preserving, we proved the following stronger result.

Corollary 1.7

Let $\Gamma$ be a topological graph with a specified closed orientation $E^{+}\left(\Gamma \right)$. Then for any cofinite equivalence relation $R$ on $\Gamma$, there exists a compatible orientation preserving cofinite equivalence relation$S$ on$\Gamma$ such that $S\subseteq R$.

Corollary 1.8

If $\Gamma$ is a compact Hausdorff totally disconnected topological graph, then its compatible cofinite equivalence relations form a fundamental system of entourages for the unique uniform structure that induces the topology of $\Gamma$ [3].

1.2 Cofinite graphs

Definition 1.9

Cofinite Graph A cofinite graph [2] is an abstract graph $\Gamma$ endowed with a Hausdorff uniformity such that the compatible cofinite entourages [2] of $\Gamma$ form a fundamental system of entourages (i.e. every entourage of $\Gamma$ contains a compatible cofinite entourage).

A group $G$ is said to act uniformly equicontinuously over a cofinite graph $\Gamma$ if and only if for each entourage $W$ over $\Gamma$ there exists an entourage [2], $V$ over $\Gamma$ such that for all $g$ in $G,(g\times g)\left[V\right]\subseteq W$, where $(g\times g)\left[V\right]=\{(g\cdot x,g\cdot y):(x,y)\in V\}$ and $g\cdot x$ is the image of $x\in \Gamma$ under the group action of $g\in G$. In this case the group action induces a (Hausdorff) uniformity over $G$ if and only if the aforesaid action is faithful.

Suppose that $G$ is a group acting faithfully and uniformly equicontinuously on a cofinite graph $\Gamma$, then the action $G\times \Gamma \to \Gamma$ is uniformly continuous. Also in that case $\widehat{G}$, the [4] profinite completion of $G$, acts on $\widehat{\Gamma }$, the [2] profinite of completion of $\Gamma$, uniformly equicontinuously. Following is an example of uniform equicontinuous group action.

Example 1.10

Let $\Gamma$ be an abstract graph with $V\left(\Gamma \right)=\{x:x\in \mathbb{Z}\}$, where $\mathbb{Z}$ is the set of all integers. Let, $E^{+}\left(\Gamma \right)=\{e_x:x\in \mathbb{Z}\},s\left(e_x\right)=x,t\left(e_x\right)=x+1$. Let, $E^{-}\left(\Gamma \right)$ be the set of all edges reversing the edges of $E^{+}\left(\Gamma \right)$, that is $E^{-}\left(\Gamma \right)=\{\overline{e_x}:x\in \mathbb{Z}\}$ and $s\left(\overline{e_x}\right)=t\left(e_x\right),t\left(\overline{e_x}\right)=s\left(e_x\right)$. Let $p$ be any prime. Then for any positive integer $n$, consider ${\Gamma }_n$ as the cycle of length $p^n$. One can say that $V\left({\Gamma }_n\right)=\{{\left[0\right]}_n,{\left[1\right]}_n,{\left[2\right]}_n...{[p^n-1]}_n\}$, where ${\left[x\right]}_n$ is the congruence class of $x$ modulo $p^n$ and $E^{+}\left({\Gamma }_n\right)=\{e_{{\left[x\right]}_n}:x\in V\left({\Gamma }_n\right)\},s\left(e_{{\left[x\right]}_n}\right)={\left[x\right]}_n,t\left(e_{{\left[x\right]}_n}\right)={[x+1]}_n$. Let $E^{-}\left({\Gamma }_n\right)$ be the set of edges reversing the edges in $E^{+}\left({\Gamma }_n\right)$, that is $E^{-}\left({\Gamma }_n\right)=\{\overline{e_{{\left[x\right]}_n}}:x\in V\left({\Gamma }_n\right)\}$ and $s\left(\overline{e_{{\left[x\right]}_n}}\right)=t\left(e_{{\left[x\right]}_n}\right),t\left(\overline{e_{{\left[x\right]}_n}}\right)=s\left(e_{{\left[x\right]}_n}\right)$. Now, consider the map of graphs $q_n:\Gamma \to {\Gamma }_n$ as $q_n\left[x\right]={\left[x\right]}_n$ and $q_n\left(e_x\right)=e_{{\left[x\right]}_n}$. Let, $R_n=Ker{q}_n=\{(\gamma ,\delta )\in \Gamma \times \Gamma :q_n\left(\gamma \right)=q_n\left(\delta \right)\}$. Then $R_n$ is a compatible equivalence relation over $\Gamma$ [2]and since there is a one-one, onto map of graphs from $\Gamma ∕R_n$ to ${\Gamma }_n$, $|\Gamma ∕R_n|<\infty$. And $I=\{R_n:n\in N\}$ is a fundamental system of entourages over $\Gamma$. The corresponding topology induced by $I$ is also Hausdorff, since for any two distinct $\gamma ,\delta \in \Gamma$, there exists sufficiently large natural number $n$ so that $R_n\left[x\right]\bigcap R_n\left[y\right]=\varphi$.Thus $\Gamma$ turns to be a cofinite graph. Consider the additive group of integers $(\mathbb{Z},+)$ and a natural group action $\mathbb{Z}\times \Gamma \mapsto \Gamma$ by translation of vertices and edges as follows: For any $g\in \mathbb{Z},x\in V\left(\Gamma \right),g.x=g+x$ and for any $e_x\in E^{+}\left(\Gamma \right),g.e_x=e_{g+x}$, for any $\overline{e_x}\in E^{-}\left(\Gamma \right),g.e_x=\overline{e_{g+x}}$. For any entourage $U$ over $\Gamma$, as $I$ is a fundamental system of entourage over $\Gamma$, there exists $n\in N$ so that $R_n\subseteq U$ and for all $g\in \mathbb{Z},(g\times g)\left[R_n\right]\subseteq R_n$. For if $x,y\in R_n$, without loss of generality let us assume that $x,y\in V\left(\Gamma \right)$. So, ${\left[x\right]}_n={\left[y\right]}_n$ which implies ${[g+x]}_n={[g+y]}_n$ and that implies $(g.x,g.y)\in R_n$. Thus the above action is uniformly equicontinuous.

2 Groups acting on cofinite graphs

Let $G$ be a group and $\Gamma$ be a cofinite graph. We say that the group $G$ acts over $\Gamma$ if and only if

(1)	For all $x$ in $\Gamma$, for all $g$ in $G,g.x$ is in $\Gamma$
(2)	For all $x$ in $\Gamma$, for all $g_1,g_2$ in $G,g_1.(g_2.x)=\left(g_1{g}_2\right).x$
(3)	For all $x$ in $\Gamma$, $1.x=x$, where $1\in G$ is the identity element of $G$.
(4)	For all $v$ in $V\left(\Gamma \right)$, for all $g$ in $G,g.v$ is in $V\left(\Gamma \right)$ and for all $e$ in $E\left(\Gamma \right)$, for all $g$ in $G,g.e$ is in $E\left(\Gamma \right)$.
(5)	For all $e$ in $E\left(\Gamma \right)$, for all $g$ in $G,g.s\left(e\right)=s(g.e),g.t\left(e\right)=t\left(ge\right),g.\left(\overline{e}\right)=\overline{g.e}$. We say, $s\left(e\right),t\left(e\right),\overline{e}$ are the source of $e$, target of $e$ and inversion of $e$ respectively, such that $s\left(\overline{e}\right)=t\left(e\right),t\left(\overline{e}\right)=s\left(e\right)$ and $\overline{\overline{e}}=e$.
(6)	There exists a $G-$invariant orientation $E^{+}\left(\Gamma \right)$ of $\Gamma$.

Note that the aforesaid group action restricted to {g} can be treated as a well defined map of graphs, $\Gamma \to \Gamma$ taking $x\mapsto g.x$.

Definition 2.1

Uniform Equicontinuous Group Action A group $G$ is said to act uniformly equicontinuously over a cofinite graph $\Gamma$, if and only if for each entourage $W$ over $\Gamma$ there exists an entourage $V$ over $\Gamma$ such that for all $g$ in $G,(g\times g)\left[V\right]$ is a subset of $W$.

Example 2.2

Let $\Gamma$ be an abstract graph with $V\left(\Gamma \right)=\{x:x\in \mathbb{Z}\}$, where $\mathbb{Z}$ is the set of all integers. Let, $E^{+}\left(\Gamma \right)=\{e_x:x\in \mathbb{Z}\},s\left(e_x\right)=x,t\left(e_x\right)=x+1$. Let, $E^{-}\left(\Gamma \right)$ be the set of all edges reversing the edges of $E^{+}\left(\Gamma \right)$, that is $E^{-}\left(\Gamma \right)=\{\overline{e_x}:x\in \mathbb{Z}\}$ and $s\left(\overline{e_x}\right)=t\left(e_x\right),t\left(\overline{e_x}\right)=s\left(e_x\right)$. Let $\mathcal{N}$ be a separating filter base [2] of finite index normal subgroups of $(\mathbb{Z},+)$, the additive group of integers. Then for any subgroup $n\mathbb{Z}\in \mathcal{N}$, consider ${\Gamma }_n$ as the cycle of length $n$. One can say that $V\left({\Gamma }_n\right)=\{{\left[0\right]}_n,{\left[1\right]}_n,{\left[2\right]}_n...{[n-1]}_n\}$, where ${\left[x\right]}_n$ is the congruence class of $x$ modulo $n$ and $E^{+}\left({\Gamma }_n\right)=\{e_{{\left[x\right]}_n}:x\in V\left({\Gamma }_n\right)\},s\left(e_{{\left[x\right]}_n}\right)={\left[x\right]}_n,t\left(e_{{\left[x\right]}_n}\right)={[x+1]}_n$. Let $E^{-}\left({\Gamma }_n\right)$ be the set of edges reversing the edges in $E^{+}\left({\Gamma }_n\right)$, that is $E^{-}\left({\Gamma }_n\right)=\{\overline{e_{{\left[x\right]}_n}}:x\in V\left({\Gamma }_n\right)\}$ and $s\left(\overline{e_{{\left[x\right]}_n}}\right)=t\left(e_{{\left[x\right]}_n}\right),t\left(\overline{e_{{\left[x\right]}_n}}\right)=s\left(e_{{\left[x\right]}_n}\right)$. Now, consider the map of graphs $q_n:\Gamma \to {\Gamma }_n$ as $q_n\left[x\right]={\left[x\right]}_n$ and $q_n\left(e_x\right)=e_{{\left[x\right]}_n}$. Let, $R_n=Ker{q}_n=\{(\gamma ,\delta )\in \Gamma \times \Gamma :q_n\left(\gamma \right)=q_n\left(\delta \right)\}$. Then $R_n$ is a compatible equivalence relation over $\Gamma$ [2] and since there is a one-one, onto map of graphs from $\Gamma ∕R_n$ to ${\Gamma }_n$, $|\Gamma ∕R_n|<\infty$. And $I=\{R_n:n\mathbb{Z}\in \mathcal{N}\}$ is a fundamental system of entourages over $\Gamma$. The corresponding topology induced by $I$ is also Hausdorff, since for any two distinct $\gamma ,\delta \in \Gamma$, there exists sufficiently large natural number $n$ so that $R_n\left[x\right]\bigcap R_n\left[y\right]=\varphi$.Thus $\Gamma$ turns to be a cofinite graph. Consider the additive group of integers $(\mathbb{Z},+)$ and a natural group action $\mathbb{Z}\times \Gamma \mapsto \Gamma$ by translation of vertices and edges as follows: For any $g\in \mathbb{Z},x\in V\left(\Gamma \right),g.x=g+x$ and for any $e_x\in E^{+}\left(\Gamma \right),g.e_x=e_{g+x}$, for any $\overline{e_x}\in E^{-}\left(\Gamma \right),g.e_x=\overline{e_{g+x}}$. For any entourage $U$ over $\Gamma$, as $I$ is a fundamental system of entourage over $\Gamma$, there exists $n\mathbb{Z}\in \mathcal{N}$ so that $R_n\subseteq U$ and for all $g\in \mathbb{Z},(g\times g)\left[R_n\right]\subseteq R_n$. For if $x,y\in R_n$, without loss of generality let us assume that $x,y\in V\left(\Gamma \right)$. So, ${\left[x\right]}_n={\left[y\right]}_n$ which implies ${[g+x]}_n={[g+y]}_n$ and that implies $(g.x,g.y)\in R_n$. Thus the above action is uniformly equicontinuous.

Lemma 2.3

If a group $G$ acts uniformly equicontinuously over a cofinite graph $\Gamma$, then there exists a fundamental system of entourages consisting of$G$-invariant compatible cofinite entourages over $\Gamma$, i.e. for any entourage $U$ over$\Gamma$ there exists a compatible cofinite entourage $R$ over $\Gamma$ such that for all $g\in G,(g\times g)\left[R\right]\subseteq R\subseteq U$.

Proof

Let $U$ be any cofinite entourage [2] over $\Gamma$. Then as $G$ acts uniformly equicontinuously over $\Gamma$, there exists a compatible cofinite entourage $S$ over $\Gamma$ such that for all $g\in G,(g\times g)\left[S\right]\subseteq U$. Choose a $G$-invariant orientation $E^{+}\left(\Gamma \right)$ of $\Gamma$. Without loss of generality, we can assume that our compatible equivalence relation $S$ on $\Gamma$ is orientation preserving i.e. whenever $(e,e^{\prime })\in R$ and $e\in E^{+}\left(\Gamma \right)$, then also $e^{\prime }\in E^{+}\left(\Gamma \right)$. Now $S\subseteq {\cup }_{g\in G}(g\times g)\left[S\right]\subseteq U$. Now if $S_0={\cup }_{g\in G}(g\times g)\left[S\right]$ and $T=〈S_0〉$, where $〈S_0〉$ is the smallest unique equivalence relation on $\Gamma$ containing $S_0$, namely, the intersection of all equivalence relations that contains $S_0$. Note that $S\subseteq T\subseteq U$. Since for all $h\in G,(h\times h)\left[S_0\right]=S_0$ and $S_0^{-1}=S_0$ it follows that $T$ is in the transitive closure of $S_0$. Let $(x,y)\in T$. Then there exists a finite sequence $x_0,x_1,\dots ,x_n$ such that $(x_i,x_{i+1})\in S_0$, for all $i=0,1,2,\dots ,n-1$ and $x=x_0,y=x_n$. Hence $(g{x}_i,g{x}_{i+1})\in S_0$, for all $i=0,1,2,\dots ,n-1$, for all $g\in G$. Thus $(g{x}_0,g{x}_n)=(gx,gy)\in T$, for all $g\in G$. Hence for all $g\in G,(g\times g)\left[T\right]\subseteq T$ and our claim that $T$ is a $G$-invariant cofinite entourage, follows. It remains to check that $T$ is compatible. Let $(x,y)\in T$. If $(x,y)\in S_0$, then there is $(t,s)\in S=S_V\cup S_E$ and $g\in G$ such that $(gt,gs)=(x,y)$. Without loss of generality let $(t,s)\in S_V$. Then $(t,s)\in V\left(\Gamma \right)\times V\left(\Gamma \right)$ which implies that $(x,y)\in T_V$. Now let $(x,y)\in T\setminus S_0$. Then there exists a finite sequence $x_0,x_1,\dots ,x_n$ such that $(x_i,x_{i+1})\in S_0$, for all $i=0,1,2,\dots ,n-1$ and $x=x_0,y=x_n$. Hence by the previous argument if $(x_0,x_1)\in T_V$ then $(x_i,x_{i+1})\in T_V$, for all $i=1,2,\dots ,n-1$. Thus $(x,y)\in T_V$. If $(x_0,x_1)\in T_E$ then $(x_i,x_{i+1})\in T_E$, for all $i=1,2,\dots ,n-1$, which implies $(x,y)\in T_E$. Let $(e_1,e_2)\in T_E$. If $(e_1,e_2)\in S_0$, then there is $(p,q)\in S$ and $g\in G$ such that $(gp,gq)=(e_1,e_2)$. Then $(s\left(p\right),s\left(q\right))\in S$. So $(s\left(e_1\right),s\left(e_2\right))$ which equals $(gs\left(p\right),gs\left(q\right))$ is in $(g\times g)\left[S\right]\subseteq S_0$ so that $(s\left(e_1\right),s\left(e_2\right))\in T$. Now let $(e_1,e_2)\in T\setminus S_0$. Then there exists a finite sequence $x_0,x_1,\dots ,x_n$ such that $(x_i,x_{i+1})\in S_0,\forall i=0,1,2,\dots ,n-1$ and $e_1=x_0,e_2=x_n$. Hence by the previous argument $(s\left(x_i\right),s\left(x_{i+1}\right))\in T,\forall i=0,1,2,\dots ,n-1$ and thus $(s\left(e_1\right),s\left(e_2\right))\in T$. Similarly, $(t\left(e_1\right),t\left(e_2\right))\in T$ and $(\overline{e_1},\overline{e_2})\in T$. Finally, to show that for any $e\in E^{+}\left(\Gamma \right),(e,\overline{e})\notin T$, if possible let $(e,\overline{e})\in T$. If $(e,\overline{e})\in S_0$, then there is $(p,q)\in S$ and $g\in G$ such that $(gp,gq)=(e,\overline{e})$. Then $\overline{e}=\overline{gp}=g\overline{p}=gq$ which implies that $\overline{p}=q$, so $(p,\overline{p})\in S$, a contradiction. Now let $(e,\overline{e})\in T\setminus S_0$. Then there exists a finite sequence $x_0,x_1,\dots ,x_n$ such that $(x_i,x_{i+1})\in S_0$, for all $i=0,1,2,\dots ,n-1$ and $e=x_0,\overline{e}=x_n$. Now let there be $(p,q)\in S$ and $g\in G$ such that $(gp,gq)=(x_0,x_1)$. Without loss of generality we may assume $(p,q)\in E^{+}\left(\Gamma \right)\times E^{+}\left(\Gamma \right)$. Then $(gp,gq)=(x_0,x_1)\in E^{+}\left(\Gamma \right)\times E^{+}\left(\Gamma \right)$. Hence $(x_i,x_{i+1})\in E^{+}\left(\Gamma \right)\times E^{+}\left(\Gamma \right)$, for all $i=1,2,\dots ,n-1$ which implies that $(e,\overline{e})\in E^{+}\left(\Gamma \right)\times E^{+}\left(\Gamma \right)$, a contradiction. Our claim follows.□

Note that in reference to Example 2.2, $I$ is in fact a fundamental system of $G$-invariant compatible cofinite entourages over $\Gamma$.

Note 2.4

Let $G$ be a group and$\Gamma$ be a cofinite graph. Let $S$ be an equivalence relation over $G$ then$S\left[g\right]=\{h\in G:(g,h)\in S\}$ is the equivalence class of $g\in G$. Similarly, if $S$ is an equivalence relation on $\Gamma$ then$S\left[\gamma \right]=\{\rho \in \Gamma :(\gamma ,\rho )\in S\}$ is the equivalence class of $\gamma \in G$. Let $G$ act on $\Gamma$. Let$R$ be a cofinite entourage. We define $N_R=\{(g_1,g_2)\in G\times G:g_{1}R\left[\gamma \right]=g_{2}R\left[\gamma \right],\forall \gamma \in \Gamma \}$, and$N_R\left[1\right]=\{g\in G:(1,g)\in N_R\}$, [4]. In the following lemmas we will show that $N_R$ is a congruence of $G$ and$N_R\left[1\right]$ is a normal subgroup of $G$ with finite index and we denote it by $N_R\left[1\right]{⊲}_{f}G$.

Lemma 2.5

$N_R\left[1\right]$ is a finite index normal subgroup of $G$ and $G∕N_R\left[1\right]$ is isomorphic with $G∕N_R$. More generally, if $N$ is a congruence on $G$, then $N\left[1\right]$ is a normal subgroup of $G$ and $G∕N\left[1\right]\cong G∕N$.

Proof

Let us first see that $N_R\left[1\right]{⊲}_{f}G$ for all $G$-invariant compatible cofinite entourage $R$ over $\Gamma$. Let $g,h\in N_R\left[1\right]$. This implies $(1,g)\in N_R$ and hence $(g,1),(1,h)\in N_R$. Thus $(g,h)\in N_R$. This implies $(g.x,h.x)$ is in $R$, for all $x\in \Gamma$ and so $(x,g^{-1}h.x)\in R$, for all $x\in \Gamma$. Hence, $(1,g^{-1}h)$ is in $N_R$ and thus $g^{-1}h\in N_R\left[1\right]$. So, $N_R\left[1\right]\le G$. For all $g\in G$, for all $x\in \Gamma ,g.x\in \Gamma$. Hence for all $k\in N_R\left[1\right],(x,k.x)\in R$, hence $(k.x,x)$ is in $R$. Thus $(kg.x,g.x)\in R$ and $(g^{-1}kg.x,g^{-1}g.x)=(g^{-1}kg.x,x)\in R$. Hence $(g^{-1}kg,1)\in N_R$. So, $g^{-1}kg\in N_R\left[1\right]$ and thus $N_R\left[1\right]⊲G$. Now let us define $\eta$ from $G∕N_R\left[1\right]$ to $G∕N_R$ via $\eta \left(g{N}_R\left[1\right]\right)=N_R\left[g\right]$. Then, $g{N}_R\left[1\right]$ is equal to $h{N}_R\left[1\right]$ if and only if $h^{-1}g\in N_R\left[1\right]$ if and only if $(1,h^{-1}g)\in N_R$ if and only if $(x,h^{-1}g.x)\in R$ if and only if $(h.x,g.x)\in R$ if and only if $(h,g)\in N_R$ if and only if $N_R\left[h\right]=N_R\left[g\right]$, for all $x$ in $\Gamma$. Thus $\eta$ is a well defined injection and hence $|G∕N_R\left[1\right]|\le |G∕N_R|<\infty$. Hence $N_R\left[1\right]{⊲}_{f}G$. It follows that $G∕N_R$ is a group and let us define $\zeta :G∕N_R\left[1\right]\to G∕N_R$ via $\zeta \left(g{N}_R\left[1\right]\right)=N_R\left[g\right]$. Then for $g_1,g_2$ in $G,g_1{N}_R\left[1\right]=g_2{N}_R\left[1\right]$ if and only if $g_2^{-1}{g}_1\in N_R\left[1\right]$ if and only if $(1,g_2^{-1}{g}_1)\in N_R$ if and only if $(x,g_2^{-1}{g}_1.x)\in R$ if and only if $(g_2.x,g_1.x)\in R$ if and only if $(g_2,g_1)\in N_R$ if and only if $N_R\left[g_2\right]$ equals $N_R\left[g_1\right]$. Hence $\zeta$ is a well defined injection. Also for all $N_R\left[g\right]$ in $G∕N_R$, there exists $g{N}_R\left[1\right]\in G∕N_R\left[1\right]$ such that $\zeta \left(g{N}_R\left[1\right]\right)=N_R\left[g\right]$. Thus $\zeta$ is surjective as well. Also for $g_1{N}_R\left[1\right],g_2{N}_R\left[1\right]\in G∕N_R\left[1\right]$, we have $\zeta \left(g_1{N}_R\left[1\right]g_2{N}_R\left[1\right]\right)=\zeta \left(g_1{g}_2{N}_R\left[1\right]\right)$ and that equals $N_R\left[g_1{g}_2\right]$ which equals $N_R\left[g_1\right]N_R\left[g_2\right]=\zeta \left(g_1{N}_R\left[1\right]\right)\zeta \left(g_2{N}_R\left[1\right]\right)$. Hence $\zeta$ is a group homomorphism and thus a group isomorphism. Also, both $G∕N_R\left[1\right],G∕N_R$, are finite discrete topological groups, so $\zeta$ is an isomorphism of cofinite groups as well.□

Lemma 2.6

Let a group $G$ act on a cofinite graph $\Gamma$ uniformly equicontinuously. Then $G$ acts on $\Gamma ∕R$ and$G∕N_R$ acts on$\Gamma ∕R$ as well, where $R$ is a$G$-invariant compatible cofinite entourage over $\Gamma$ and $\Gamma ∕R$ is the quotient graph of $\Gamma$ with respect to $R$. If$I$ is a fundamental system of $G$-invariant compatible cofinite entourages over $\Gamma$, then $\{N_R\mid R\in I\}$ forms a fundamental system of cofinite congruences [5] for some uniformity over$G$.

Proof

Let $R$ be a $G$-invariant compatible cofinite entourage over $\Gamma$. Let us define a group action $G\times \Gamma ∕R\to \Gamma ∕R$ via $g.R\left[x\right]=R[g.x]$, for all $g\in G$, for all $x\in \Gamma$. Now let $R\left[x\right]=R\left[y\right]$ so $(x,y)\in R$ which implies that $(g.x,g.y)\in R$. Then $R[g.x]=R[g.y]$. Hence the induced group action is well defined.

Let us now consider the group action $G∕N_R\times \Gamma ∕R\to \Gamma ∕R$, defined via $N_R\left[g\right].R\left[x\right]=R[g.x]$, for all $x\in \Gamma$, for all $g\in G$. Now let $(N_R\left[g\right],R\left[x\right])=(N_R\left[h\right],R\left[y\right])$ which implies that $(g,h)\in N_R,(x,y)$ is in $R$. Then $(g.x,h.x)\in R$, as $h^{-1}\in G,(h^{-1}g.x,h^{-1}h.x)\in R$. So $(h^{-1}g.x,y)\in R$. Thus $(g.x,h.y)\in R$ which implies that $R[g.x]$ equals $R[h.y]$. Hence the induced group action is well defined. Let us now show that $N_R$ is an equivalence relation over $G$, for all $G$-invariant compatible cofinite entourage $R$ over $\Gamma$.

(1)	for all $g\in G$, for all $x\in \Gamma ,(g.x,g.x)\in R$. Hence $(g,g)\in N_R$, for all $g\in G$ which implies that $D\left(G\right)\subseteq N_R$.
(2)	Now $(g,h)\in N_R\iff (g.x,h.x)\in R$, for all $x\in \Gamma$ $\iff (h.x,g.x)\in R$, for all $x\in \Gamma$. $\iff (h,g)\in N_R$. Thus $N_R^{-1}=N_R$.
(3)	Let $(g,h),(h,k)\in N_R$. This implies $(g.x,h.x),(h.x,k.x)$ is in $R,\forall x\in \Gamma$. Hence $(g.x,k.x)\in R$, for all $x\in \Gamma$. So $(g,k)\in N_R$ which implies that ${\left(N_R\right)}^2\subseteq N_R$.

Also we now check that $N_R$ is a congruence over $G$. For, let us take $(g_1,g_2),(g_3,g_4)\in N_R$. Then for all $x\in \Gamma ,(g_1.x,g_2.x),(g_3.x,g_4.x)\in R$; for all $x\in \Gamma ,g_3.x\in \Gamma$ and so $(g_1{g}_3.x,g_2{g}_3.x)\in R$ and $(g_2{g}_3.x,g_2{g}_4.x)$ is in $R$, since $R$ is $G$-invariant. Thus $(g_1{g}_3.x,g_2{g}_4.x)\in R$, for all $x\in \Gamma$ so that $(g_1{g}_3,g_2{g}_4)\in N_R$. Thus our claim follows. Let us now show that $G∕N_R$ is finite. Furthermore, define $g:\Gamma ∕R\to \Gamma ∕R$ as $g$ maps $\left(R\left[x\right]\right)$ into $R[g.x]$. Now, $R\left[x\right]=R\left[y\right]⟺(x,y)\in R$ if and only if $(g.x,g.y)\in R⟺R[g.x]=R[g.y]$. Hence the map $g$ is a well defined injection. Now for all $R\left[x\right]\in \Gamma ∕R$ there exists $g^{-1}R\left[x\right]\in \Gamma ∕R$ such that $g\left(g^{-1}R\left[x\right]\right)$ equals $R\left[x\right]$. Hence $g\in Sym(\Gamma ∕R)$, where $Sym(\Gamma ∕R)$ is the collection of all graph isomorphisms from $\Gamma ∕R\to \Gamma ∕R$, [2]. Now let us define a map $\theta :G∕N_R\to Sym(\Gamma ∕R)$ via $\theta \left(N_R\left[g\right]\right)=g$. Now $N_R\left[g_1\right]$ equals $N_R\left[g_2\right]$ if and only if $(g_1,g_2)\in N_R$ if and only if $(g_1.x,g_2.x)\in R$ for all $x\in \Gamma$. Hence $(g_1.x,g_2.x)\in R$ if and only if $R[g_1.x]=R[g_2.x]$ if and only if $g_1\left(R\left[x\right]\right)=g_2\left(R\left[x\right]\right)$ if and only if $g_1=g_2$ in $Sym(\Gamma ∕R)$. Hence $\theta$ is a well defined injection. Thus $|G∕N_R|\le \left|Sym(\Gamma ∕R)\right|<\infty$ as $|\Gamma ∕R|<\infty$. So, next we would like to show that $\{N_R\mid R\in I\}$ forms a fundamental system of cofinite congruences over $G$.

(1)	$D\left(G\right)\subseteq N_R$, for all $R\in I$, as $N_R$ is reflexive.
(2)	Now for some $R,S\in I,(g_1,g_2)\in N_R\bigcap N_S$ if and only if $(g_1.x,g_2.x)\in R\bigcap S$, for all $x\in \Gamma \iff (g_1,g_2)\in N_{R\bigcap S}$. Thus $N_R\bigcap N_S=N_{R\bigcap S}$.
(3)	For all $N_R,N_R^2=N_R$, as $N_R$ is transitive.
(4)	For all $N_R,N_R^{-1}=N_R$, as $N_R$ is symmetric.

Hence our claim follows.□

Note 2.7

Let us refer back to Example 2.2 and define a group action$\mathbb{Z}\times {\Gamma }_n\mapsto {\Gamma }_n$ as following $g.{\left[x\right]}_n={[g+x]}_n$, for any $x\in V\left({\Gamma }_n\right),g.e_{{\left[x\right]}_n}=e_{{[g+x]}_n},g.\overline{e_{{\left[x\right]}_n}}=\overline{e_{g+x}}$, for any $\overline{e_x}\in E^{-}\left({\Gamma }_n\right)$. Thus for any $n$, where$n\mathbb{Z}\in \mathcal{N},\mathbb{Z}∕N_{R_n}$ is isomorphic to $\mathbb{Z}∕n\mathbb{Z}$.

Definition 2.8

We say a group $G$ acts on a cofinite graph $\Gamma$ faithfully, if for all $g$ in $G\setminus \left\{1\right\}$ there exists $x$ in $\Gamma$ such that $gx$ is not equal to $x$ in $\Gamma$.

Lemma 2.9

The induced uniform topology over $G$ as inLemma 2.6 is Hausdorff if and only if $G$ acts faithfully over $\Gamma$.

Proof

Let us first assume that $G$ acts faithfully over $\Gamma$. Now let $g\ne h$ in $G$. Then $h^{-1}g\ne 1$. So there exists $x\in \Gamma$ such that $h^{-1}g.x\ne x$ implying that $g.x\ne h.x$. Then there exists a $G$-invariant compatible cofinite entourage $R$ over $\Gamma$ such that $(g.x,h.x)\notin R$, as $\Gamma$ is Hausdorff. Hence $(g,h)\notin N_R$. Thus $G$ is Hausdorff.

Conversely, let us assume that $G$ is Hausdorff and let $g\ne 1$ in $G$. Then there exists some $G$-invariant compatible cofinite entourage $R$ over $\Gamma$ such that $(1,g)\notin N_R$. Hence there exists $x\in \Gamma$ such that $(x,g.x)\notin R$. Hence $R\left[x\right]\ne R[g.x]$ so that $x\ne g.x$. Our claim follows.□

Lemma 2.10

Suppose that $G$ is a group acting uniformly equicontinuously on a cofinite graph$\Gamma$ and give$G$ the induced uniformity as inLemma 2.6. Then the action $G\times \Gamma \to \Gamma$ is uniformly continuous.

Proof

Let $R$ be a $G$-invariant cofinite entourage over $\Gamma$. If $I$ is a fundamental system of $G$-invariant compatible cofinite entourages over $\Gamma$. Then $\{N_R\times R:R\in I\}$ ia a fundamental system of entourage for a uniform structure over $G\times \Gamma$, [2]. Now let $((g,x),(h,y))\in N_R\times R$, i.e. $(g,h)\in N_R,(x,y)\in R$. Now $x$ in $\Gamma$ and $(gx,hx)\in R$ this implies $(h^{-1}gx,x)\in R$. We have $(h^{-1}gx,y)\in R$ and hence $(gx,hy)\in R$. Thus our claim.□

Let us define a directed order ‘$\le$’ on $I$, a fundamental system of $G$-invariant entourages on a cofinite graph $\Gamma$ as in Lemma 2.6. We say, $R\le S$ in $I$, then $S\subseteq R$. Let $(g_1,g_2)\in N_S$. Then $(g_{1}x,g_{2}x)\in S$, for all $x\in \Gamma$ and hence $(g_{1}x,g_{2}x)\in R$, for all $x\in \Gamma$ which implies $(g_1,g_2)\in N_R$. Thus $N_S\subseteq N_R$. For all $R\le S$, in $I$, let us define ${\psi }_{RS}:G∕N_S\to G∕N_R$ via ${\psi }_{RS}\left(N_S\left[g\right]\right)=N_R\left[g\right]$. Then ${\psi }_{RS}$ is a well defined uniformly continuous group isomorphism, as each of $G∕N_R,G∕N_S$ is finite discrete groups. If $R=S$, then ${\psi }_{RR}=i{d}_{G∕N_R}$. And if $R\le S\le T$, then ${\psi }_{RS}{\psi }_{ST}={\psi }_{RT}$. Then $\{G∕N_R\mid R\in I,{\psi }_{RS},R\le S\in I\}$, forms an inverse system of finite discrete groups. Let $\widehat{\Gamma }={\underleftarrow{lim}}_{R\in I}\Gamma ∕R$ and $\widehat{G}={\underleftarrow{lim}}_{R\in I}G∕N_R$, where ${\psi }_R:\widehat{G}\to G∕N_R$ is the corresponding canonical projection map, [2]. Now if $I_1,I_2$ are two fundamental systems of $G$-invariant cofinite entourages over $\Gamma$, clearly $I_1,I_2$ will form fundamental systems of cofinite congruences, for two induced uniformities, over $G$. Now let $N_{R_1}$ be a cofinite congruence over $G$ for some $R_1\in I_1$. Then there exists a $R_2$, cofinite entourage over $\Gamma$, such that $R_2\in I_2$ and $R_2\subseteq R_1$. Hence $N_{R_2}\subseteq N_{R_1}$. Now let $N_{S_2}$ be a cofinite congruence over $G$ for some $S_2\in I_2$. Then there exists $S_1$, cofinite entourage over $\Gamma$, such that $S_1\in I_1$ and $S_1\subseteq S_2$. Hence $N_{S_1}\subseteq N_{S_2}$. Thus any cofinite congruence corresponding to the directed set $I_1$ is a cofinite congruence corresponding to the directed set $I_2$ and vice versa. Thus the two induced uniform structures over $G$ are equivalent and so the completion of $G$ with respect to the induced uniformity, from the cofinite graph $\Gamma$, is unique up to both algebraic and topological isomorphism.

Theorem 2.11

If $G$ acts on$\Gamma$, as in Lemma 2.6, faithfully then $\widehat{G}$ acts on $\widehat{\Gamma }$ uniformly equicontinuously.

Proof

Let a group $G$ act on $\Gamma$ uniformly equicontinuously. We fix a $G$-invariant orientation $E^{+}\left(\Gamma \right)$ of $\Gamma$. By Lemma 2.10 the action is uniformly continuous as well. Let $\chi :G\times \Gamma \to \Gamma$ be this group action. Now since $\Gamma$ is topologically embedded in $\widehat{\Gamma }$ by the inclusion map, say, $i$, the map $i\circ \chi :G\times \Gamma \to \widehat{\Gamma }$ is a uniformly continuous. Then there exists a unique uniformly continuous map $\widehat{\chi }:\widehat{G}\times \widehat{\Gamma }\to \widehat{\Gamma }$ that extends $\chi$. We claim that $\widehat{\chi }$ is the required group action. We can take $\widehat{\Gamma }=\underleftarrow{lim}\Gamma ∕R$ and $\widehat{G}=\underleftarrow{lim}G∕N_R$, where $R$ runs throughout all $G$-invariant compatible cofinite entourages of $\Gamma$ that are orientation preserving. Then $\widehat{G}\times \widehat{\Gamma }=\underleftarrow{lim}(G∕N_R\times \Gamma ∕R)\cong \underleftarrow{lim}(G∕N_R\times \Gamma ∕R)$ [6] and we define a group action of $G$ over $\Gamma$ coordinatewise as follows ${\left(N_R\left[g_R\right]\right)}_R.{\left(R\left[x_R\right]\right)}_R={\left(R[g_R.x_R]\right)}_R$. If possible let, $({\left(N_R\left[g_R\right]\right)}_R,{\left(R\left[x_R\right]\right)}_R)=({\left(N_R\left[h_R\right]\right)}_R,{\left(R\left[y_R\right]\right)}_R)$. So, $N_R\left[g_R\right]$ equals $N_R\left[h_R\right]$ and $R\left[x_R\right]=R\left[y_R\right],\forall R\in I,(g_R,h_R)\in N_R$ and $(x_R,y_R)\in R$. This implies that $(g_R.x_R,h_R.x_R)\in R$ which further ensures that $(h_R^{-1}{g}_R.x_R,x_R)\in R$. Then $(h_R^{-1}{g}_R.x_R,y_R)\in R$ and $(g_R.x_R,h_R.y_R)\in R$. Hence ${\left(R[g_R.x_R]\right)}_R={\left(R[h_R.y_R]\right)}_R$. So, the action is well defined. Let $g={\left(N_R\left[g_R\right]\right)}_R$ and $h={\left(N_R\left[h_R\right]\right)}_R$ in $\widehat{G}$, $x={\left(R\left[x_R\right]\right)}_R\in \widehat{\Gamma }$. Now $h.(g.x)=h.{\left(R[g_R.x_R]\right)}_R={\left(R[h_R{g}_R.x_R]\right)}_R$ which then equals ${\left(N_R\left[h_R{g}_R\right]\right)}_R.x=\left(hg\right).x$. Hence the action is associative. Now ${\left(N_R\left[1\right]\right)}_R.{\left(R\left[x_R\right]\right)}_R={\left(R\left[1x_R\right]\right)}_R={\left(R\left[x_R\right]\right)}_R$. Furthermore for all vertex $v={\left(R\left[v_R\right]\right)}_R\in V\left(\widehat{\Gamma }\right)$ and for all $g={\left(N_R\left[g_R\right]\right)}_R\in \widehat{G}$ one can say that $g.v={\left(R[g_R.v_R]\right)}_R\in V\left(\widehat{\Gamma }\right)$ as each $g_R.v_R\in V\left(\Gamma \right)$. Similarly, for all $e={\left(R\left[e_R\right]\right)}_R$ in $E\left(\widehat{\Gamma }\right)$ and for all $g={\left(N_R\left[g_R\right]\right)}_R$ in $\widehat{G}$, $g.e={\left(R\left[g_R{e}_R\right]\right)}_R$ in $E\left(\widehat{\Gamma }\right)$. For all $e={\left(R\left[e_R\right]\right)}_R$ in $E\left(\widehat{\Gamma }\right)$, for all $g={\left(N_R\left[g_R\right]\right)}_R$ in $\widehat{G}$, we have $s(g.e)=s\left({\left(R\left[g_R{e}_R\right]\right)}_R\right)$ and so ${\left(R\left[g_{R}s\left(e_R\right)\right]\right)}_R$ equals $(g.{\left(R\left[s\left(e_R\right)\right]\right)}_R)$ and that equals $g.s\left(e\right)$. Hence the properties $t(g.e)=g.t\left(e\right)$ and $\overline{g.e}=g.\overline{e}$ follow similarly. Finally, let $E^{+}\left(\widehat{\Gamma }\right)$ consist of all the edges ${\left(R\left[e_R\right]\right)}_R$, where $e_R\in E^{+}\left(\Gamma \right)$. Since each $R$ is orientation preserving, it follows that $E^{+}\left(\widehat{\Gamma }\right)$ is an orientation of $\widehat{\Gamma }$. Since $E^{+}\left(\Gamma \right)$ is $G$-invariant, we see that $E^{+}\left(\widehat{\Gamma }\right)$ is $\widehat{\Gamma }$-invariant. Hence this is a well defined group action. Also for all $g\in G$, and $x\in \Gamma$, ${\left(N_R\left[g\right]\right)}_R.{\left(R\left[x\right]\right)}_R$ equals ${\left(R[g.x]\right)}_R$ which equals $g.x$ in $\Gamma$, (please see [6], for any further clarification on how to embed $G$ in $\widehat{G}$ and $\Gamma$ in $\widehat{\Gamma }$. We use the notations ${\left(N_R\left[g\right]\right)}_R$ and ${\left(R\left[x\right]\right)}_R,R{\left[gx\right]}_R$ to refer to the $R$th coordinates of $g$ and $x,gx$ in $\widehat{G}$ and $\widehat{\Gamma }$, respectively). Thus the restriction of this group action agrees with the group action $\chi$. Now $\{R\mid R\in I\}$ is a fundamental system of cofinite entourages over $\Gamma$, and $\{N_R\mid R\in I\}$ is a fundamental system of cofinite congruences over $G$. Hence $\{\overline{R}\mid R\in I\}$ is a fundamental system of cofinite entourages over $\widehat{\Gamma }$ and $\{\overline{N_R}\mid R\in I\}$ is a fundamental system of cofinite congruences over $\widehat{G}$ respectively, where $\overline{R}$ is the topological closure of $R$ in $\Gamma \times \Gamma$. Let us now see that the aforesaid group action is uniformly continuous. For let us consider the group action $G∕N_R\times \Gamma ∕R\to \Gamma ∕R$ defined via $N_R\left[g\right]R\left[x\right]=R[g.x]$, which is uniformly continuous as both $G∕N_R\times \Gamma ∕R$ and $\Gamma ∕R$ are finite discrete uniform topological spaces. Hence the group action, $\widehat{G}\times \widehat{\Gamma }\to \widehat{\Gamma }$ is uniformly continuous. Thus the aforesaid group action is our choice of $\widehat{\chi }$, by the uniqueness of $\widehat{\chi }$, [2]. So the restriction of the aforesaid action $\left\{\widehat{g}\right\}\times \widehat{\Gamma }\to \widehat{\Gamma }$ is a uniformly continuous map of graphs, for all $\widehat{g}\in \widehat{G}$. We check that for all $(x,y)\in R$ and for all $\widehat{g}\in \widehat{G}$ the ordered pair $(\widehat{g}.x,\widehat{g}.y)\in \overline{R}$. For, let $\widehat{g}={\left(N_R\left[g_R\right]\right)}_R\in \widehat{G}$ and for $x,y\in \Gamma ,({\left(R\left[x\right]\right)}_R,{\left(R\left[y\right]\right)}_R)\in R$. Now $\overline{R}\left[{\left(R[g_R.x]\right)}_R\right]=\overline{R}[g_R.x]=\overline{R}[g_R.y]=\overline{R}\left[{\left(R[g_R.y]\right)}_R\right]$. So, $({\left(N_R\left[g_R\right]\right)}_R\cdot {\left(R\left[x\right]\right)}_R,{\left(N_R\left[g_R\right]\right)}_R\cdot {\left(R\left[y\right]\right)}_R)\in \overline{R}$. This implies $(\widehat{g}\times \widehat{g})\left[R\right]$ is a subset of $\overline{R}$. Thus for all $\widehat{g}\in \widehat{G}$ we observe that $(\widehat{g}\times \widehat{g})\left[\overline{R}\right]$ is a subset of $\overline{\widehat{g}\times \widehat{g}\left[R\right]}$ which is a subset of $\overline{\overline{R}}=\overline{R}$. Hence $\overline{R}$ is $\widehat{G}$ invariant.□

Thus ${\Phi }_1=\{N_{\overline{R}}\mid R\in I\}$ and ${\Phi }_2=\{\overline{N_R}\mid R\in I\}$ form fundamental systems of cofinite congruences over $\widehat{G}$. Let ${\tau }_{{\Phi }_1},{\tau }_{{\Phi }_2}$ be the topologies induced by ${\Phi }_1,{\Phi }_2$ respectively.

Theorem 2.12

The uniformities on $\widehat{G}$ obtained by ${\Phi }_1$ and ${\Phi }_2$ are equivalent.

Proof

Let us first show that $N_{\overline{R}}\cap G\times G=N_R$. For, let $(g,h)\in N_R$. Then for all $x\in \Gamma ,(g.x,h.x)\in R\subseteq \overline{R}$. Now let ${\left(R\left[x_R\right]\right)}_R\in \widehat{\Gamma }$. Then $\overline{R}\left[g{\left(R\left[x_R\right]\right)}_R\right]=\overline{R}[g.x_R]=\overline{R}[h.x_R]=\overline{R}\left[h{\left(R\left[x_R\right]\right)}_R\right]$ which implies that $(g,h)\in N_{\overline{R}}\cap G\times G$. Thus, $N_R\subseteq N_{\overline{R}}\cap G\times G$. Again, if $(g,h)$ belongs to $N_{\overline{R}}\cap G\times G$, then for all $x\in \Gamma \subseteq \widehat{\Gamma }$, and so $(g.x,h.x)\in \overline{R}\cap \Gamma \times \Gamma =R$ and this implies $(g,h)\in N_R$. Our claim follows. Then as uniform subgroups $(G,{\tau }_{{\Phi }_1})\cong (G,{\tau }_{{\Phi }_2})$, both algebraically and topologically, their corresponding completions $(\widehat{G},{\tau }_{{\Phi }_1})\cong (\widehat{G},{\tau }_{{\Phi }_2})$, both algebraically and topologically. Since for all $S\in I$, ${\psi }_S:G\to G∕N_S$ is a uniform continuous group homomorphism and $G∕N_S$ is discrete, there exists a unique uniform continuous extension of ${\psi }_S$, namely, $\widehat{{\psi }_S}:\widehat{G}\to G∕N_S$. Let us define ${\lambda }_S:\widehat{G}\to G∕N_S$ via ${\lambda }_S\left(g\right)=N_S\left[g_S\right]$, where $g={\left(N_R\left[g_R\right]\right)}_R$, [6]. Now let $g={\left(N_R\left[g_R\right]\right)}_R,h={\left(N_R\left[h_R\right]\right)}_R\in \widehat{G}$ be such that $g=h$ which implies that $N_S\left[g_S\right]=N_S\left[h_S\right]$ and hence ${\lambda }_S$ is well defined. Now let $(g,h)\in N_{\overline{S}}$. First of all $N_{\overline{S}}\left[g_S\right]=N_{\overline{S}}\left[g\right]=N_{\overline{S}}\left[h\right]=N_{\overline{S}}\left[h_S\right]$. So, $(g_S,h_S)\in N_{\overline{S}}\bigcap G\times G=N_S$. Hence $N_S\left[g_S\right]=N_S\left[h_S\right]$ which implies that ${\lambda }_S\left(g\right)={\lambda }_S\left(h\right)$, so $({\lambda }_S\left(g\right),{\lambda }_S\left(h\right))\in D(G∕N_R)$. Thus $N_{\overline{S}}$ is a subset of ${({\lambda }_S\times {\lambda }_S)}^{-1}D(G∕N_R)$. Hence ${\lambda }_S$ is uniformly continuous. Now for all $g,h\in \widehat{G},{\lambda }_S\left(gh\right)=N_S\left[g_S{h}_S\right]=N_S\left[g_S\right]N_S\left[h_S\right]={\lambda }_S\left(g\right){\lambda }_S\left(h\right)$ and for all $g\in G,{\lambda }_S\left(g\right)={\lambda }_S\left({\left(N_R\left[g\right]\right)}_R\right)=N_S\left[g\right]={\psi }_S\left(g\right)$. Thus ${\lambda }_S$ is a well defined uniformly continuous group homomorphism that extends ${\psi }_S$. Then by the uniqueness of the extension, $\widehat{{\psi }_S}={\lambda }_S$. Now $N_{\overline{S}}$ is a closed subspace of $\widehat{G}$, then $\overline{N_{\overline{S}}\cap G\times G}=\overline{N_S}$ which implies that $\overline{N_S}$ is a subset of $\overline{N_{\overline{S}}}$ which equals $N_{\overline{S}}$. Let us define $\theta$ from $\widehat{G}∕N_{\overline{S}}$ to $G∕N_S$ as $\theta$ takes $N_{\overline{S}}\left[g\right]$ into $N_S\left[g_S\right]$, where $g={\left(N_R\left[g_R\right]\right)}_R$. Now $N_{\overline{S}}\left[g\right]=N_{\overline{S}}\left[h\right]$ in $\widehat{G}∕N_{\overline{S}}$ will imply $(g_S,h_S)$ is in $N_{\overline{S}}$ and this implies for all $x$ in $\Gamma$ the ordered pair $(g_{S}x,h_{S}x)$ is in $\overline{S}\bigcap \Gamma \times \Gamma$ which is eventually equal to $S$. Thus $(g_S,h_S)\in N_S$. Then $\theta \left(N_{\overline{S}}\left[g\right]\right)=N_S\left[g_S\right]$ which is equal to $N_S\left[h_S\right]$ and that equals $\theta \left(N_{\overline{S}}\left[h\right]\right)$. Hence $\theta$ is well defined. On the other hand let $N_{\overline{S}}\left[g\right]$, $N_{\overline{S}}\left[h\right]$ be such that $\theta \left(N_{\overline{S}}\left[g\right]\right)$ equals $\theta \left(N_{\overline{S}}\left[h\right]\right)$. Thus $N_S\left[g_S\right]=N_S\left[h_S\right]$ implies that $(g_S,h_S)\in N_S\subseteq N_{\overline{S}}$. Hence $N_{\overline{S}}\left[g\right]=N_{\overline{S}}\left[g_S\right]=N_{\overline{S}}\left[h_S\right]=N_{\overline{S}}\left[h\right]$. So, $\theta$ is injective as well. Also for all $N_S\left[g\right]\in G∕N_S$ there exists $N_{\overline{S}}\left[g\right]\in \widehat{G}∕N_{\overline{S}}$ such that $\theta \left(N_{\overline{S}}\left[g\right]\right)=N_S\left[g\right]$. So $\theta$ is surjective. Finally, $\theta \left(N_{\overline{S}}\left[g\right]N_{\overline{S}}\left[h\right]\right)$ equals $\theta \left(N_{\overline{S}}\left[gh\right]\right)$ and that equals $N_S\left[g_S{h}_S\right]$ which is $N_S\left[g_S\right]N_S\left[h_S\right]$ and finally that equals $\theta \left(N_{\overline{S}}\left[g\right]\right)\theta \left(N_{\overline{S}}\left[h\right]\right)$. So $\theta$ is a well defined group isomorphism, both algebraically and topologically. Hence $\widehat{G}∕N_{\overline{S}}\cong G∕N_S\cong \widehat{G}∕\overline{N_S}$ which implies that $|\widehat{G}∕N_{\overline{S}}\left[1\right]|$ is equal to $|\widehat{G}∕\overline{N_S}\left[1\right]|$. But since $\overline{N_S}\subseteq N_{\overline{S}}$ one obtains $\overline{N_S}\left[1\right]\le N_{\overline{S}}\left[1\right]\le \widehat{G}$ and thus $|\widehat{G}∕N_{\overline{S}}\left[1\right]||N_{\overline{S}}\left[1\right]:\overline{N_S}\left[1\right]|$ equals $|\widehat{G}∕\overline{N_S}\left[1\right]|$. Hence $|N_{\overline{S}}\left[1\right]:\overline{N_S}\left[1\right]|=1$ which implies that $N_{\overline{S}}\left[1\right]=\overline{N_S}\left[1\right]$ and thus $N_{\overline{S}}=\overline{N_S}$ as each of them is congruences. Thus our claim.□

Note 2.13

Thus referring back to Example 2.2, the action$\mathbb{Z}\times \Gamma \mapsto \Gamma$ has a unique uniform equicontinuous extension from $\widehat{\mathbb{Z}}\times \widehat{\Gamma }\mapsto \widehat{\Gamma }$, where $\widehat{\Gamma }=\underleftarrow{lim}\Gamma ∕R_n,\widehat{\mathbb{Z}}=\underleftarrow{lim}\mathbb{Z}∕N_{R_n}$ are the respective profinite completions of $\Gamma$ and $\mathbb{Z}$.