Automorphism group of the commuting graph of $2\times 2$ matrix ring over $\mathbb{Z}_{p^{s}}$

1.   Introduction

Let $R$ be a ring with identity, and let $C(R)$ be the center of $R$. The commuting graph $\Gamma(R)$ of $R$ is the graph associated to $R$ whose vertices are the elements of $R\setminus C(R)$ such that distinct vertices $A$ and $B$ are adjacent if and only if $AB = BA$. For the purpose of investigating the structures of a group or a ring, there are many associated graphs that have been studied extensively. Let $M_{n}(F)$ denote the ring of $n\times n$ matrices over $F$, where $F$ is a field and $n\geq 2$ an arbitrary integer. In ^[1], if $F$ is a finite field and $\Gamma(R)\cong\Gamma(M_{n}(F))$, then $\lvert R\rvert = \lvert M_{n}(F)\rvert$. Furthermore, if $F$ is a prime field and $n = 2$, then $R\cong M_{2}(F)$. In ^[2], this result still holds if it is just assumed that $F$ is a finite field. There are also some graph-theoretic properties of the commuting graphs that have been investigated, such as connectivity and domination number. In ^[3], Akbari et al. showed that $\Gamma(M_{n}(F))$ is a connected graph if and only if every field extension of $F$ of degree $n$ contains a proper intermediate field. Also it is shown that for two fields $F$ and $E$ and integers $n, m\geq2$, if $\Gamma(M_{n}(F))\cong\Gamma(M_{m}(E))$, then $n = m$ and $\lvert F\rvert = \lvert E\rvert$.

The commuting graph of a finite group $\Delta(G)$ is the graph whose vertex set is $G$ with $x, y \in G$, $x \neq y$, joined by an edge whenever $xy = yx$, where $G$ is a finite group. The graph $\Delta(G)$ has been studied in ^[4,5,6,7]. The set of all automorphisms of a graph forms a group known as the graph's automorphism group. The automorphism group of a graph describes its symmetries. In ^[6], it is proved that the automorphism group of $\Delta(G)$ is abelian if and only if $\lvert G\rvert \leq 2$. With the wreath product, Mirzargar et al. ^[7] determined the automorphism group of $\Delta(G)$, where $G$ is an AC-group. In ^[8], it is proved that the automorphism group of $\Gamma(M_{2}(F))$ is a direct product of symmetric groups, where $F$ is a finite field. In this paper, motivated by these works, we extend the finite field to the ring of integers modulo $p^{s}$, and we completely determine the automorphism group of $\Gamma(M_{2}(\mathbb{Z}_{p^{s}}))$, where $\mathbb{Z}_{p^{s}}$ is the ring of integers modulo $p^{s}$, $p$ is a prime and $s$ is a positive integer. This paper is organized as follows. In section 2, we give some preliminaries, notation, lemmas and definition of the wreath product. In section 3, we show that the automorphism group of $\Gamma(M_{2}(\mathbb{Z}_{p^{s}}))$ is a subgroup of a direct product of some wreath products, and we completely characterize it in Theorem 3.8.

2.   Preliminaries and notation

In this paper, let $M_{2}(\mathbb{Z}_{p^{s}})$ denote the $2\times 2$ matrix ring over $\mathbb{Z}_{p^{s}}$, we write it $R$ for short. Let $E_{ij}$ denote the matrix in $R$ having $1$ in its $(i, j)$ entry and zeros elsewhere, and let $E$ denote the identity matrix. It is well known that $C(R) = \{aE\mid a\in \mathbb{Z}_{p^{s}} \}$. For $A\in R$, $C_{R}(A) = \{B\in R\mid AB = BA \}$ is called the centralizer of $A$ in $R$. For the ring $R$, let us denote by $U(R)$ and $D(R)$ the unit group and the zero divisor set of $R$ respectively. The commuting graph of $R$ is the graph with vertices $R\setminus C(R)$, and distinct vertices $A$ and $B$ are adjacent if and only if $AB = BA$. In a graph $G$, if $x$ is adjacent to $y$ (denoted by $[x, y]$), then we say that $x$ is a neighbor of $y$ or that $y$ is a neighbor of $x$. Let $N(x)$ denote the neighbors of $x$ in $G$. A graph automorphism of a graph $G$ is a bijection on vertex set (denoted by $V(G)$) which preserves adjacency. For $a\in \mathbb{Z}_{p^{s}}$, let $\langle a \rangle$ be the ideal of $\mathbb{Z}_{p^{s}}$ generated by $a$, we will denote by ${\rm Ann}(a)$ the set $\{b\in\mathbb{Z}_{p^{s}}\mid ab = 0 \}$, and by ${\rm Ass}(a)$ the set $\{ua\mid u\in U(\mathbb{Z}_{p^{s}}) \}$. Write $T = \{0, 1, \cdots, p-1\}\subseteq\mathbb{Z}_{p^{s}}$. The subset of $T$ consisting of all non-zero elements is denoted by $T^{*}$. Let us denote by $S_{n}$ the symmetric group of degree $n$. For a set $D$, we will denote by $\lvert D\rvert$ the size of $D$, and by $S_{D}$ the symmetric group on $D$.

Lemma 2.1. [9,p. 328] Every non-zero element in $\mathbb{Z}_{p^{s}}$ can be written uniquely as

where $t_{i}\in T$, $i\in\{0, 1, \cdots, s-1\}$. Furthermore, $\lvert \langle p^{i}\rangle\rvert = p^{s-i}$, $\lvert {\rm Ass}(p^{i}) \rvert = p^{s-i}-p^{s-i-1}$, and ${\rm Ann}(p^{i}) = \langle p^{s-i}\rangle$.

Definition 2.2. [10,p. 172] Let $D$ and $Q$ be groups, let $\Omega$ be a finite $Q$-set, and let $K = \prod_{\omega\in \Omega}D_{\omega}$, where $D_{\omega}\cong D$ for all $\omega\in \Omega$. Then the $wreath$ $product$ of $D$ by $Q$, denoted by $D\wr_{\Omega} Q$, is the semidirect product of $K$ by $Q$, where $Q$ acts on $K$ by $q\cdot (d_{\omega}) = (d_{q\omega})$ for $q\in Q$ and $(d_{\omega})\in \prod_{\omega\in \Omega}D_{\omega}$.

Lemma 2.3. ([10,p. 178] or [11,Theorem 2.1.6]) Let $X = B_{1} \cup\cdots\cup B_{m}$ be a partition of a set $X$ in which each $B_{i}$ has $k$ elements. If $G = \{g\in S_{X}|$ for each $i$, there is $j$ with $g(B_{i}) = B_{j}\}$, then $G\cong S_{k}\wr_{\Omega_{m}} S_{m}$, where $\Omega_{m} = \{1, 2, \cdots, m\}$.

Let $X = \bigcup_{i_{1} = 1}^{m_{1}}B_{i_{1}}$ be a partition of a set $X$ in which each $B_{i_{1}}$ has same size. Let $B_{i_{1}} = \bigcup_{i_{2} = 1}^{m_{2}}B_{i_{1}, i_{2}}$ be a partition of a set $B_{i_{1}}$ in which each $B_{i_{1}, i_{2}}$ has same size, where $i_{1} = 1, 2, \cdots, m_{1}$. Continuing in this way we obtain partitions

of $X$ in which each $B_{i_{1}, \cdots, i_{j}}$ has same size for $j = 1, \cdots, k$. With this notation, by Lemma 2.3, we have the following:

Corollary 2.4. ([12,p. 93] or [11,Theorem 2.1.15]) Let $G$ be the largest subgroup of $S_{X}$ preserving above partitions and $\lvert B_{i_{1}, \cdots, i_{k}}\rvert = m_{k+1}$. Then $G = \{g\in S_{ X}|$ for each $i_{j}$, there is $i_{j}'$ with $g(B_{i_{1}, \cdots, i_{j}}) = B_{i_{1}', \cdots, i_{j}'},$ $j = 1, \cdots, k \}$. Moreover, $G\cong (\cdots(S_{m_{k+1}}\wr_{\Omega_{m_{k}}} S_{m_{k}})\wr\cdots\wr_{\Omega_{m_{2}}} S_{m_{2}})\wr_{\Omega_{m_{1}}} S_{m_{1}}$, where $\Omega_{m_{i}} = \{1, 2, \cdots, m_{i}\}$ for $i = 1, 2, \cdots, k+1$.

With the associativity of the wreath product (see [10,Theorem 7.26]), we will simply write $(\cdots(S_{m_{k+1}}\wr_{\Omega_{m_{k}}} S_{m_{k}})\wr\cdots\wr_{\Omega_{m_{2}}} S_{m_{2}})\wr_{\Omega_{m_{1}}} S_{m_{1}}$ as $S_{m_{k+1}}\wr S_{m_{k}}\wr\cdots\wr S_{m_{2}}\wr S_{m_{1}}$. In [11,p. 68], the iterated wreath product $S_{m_{k+1}}\wr S_{m_{k}}\wr\cdots\wr S_{m_{2}}\wr S_{m_{1}}$ consists of all $f_{k+1}\wr f_{k}\wr\cdots\wr f_{2}\wr f_{1}$, where $f_{1}\in S_{m_{1}}$ and

$j = 2, 3, \cdots, k+1$, with the action on $\prod_{j = 1}^{k+1}\Omega_{m_{j}}$ defined by

where $y_{1} = f_{1}(x_{1})$ and $y_{j} = g_{j, y_{1}, y_{2}, \cdots, y_{j-1} }(x_{j})$, $j = 2, 3, \cdots, k+1$ for all $(x_{1}, x_{2}, \cdots, x_{k+1})\in\prod_{j = 1}^{k+1}\Omega_{m_{j}}$ and $f_{k+1}\wr f_{k}\wr\cdots\wr f_{2}\wr f_{1}\in S_{m_{k+1}}\wr S_{m_{k}}\wr\cdots\wr S_{m_{2}}\wr S_{m_{1}}$.

3.   Automorphisms of $\Gamma(R)$

Let $R_{0}$ denote the set $\{aE_{11}+bE_{12}+cE_{21}+dE_{22}\mid a-d\in U(\mathbb{Z}_{p^{s}})\; \text{or}\; b\in U(\mathbb{Z}_{p^{s}})\; \text{or}\; c\in U(\mathbb{Z}_{p^{s}}) \}$. Then $R\setminus R_{0} = \{aE_{11}+bE_{12}+cE_{21}+dE_{22}\mid a-d\in D(\mathbb{Z}_{p^{s}}), \; b\in D(\mathbb{Z}_{p^{s}})\; \text{and}\; c\in D(\mathbb{Z}_{p^{s}}) \}$. Since $\lvert D(\mathbb{Z}_{p^{s}})\rvert = p^{s-1}$, an easy computation shows that $\lvert R\setminus R_{0} \rvert = p^{4s-3}$. Therefore $\lvert R_{0}\rvert = p^{4s}-p^{4s-3}$. For $A, B\in R$, we write $A\sim B$ if there exist $a\in U(\mathbb{Z}_{p^{s}})$ and $b\in \mathbb{Z}_{p^{s}}$ such that $A = aB+bE$. A trivial verification shows that $\sim$ is an equivalence relation on $R$. Set $[A] = \{ B\in R\mid B\sim A\}$. It follows immediately that $[A]$ is the equivalence class of $A$ on $R$ under the equivalence relation of $\sim$.

Lemma 3.1. Every equivalence class in $R_{0}$ has size $p^{2s}-p^{2s-1}$. Moreover, there are $p^{2s}+p^{2s-1}+p^{2s-2}$ distinct equivalence classes in $R_{0}$.

Proof. Assume that $A = aE_{11}+bE_{12}+cE_{21}+dE_{22}\in R_{0}$, where $a-d\in U(\mathbb{Z}_{p^{s}})\; \text{or}\; b\in U(\mathbb{Z}_{p^{s}})\; \text{or}\; c\in U(\mathbb{Z}_{p^{s}})$. Let $A_{1} = a_{1}A+b_{1}E$ and $A_{2} = a_{2}A+b_{2}E\in [A]$, where $a_{1}, a_{2}\in U(\mathbb{Z}_{p^{s}})$ and $b_{1}, b_{2}\in \mathbb{Z}_{p^{s}}$. We claim that if $a_{1}\neq a_{2}$ or $b_{1}\neq b_{2}$, then $A_{1}\neq A_{2}$. If $a_{1} = a_{2}$ and $b_{1}\neq b_{2}$, then $A_{1}-A_{2} = (b_{1}-b_{2})E$. It is clear that $A_{1}\neq A_{2}$. If $a_{1}\neq a_{2}$ and $b_{1} = b_{2}$, then $A_{1}-A_{2} = ((a_{1}-a_{2})(a-d)+(a_{1}-a_{2})d)E_{11}+(a_{1}-a_{2})bE_{12}+(a_{1}-a_{2})cE_{21}+(a_{1}-a_{2})dE_{22}$. If $(a_{1}-a_{2})d = 0$, then $(a_{1}-a_{2})(a-d)\neq 0$ or $(a_{1}-a_{2})b\neq 0$ or $(a_{1}-a_{2})c\neq 0$ (i.e., $A_{1}\neq A_{2}$), since $a_{1}-a_{2}\neq 0$, $a-d\in U(\mathbb{Z}_{p^{s}})\; \text{or}\; b\in U(\mathbb{Z}_{p^{s}})\; \text{or}\; c\in U(\mathbb{Z}_{p^{s}})$. If $(a_{1}-a_{2})d\neq 0$, then it is obvious that $A_{1}\neq A_{2}$. If $a_{1}\neq a_{2}$ and $b_{1}\neq b_{2}$, then $A_{1}-A_{2}$ = $((a_{1}-a_{2})(a-d)+(a_{1}-a_{2})d$+$b_{1}-b_{2}$)$E_{11}$+$(a_{1}-a_{2})bE_{12}+(a_{1}-a_{2})cE_{21}$+$((a_{1}-a_{2}$)$d+b_{1}-b_{2})E_{22}$. Similarly, we have $A_{1}\neq A_{2}$. It is well known that $\lvert U(\mathbb{Z}_{p^{s}})\rvert = p^{s}-p^{s-1}$. So $\lvert [A]\rvert = p^{2s}-p^{2s-1}$.

It is easily seen that if $A\in R_{0}$, then $[A]\subseteq R_{0}$. This fact makes it obvious that $R_{0}$ is the disjoint union of some equivalence classes. Since $\lvert R_{0}\rvert = p^{4s}-p^{4s-3}$, there are exactly $p^{2s}+p^{2s-1}+p^{2s-2}$ equivalence classes in $R_{0}$.

In fact, a trivial verification shows that the set of equivalence class representatives in $R_{0}$ is

We denote this set by $P_{0}$. By Lemma 3.1, we can write

It is immediate that $R_{0} = \bigcup_{i_{0} = 1}^{\vert P_{0}\rvert}[A_{0, i_{0}}].$

Let $j\in \{1, 2, \cdots, s-1\}$. Set $P_{j} = p^{j}P_{0}$. Since $\mathbb{Z}_{p^{s}}$ is a principal ideal ring,

From Lemma 3.1, $\lvert P_{j}\rvert = p^{2s-2j}+p^{2s-2j-1}+p^{2s-2j-2}$. Write $P_{j} = \{A_{j, 1}, A_{j, 2}, \cdots, A_{j, \lvert P_{j}\rvert}\}.$ Set

Accordingly, there are seven forms in $\bigcup_{j = 0}^{s-1}P_{j}$. For example, let $j, k\in \{0, 1, \cdots, s-1\}$, if $A_{j, i_{j}} = p^{j}E_{11}+a_{1}E_{12}+b_{1}E_{21}$, $A_{k, i_{k}} = a_{2}E_{11}+p^{k}E_{12}+b_{2}E_{21}$, where $a_{1}, b_{1}\in \langle p^{j+1}\rangle$, $a_{2}, b_{2}\in \langle p^{k+1}\rangle$, then we say that $A_{j, i_{j}}$ and $A_{k, i_{k}}$ have different forms.

Lemma 3.2. Let $R_{j} = \bigcup_{i_{j} = 1}^{\lvert P_{j}\rvert}[A_{j, i_{j}}],$ where $j = 0, 1, \dots, s-1$. Then

is a partition of $R$.

Proof. By the definition of $C(R)$, we have $C(R)\cap R_{j} = \varnothing$ for all $j\in \{0, 1, \cdots, s-1 \}$. By construction, $C(R)\nsubseteq R_{0}$ and hence $C(R)\nsubseteq R_{j}$ for $j\in \{1, 2, \cdots, s-1\}$. Let $A_{j, i_{j}}\in P_{j}$. Then $A_{j, i_{j}} = p^{j}A_{0, i_{0}}$ for a certain $A_{0, i_{0}}\in P_{0}$. Consequently, $[A_{j, i_{j}}] = [p^{j}A_{0, i_{0}}]$=$\{ap^{j}A_{0, i_{0}}$+$bE\mid a\in U(\mathbb{Z}_{p^{s}})\; \text{and}\; b\in \mathbb{Z}_{p^{s}}\}$=$\{aA_{0, i_{0}}$+$bE\mid a\in {\rm Ass}(p^{j})\; \text{and}\; b\in \mathbb{Z}_{p^{s}}\}$. By Lemma 2.1, in much the same way as Lemma 3.1, the size of an equivalence class in $R_{j}$ is $p^{2s-j}-p^{2s-j-1}$. It follows that $\lvert R_{j} \rvert = p^{4s-3j}-p^{4s-3j-3}$. Then

It remains to prove that $R_{j_{1}}\cap R_{j_{2}} = \varnothing$ for all $j_{1}\neq j_{2}\in \{0, 1, \cdots, s-1\}$. Assume that $A\in R_{j_{1}}\cap R_{j_{2}}\neq\varnothing$. Then there exist $a_{1}, a_{2}\in U(\mathbb{Z}_{p^{s}})$, $b_{1}, b_{2}\in \mathbb{Z}_{p^{s}}$, $A_{j_{1}, i_{j_{1}}}\in P_{j_{1}}$ and $A_{j_{2}, i_{j_{2}}}\in P_{j_{2}}$ such that $A = a_{1}A_{j_{1}, i_{j_{1}}}+b_{1}E = a_{2}A_{j_{2}, i_{j_{2}}}+b_{2}E$. It implies that $A_{j_{1}, i_{j_{1}}} = a_{1}^{-1}a_{2}A_{j_{2}, i_{j_{2}}}+a_{1}^{-1}(b_{2}-b_{1})E$. Since the $(2, 2)$ entries of $A_{j_{1}, i_{j_{1}}}$ and $A_{j_{2}, i_{j_{2}}}$ are equal to $0$, $a_{1}^{-1}(b_{2}-b_{1}) = 0$. Thus, $A_{j_{1}, i_{j_{1}}} = a_{1}^{-1}a_{2}A_{j_{2}, i_{j_{2}}}$. Suppose that $A_{j_{1}, i_{j_{1}}} = p^{j_{1}}E_{11}+\star p^{j_{1}+1}E_{12}+\star E_{21}$ and $A_{j_{2}, i_{j_{2}}} = p^{j_{2}}E_{11}+\star p^{j_{2}+1}E_{12}+\star E_{21}$. We thus get $j_{1} = j_{2}$. This contradicts our assumption $j_{1}\neq j_{2}$. Similarly, we obtain contradictions in the other cases of $A_{j_{1}, i_{j_{1}}}$ and $A_{j_{2}, i_{j_{2}}}$. This completes the proof.

Lemma 3.3. Let $A\in[A_{j, i_{j}}]$, $B\in[A_{k, i_{k}}]$, where $j, k\in\{0, 1, \cdots, s-1\}$, $A_{j, i_{j}}\in P_{j}$ and $A_{k, i_{k}}\in P_{k}$.

(i) Let $j+k\leq s-1$. Then $AB = BA$ if and only if $p^{k}A_{j, i_{j}} = p^{j}A_{k, i_{k}}$.

(ii) Let $j+k > s-1$. Then $AB = BA$.

Proof. It is easily seen that $AB = BA$ if and only if $A_{j, i_{j}}A_{k, i_{k}} = A_{k, i_{k}}A_{j, i_{j}}$.

(i) Suppose that $A_{j, i_{j}} = p^{j}E_{11}+a_{1}E_{12}+b_{1}E_{21}$, $A_{k, i_{k}} = a_{2}E_{11}+p^{k}E_{12}+b_{2}E_{21}$, where $a_{1}, b_{1}\in \langle p^{j+1}\rangle$, $a_{2}, b_{2}\in \langle p^{k+1}\rangle$. Then $A_{j, i_{j}}A_{k, i_{k}} = \star E_{11}+p^{j+k}E_{12}+\star E_{21}$, $A_{k, i_{k}}A_{j, i_{j}} = \star E_{11}+\star p^{j+k+2}E_{12}+\star E_{21}$. Obviously, $A_{j, i_{j}}A_{k, i_{k}}\neq A_{k, i_{k}}A_{j, i_{j}}$. By similar arguments, it is easy to check that $A_{j, i_{j}}A_{k, i_{k}}\neq A_{k, i_{k}}A_{j, i_{j}}$ when $A_{j, i_{j}}$ and $A_{k, i_{k}}$ have different forms.

Without loss of generality we assume that $j\geq k$. Now suppose that $A_{j, i_{j}}A_{k, i_{k}} = A_{k, i_{k}}A_{j, i_{j}}$, where $A_{j, i_{j}} = p^{j}E_{11}+a_{1}E_{12}+b_{1}E_{21}$, $A_{k, i_{k}} = p^{j}E_{11}+a_{2}E_{12}+b_{2}E_{21}$, $a_{1}, b_{1}\in \langle p^{j+1}\rangle$, $a_{2}, b_{2}\in \langle p^{k+1}\rangle$. By Lemma 2.1, we can assume that $a_{1} = \sum_{i = j+1}^{s-1}r_{i}p^{i}$, $b_{1} = \sum_{i = j+1}^{s-1}t_{i}p^{i}$, $a_{2} = \sum_{i = k+1}^{s-1}u_{i}p^{i}$ and $b_{2} = \sum_{i = k+1}^{s-1}v_{i}p^{i}$, where $r_{i}, t_{i}, u_{i}, v_{i}\in T$. Since $A_{j, i_{j}}A_{k, i_{k}} = A_{k, i_{k}}A_{j, i_{j}}$, it is obvious that $r_{j+1} = u_{k+1}$, $r_{j+2} = u_{k+2}$, $\cdots$, $r_{s-k-1} = u_{s-j-1}$, and $t_{j+1} = v_{k+1}$, $t_{j+2} = v_{k+2}$, $\cdots$, $t_{s-k-1} = v_{s-j-1}$. It is immediately that $p^{k}A_{j, i_{j}} = p^{j}A_{k, i_{k}}$. In other cases we conclude similarly that $p^{k}A_{j, i_{j}} = p^{j}A_{k, i_{k}}$.

Conversely, suppose that $p^{k}A_{j, i_{j}} = p^{j}A_{k, i_{k}}$. An easy computation shows that it occurs only when $A_{j, i_{j}}$ and $A_{k, i_{k}}$ have same form. Assume that $A_{j, i_{j}} = p^{j}E_{11}+a_{1}E_{12}+b_{1}E_{21}$, $A_{k, i_{k}} = p^{j}E_{11}+a_{2}E_{12}+b_{2}E_{21}$ with $a_{1} = \sum_{i = j+1}^{s-1}r_{i}p^{i}$, $b_{1} = \sum_{i = j+1}^{s-1}t_{i}p^{i}$, $a_{2} = \sum_{i = k+1}^{s-1}u_{i}p^{i}$ and $b_{2} = \sum_{i = k+1}^{s-1}v_{i}p^{i}$, where $r_{i}, t_{i}, u_{i}, v_{i}\in T$. Since $p^{k}A_{j, i_{j}} = p^{j}A_{k, i_{k}}$, it is easy to check that $r_{j+1} = u_{k+1}$, $r_{j+2} = u_{k+2}$, $\cdots$, $r_{s-k-1} = u_{s-j-1}$, and $t_{j+1} = v_{k+1}$, $t_{j+2} = v_{k+2}$, $\cdots$, $t_{s-k-1} = v_{s-j-1}$. It is clear that $A_{j, i_{j}}A_{k, i_{k}} = A_{k, i_{k}}A_{j, i_{j}}$. The proof for other cases is similar.

(ii) If $j+k > s-1$, then $A_{j, i_{j}}A_{k, i_{k}} = 0 = A_{k, i_{k}}A_{j, i_{j}}$. Therefore, $AB = BA$.

For fixed $j, k\in \{0, 1, \cdots, s-1\}$ and $i_{k}\in \{1, 2, \cdots, \lvert P_{k}\rvert \}$, set

By Lemma 3.3, we have the following proposition.

Proposition 3.4. Let $A\in [A_{k, i_{k}}]$, where $k\in\{0, 1, \cdots, s-1\}$ and $A_{k, i_{k}}\in P_{k}$.

(i) $C_{R}(A) = \bigcup\limits_{j = 0}^{s-1}[p^{j}A_{0, i_{0}}]\bigcup C(R)$.

(ii) Let $0 < k\leq s-1$. Then $C_{R}(A) = \bigcup\limits_{j = 0}^{s-k-1} R_{j}^{k, i_{k}}\bigcup\limits_{j = s-k}^{s-1}R_{j}\bigcup C(R)$.

For fixed $k, j\in \{0, 1, \cdots, s-1\}$, $k\geq j$, $i_{k}\in \{1, 2, \cdots, \lvert P_{k}\rvert\},$ $i_{k+1}\in \{1, 2, \cdots, \lvert P_{k+1}\rvert\},$ $\cdots,$ $i_{s-1}\in \{1, 2, \cdots, \lvert P_{s-1}\rvert\}$, if $p^{s-1-k}A_{k, i_{k}} = p^{s-1-(k+1)}A_{k+1, i_{k+1}} = \cdots = p^{0}A_{s-1, i_{s-1}}$, then set

Since $p^{s-1-j}P_{j} = p^{s-1-(j+1)}P_{j+1} = \cdots = P_{s-1}$,

Lemma 3.5. Let $0\leq j\leq k\leq s-1$, $A_{k, i_{k}}\in P_{k}$, $A_{k+1, i_{k+1}}\in P_{k+1}$, $\cdots$, $A_{s-1, i_{s-1}}\in P_{s}$ and $p^{s-1-k}A_{k, i_{k}} = p^{s-1-(k+1)}A_{k+1, i_{k+1}} = \cdots = p^{0}A_{s-1, i_{s-1}}$. Then the number of equivalence classes in $R_{j, i_{s-1}, \cdots, i_{k+1}, i_{k}}$ is $p^{2(k-j)}$.

Proof. From the construction of $P_{j}$ and $P_{k}$, we know that $p^{k-j}P_{j} = p^{k}P_{0} = P_{k}$. Define two maps $f:\langle p^{j+1}\rangle\rightarrow \langle p^{k+1}\rangle$ by $\sum_{i = j+1}^{s-1}t_{i}p^{i}\mapsto\sum_{i = j+1}^{s-k+j-1}t_{i}p^{i+k-j}$ and $g:{\rm Ass}(p^{j})\rightarrow {\rm Ass}(p^{k})$ by $\sum_{i = j}^{s-1}t_{i}p^{i}\mapsto\sum_{i = j}^{s-k+j-1}t_{i}p^{i+k-j}$, where $t_{j}\in T^{*}$, $t_{i}\in T$, $i = j+1, j+2, \cdots, s-1$. Clearly, $f$, $g$ are surjective, and we have ${\rm ker}(f) = \{\sum_{i = s-k+j}^{s-1}t_{i}p^{i}\mid t_{i}\in T, \; i = s-k+j, s-k+j+1, \cdots, s-1 \} = \langle p^{s-k+j}\rangle$ and ${\rm ker}(g) = \{p^{j}+\sum_{i = s-k+j}^{s-1}t_{i}p^{i}\mid t_{i}\in T, \; i = s-k+j, s-k+j+1, \cdots, s-1\}$. By Lemma 2.1 and $\lvert T\rvert = p$, $\lvert{\rm ker}(f)\rvert = \lvert{\rm ker}(g)\rvert = p^{k-j}$. Then the size of the inverse image of each element in $\langle p^{k+1}\rangle$ and ${\rm Ass}(p^{k})$ under $f$ and $g$ is $p^{k-j}$ respectively. Moreover, it is evident that the number of solutions of $p^{k-j}X = A_{k, i_{k}}$ in $P_{j}$ is $p^{2(k-j)}$. In fact, the number of equivalence classes in $R_{j, i_{s-1}, \cdots, i_{k+1}, i_{k}}$ is equal to the number of solutions of $p^{k-j}X = A_{k, i_{k}}$ in $P_{j}$, which completes the proof.

From Lemma 3.5, $\lvert N_{k-1}^{i_{k}}\rvert = p^{2}$ for all $k\in\{1, 2, \cdots, s-1\}$ and $i_{k}\in\{1, 2, \cdots \lvert P_{k}\rvert \}$. Recall that $\Omega_{p^{2}} = \{1, 2, \cdots, p^{2} \}$. It is easily seen that there exists a unique map $\varphi_{i_{k}}:N_{k-1}^{i_{k}}\rightarrow \Omega_{p^{2}}$ such that for $i, j\in N_{k-1}^{i_{k}}$, if $i < j$, then $\varphi_{i_{k}}(i) < \varphi_{i_{k}}(j)$. Let $i_{k}'\in\{1, 2, \cdots \lvert P_{k}\rvert \}$. Define a map

by $i\mapsto j$ if $\varphi_{i_{k}}(i) = \varphi_{i_{k}'}(j)$.

Corollary 3.6. Let $R = M_{2}(\mathbb{Z}_{p^{s}})$, with $p$ prime and $s$ positive integer. Let $A, B\in R$. Then $C_{R}(A) = C_{R}(B)$ if and only if $[A] = [B]$.

Proof. If $A, B\in C(R)$, it is obviously that $C_{R}(A) = R = C_{R}(B)$ if and only if $[A] = C(R) = [B]$. If $A\in C(R)$ and $B\notin C(R)$, it is clear that $C_{R}(A) = R\neq C_{R}(B)$. Similarly, if $A\notin C(R)$ and $B\in C(R)$, then $C_{R}(A)\neq C_{R}(B)$.

Now let $A, B\in R\setminus C(R)$. Suppose that $C_{R}(A) = C_{R}(B)$, where $A\in[A_{j, i_{j}}]$, $B\in[A_{k, i_{k}}]$, $j, k\in \{0, 1, \cdots, s-1 \}$. We claim that $j = k$ and $i_{j} = i_{k}$. If $j = 0$ and $k\neq 0$, by Proposition 3.4, we know that $C_{R}(A)\neq C_{R}(B)$, a contradiction. Similarly, if $j\neq 0$ and $k = 0$, then $C_{R}(A)\neq C_{R}(B)$, a contradiction. If $0 < j\neq k\leq s-1$, then $\bigcup_{l = s-j}^{s-1}R_{l}\neq \bigcup_{l = s-k}^{s-1}R_{l}$. By Proposition 3.4 (ii), $C_{R}(A) = \bigcup_{l = 0}^{s-j-1}R_{l}^{j, i_{j}}\bigcup_{l = s-j}^{s-1}R_{l}\neq \bigcup_{l = 0}^{s-k-1}R_{k}^{k, i_{k}}\bigcup_{l = s-k}^{s-1}R_{l} = C_{R}(B)$, a contradiction. If $j = k = 0$ and $i_{j}\neq i_{k}$, then $[A_{0, i_{j}}]\neq [A_{0, i_{k}}]$. By Proposition 3.4 (i), $C_{R}(A) = [A_{0, i_{j}}]\bigcup_{l = 1}^{s-1}[p^{l}A_{0, i_{j}}]\neq [A_{0, i_{k}}]\bigcup_{l = 1}^{s-1}[p^{l}A_{0, i_{k}}] = C_{R}(B)$, a contradiction. If $0 < j = k\leq s-1$ and $i_{j}\neq i_{k}$, then $A_{j, i_{j}}\neq A_{j, i_{k}}$. Thus, by the proof of Lemma 3.5, $R_{0}^{j, i_{j}} = R_{0, i_{s-1}, \cdots, i_{j}}\neq R_{0, i_{s-1}, \cdots, i_{k}} = R_{0}^{j, i_{k}}$. Furthermore, $C_{R}(A) = R_{0}^{j, i_{j}}\bigcup_{l = 1}^{s-j-1} R_{l}^{j, i_{j}}\bigcup_{l = s-j}^{s-1}R_{l}\neq R_{0}^{j, i_{k}}\bigcup_{l = 1}^{s-j-1} R_{l}^{j, i_{k}}\bigcup_{l = s-j}^{s-1}R_{l} = C_{R}(B)$ by Proposition 3.4 (ii), a contradiction. Therefore $j = k$ and $i_{j} = i_{k}$ as claimed. This means that $A_{j, i_{j}} = A_{k, i_{k}}$ (i.e. $[A] = [B]$). The converse is straightforward.

Corollary 3.7. Let $R = M_{2}(\mathbb{Z}_{p^{s}})$, with $p$ prime and $s$ positive integer. If $f\in Aut(\Gamma(R))$, then $f(R_{j}) = R_{j}$ for $j\in \{0, 1, \cdots, s-1 \}$, where $R_{j}$ is as defined in $(3.1)$.

Proof. For $j = 0, 1, \cdots, s-1$, if $A\in R_{j}$, then $\lvert C_{R}(A)\setminus C(R) \rvert = p^{2s+2j}-p^{s}$ by Proposition 3.4 and the proof of Lemma 3.5. This means that if $A\in R_{j}$, $B\in R_{k}$ and $j\neq k$, then $\lvert N(A) \rvert\neq \lvert N(B) \rvert$, where $j, k\in \{0, 1, \cdots, s-1 \}$. Since automorphisms of a graph must preserve the number of neighbors of vertices, $f(R_{j}) = R_{j}$, where $j\in \{0, 1, \cdots, s-1 \}$.

Recall that a graph automorphism of a graph $G$ is a bijection on vertex set which preserves adjacency. If $\lvert V(G)\rvert = n$, then in the obvious way ${\rm Aut}(G)$ is isomorphic to a subgroup of $S_{n}$. Specifically, ${\rm Aut}(G) = \{f\in S_{n}\mid for\; all\; x, y\in V(G), \; [x, y] \Leftrightarrow [f(x), f(y)]\}$. It is easy to show that ${\rm Aut}(G) = \{f\in S_{n}\mid for\; all \; x\in V(G), \; f(N(x)) = N(f(x))\}$. For $\Gamma(R)$, $N(A) = C_{R}(A)\setminus \{C(R)\cup A\}$. This means that ${\rm Aut}(\Gamma(R)) = \{f\in S_{\sum_{j = 0}^{s-1}\lvert R_{j}\rvert}\mid for\; all \; A\in V(\Gamma(R)), \; f(N(A)) = N(f(A)) \}$.

We now prove our main result about the automorphism group of the commuting graph of $M_{2}(\mathbb{Z}_{p^{s}})$. To state it, we need to define a group. For each $j\in \{0, 1, \cdots, s-1 \}$ denote

Let $G$ be a subset of $\prod_{j = 0}^{s-1}G_{s-1-j}$ and define:

The multiplication law of the iterated wreath product is defined in [11,p. 68], the proof that $G$ is a subgroup of $\prod_{j = 0}^{s-1}G_{s-1-j}$ is routine.

Theorem 3.8. Let $R = M_{2}(\mathbb{Z}_{p^{s}})$, with $p$ prime and $s$ positive integer. Then ${\rm Aut}(\Gamma(R))\cong G$, where $G$ is a group defined in $(3.3)$.

Proof. By Lemma 3.2 and Corollary 3.7, ${\rm Aut}(\Gamma(R))$ is isomorphic to a subgroup of $\prod_{j = 0}^{s-1}S_{ R_{j}}.$ So $f\in {\rm Aut}(\Gamma(R))$ can be written as a product $\prod_{j = 0}^{s-1}f_{j}$, where $f_{j}\in S_{ R_{j}}$. We claim that

where $j = 0, 1, \cdots, s-1$.

Let $j\in \{1, \cdots, s-1\}$ and $(\cdots, f_{j}, \cdots) = f\in{\rm Aut}(\Gamma(R))$. Assume that $A\in [A_{j, i_{j}}]$, $B\in [A_{j, i_{j}'}]$ with $f_{j}(A) = B$. By Proposition 3.4 (ii) and $f(N(A)) = N(f(A))$,

Then $f(R_{0}^{j, i_{j}}) = R_{0}^{j, i_{j}'}$ by Corollary 3.7. It is immediate that $f([A_{s-1, i_{s-1}}]) = [A_{s-1, i_{s-1}'}]$ by Proposition 3.4 (i), where $[A_{s-1, i_{s-1}}] = p^{s-1}R_{0}^{j, i_{j}}$, $[A_{s-1, i_{s-1}'}] = p^{s-1}R_{0}^{j, i_{j}'}$. Since Proposition 3.4 (ii) and $f(N([A_{s-1, i_{s-1}}])) = N(f([A_{s-1, i_{s-1}}]))$,

Thus $f(R_{0}^{s-1, i_{s-1}}) = R_{0}^{s-1, i_{s-1}'}$. It is evident that $f(p^{j}R_{0}^{s-1, i_{s-1}}) = p^{j}R_{0}^{s-1, i_{s-1}'}$ by Proposition 3.4 (i), i.e.,

Similarly, we have

where $i_{s-2}, i_{s-2}'\in \{1, \cdots, \lvert P_{s-2}\rvert \}$, $\cdots$, $i_{j+2}, i_{j+2}'\in \{1, \cdots, \lvert P_{j+2}\rvert \}$, $i_{j+1}, i_{j+1}'\in \{1, \cdots, \lvert P_{j+1}\rvert \}$ with

Obviously,

Hence, for $i_{s-1}\in\{1, 2, \cdots, \lvert P_{s-1}\rvert \}$, $i_{s-2}\in N_{s-2}^{i_{s-1}}$, $\cdots$, $i_{j+1}\in N_{j+1}^{i_{j+2}}$, $i_{j}\in N_{j}^{i_{j+1}}$, there are $i_{s-1}'\in\{1, 2, \cdots, \lvert P_{s-1}\rvert \}$, $i_{s-2}'\in N_{s-2}^{i_{s-1}'}$, $\cdots$, $i_{j+1}'\in N_{j+1}^{i_{j+2}'}$, $i_{j}'\in N_{j}^{i_{j+1}'}$ such that

By Lemma 3.5, $\lvert N_{k}^{i_{k+1}} \rvert = p^{2}$, $k = j, j+1, \cdots, s-2.$ In the proof of Lemma 3.2, we know that $\vert [A] \rvert = p^{2s-j}-p^{2s-j-1}$ for $A\in R_{j}$. Therefore $\{f_{j}\in S_{R_{j}}\mid (\cdots, f_{j}, \cdots) = f\in{\rm Aut}(\Gamma(R))\}\cong G_{s-1-j}$ by Corollaries 2.4 and 3.6. The proof for $j = 0$ is similar.

From the above proof, it follows that ${\rm Aut}(\Gamma(R))$ is a subgroup of $\prod_{j = 0}^{s-1}G_{s-1-j}$. Let $j\in \{0, 1, \cdots, s-2 \}$. Let $\phi_{j}$ be an isomorphism between $\{f_{j}\in S_{ R_{j}}\mid (\cdots, f_{j}, \cdots) = f\in{\rm Aut}(\Gamma(R))\}$ and $G_{s-1-j}$. Suppose that $(\cdots, f_{j}, f_{j+1}, \cdots) = f\in {\rm Aut}(\Gamma(R))$, where $\phi_{j}(f_{j}) = h_{j}\wr g_{j}\wr \cdots\wr g_{s-2}\wr g_{s-1}\in G_{s-1-j}$. As defined in (2.1), $g_{s-1}\in S_{p^{2}+p+1}$,

$k = s-2, s-3, \cdots, j$, and

As the action defined in (2.2), we define $f_{j}(R_{j, i_{s-1}}) = R_{j, y_{s-1}}$,

where $y_{s-1} = g_{s-1}(i_{s-1})$, $y_{k} = g_{k, y_{s-1}, \cdots, y_{k+2}, y_{k+1 }}(\varphi_{k}^{y_{k+1}}(i_{k}))$, $\varphi_{k}^{y_{k+1}}$ is defined in (3.2), $k = s-2, s-3, \cdots, j$ and $f_{j}(aA_{j, i_{j}}+bE) = h_{j, y_{s-1}, \cdots, y_{j+1}, y_{j}}(aA_{j, y_{j}}+bE)$ for all $a\in {\rm Ass}(p^{j})$, $b\in\mathbb{Z}_{p^{s}}$. Suppose that $\phi_{j+1}(f_{j+1}) = h_{j+1}\wr g_{j+1}'\wr \cdots\wr g_{s-2}'\wr g_{s-1}'\in G_{s-1-(j+1)}$. We next claim that $g_{j+1} = g_{j+1}'$, $g_{j+2} = g_{j+2}'$, $\cdots$, $g_{s-1} = g_{s-1}'$. If there exists $k\in \{j+1, j+2, \cdots, s-1 \}$ such that $g_{j+1} = g_{j+1}'$, $\cdots$, $g_{k-1} = g_{k-1}'$, $g_{k}\neq g_{k}'$, $g_{k+1} = g_{k+1}'$, $\cdots$, $g_{s-1} = g_{s-1}'$, then there exist $i_{s-1}\in\{1, 2, \cdots, p^{2}+p+1 \}, \cdots, i_{k+1}\in N_{k+1}^{i_{k+2}}, i_{k}\in N_{k}^{i_{k+1}}$ such that $y_{k}\neq y_{k}'$, where $y_{k}, y_{k}'$ are defined above. Assume that $f_{j}(R_{j, i_{s-1}, \cdots, i_{k+1}, i_{k}}) = R_{j, y_{s-1}, \cdots, y_{k+1}, y_{k}}$ and $f_{j+1}(R_{j+1, i_{s-1}, \cdots, i_{k+1}, i_{k}}) = R_{j+1, y_{s-1}, \cdots, y_{k+1}, y'_{k}}$. By Proposition 3.4 (i) and $f(N(A)) = N(f(A))$ for all $A\in R\setminus C(R)$, $f_{0}(R_{0, i_{s-1}, \cdots, i_{k}}) = R_{0, y_{s-1}, \cdots, y_{k+1}, y_{k}}$ and $f_{0}(R_{0, i_{s-1}, \cdots, i_{k}}) = R_{0, y_{s-1}, \cdots, y_{k+1}, y_{k}'}$. Since $y_{k}\neq y_{k}'$, $R_{0, y_{s-1}, \cdots, y_{k+1}, y_{k}}\neq R_{0, y_{s-1}, \cdots, y_{k+1}, y_{k}'}$, i.e., $f_{0}(R_{0, i_{s-1}, \cdots, i_{k}})\neq f_{0}(R_{0, i_{s-1}, \cdots, i_{k}})$, which is impossible. By this claim, we know that $f\in {\rm Aut}(\Gamma(R))$ can be written as $(h_{0}\wr g_{0}\wr \cdots\wr g_{s-2}\wr g_{s-1}, h_{1}\wr g_{1}\wr \cdots\wr g_{s-2}\wr g_{s-1}, \cdots, h_{s-1}\wr g_{s-1})$, where $h_{j}\wr g_{j}\wr \cdots\wr g_{s-2}\wr g_{s-1}\in G_{s-1-j}, \; j = 0, 1, \dots, s-1.$ Therefore ${\rm Aut}(\Gamma(R))\cong G$.

4.   Conclusions

In this paper, we show that the automorphism group of $\Gamma(M_{2}(\mathbb{Z}_{p^{s}}))$ is a subgroup of a direct product of some wreath products, and we completely characterize it in Theorem 3.8.

Acknowledgments

The author wishes to express his thanks to the referees for their time and comments.

Conflict of interest

The author declares no conflicts of interest in this paper.