The diameter of proper power graphs of alternating groups

TL;DR

This paper investigates the diameter of the proper power graph of alternating groups, establishing bounds between 6 and 11 for large n, and describes short paths within these graphs.

Contribution

It improves the bounds on the diameter of proper power graphs of alternating groups for large n and characterizes short paths in these graphs.

Findings

Diameter of An is between 6 and 11 for n ≥ 51.

Proper power graphs are connected for large n.

Short paths in these graphs are characterized.

Abstract

The power graph of finite group G is a simple graph whose vertex set is G and two distinct elements a and b are adjacent if and only if one of them is a power of the other. The proper power graph of G is a graph which is obtained by deleting the identity vertex (the identity element of G). In this paper, we improve the diameter bound of proper power graph of alternating group of degree n which the graph is connected. We show that the diameter of An is between 6 and 11, if the n at least 51. We also describe a number of short paths in these power graphs.

Equations

$$\sigma =(z_{11},z_{21},z_{31},z_{41},z_{51},z_{12},z_{22},z_{32},z_{42},z_{52},z_{13},z_{23},z_{33},z_{43},z_{53}).$$

Full text

Keywords: Alternating groups, Diameter, Proper power graph.

MSC(2010): 20D06, 05C25.

1 Introduction

In a graph $\Gamma$, a path is an alternating sequence of vertices and edges which begins and ends at a pair of vertices such that consecutive terms are incident. When every pair of vertices are joined by a path, the graph $\Gamma$ is connected. The number of edges in the path is the length of a path. The length of the shortest path connecting between two vertices $x$ and $y$ in a connected graph is assigned to as its the distance, which is denoted by $d(x,y)$. The maximum distance between all pairs of vertices of $\Gamma$ is assigned to as its the diameter of a connected graph $\Gamma$, which is denoted by $diam(\Gamma)$.

In a group $G$, the order of $x$ in $G$, denoted $o(x)$, is the smallest positive integer such that $x^{o(x)}=1$, where $1$ is the identity element of $G$. The power graph of a group $G$ is the simple graph ${\mathcal{P}}(G)$, with vertex-set $G$ and vertices $x$ and $y$ are adjacent, denoted $x\sim y$, if and only if $x\neq y$ and either $y=x^{m}$ or $x=y^{m}$ for some positive integer $m$. The identity element $1$ of a finite group $G$ is adjacent to every other vertex $x$ since $x^{o(x)}=1$. The proper power graph of $G$, denoted ${\mathcal{P}}^{*}(G)$, is the graph obtained from ${\mathcal{P}}(G)$ by deleting the vertex $1$. While the power graph of any group is connected, the proper power graph may not be.

Kelarev and Quinn [6] introduced the power digraphs for semigroups. Chakrabarty et al. [1] defined the (undirected) power graphs of semigroups. Chattopadhyay and Panigrahi [2] considered connectivity and planarity of power graphs of finite cyclic, dihedral and dicyclic groups. Moghaddamfar et al. [8] studied some properties of the power graph ${\mathcal{P}}(G)$ and the subgraph ${\mathcal{P}}^{*}(G)$. In [5], Jafari investigated connectivity, diameter and clique number of proper power graphs. In [4], Doostabadi and Farrokhi addressed the connectedness and diameter of the proper power graph of a alternating group.

Theorem 1.1**.**

([4], Theorem 4.7.) Let $G=A_{n}$ be an alternating group $(n\geq 3)$. If $n$, $n-1$, $n-2$, $n/2$, $(n-1)/2$, $(n-2)/2$ are not primes, then ${\mathcal{P}}^{*}(G)$ is connected and $diam({\mathcal{P}}^{*}(G))\leq 22$.

In this paper, we improve the diameter bound of Theorem  1.1. We show that $6\leq diam({\mathcal{P}}^{*}(A_{n}))\leq 11$ for $n>51$. We also describe a number of short paths in these power graphs.

Throughout this paper, we assume that $n>51$ is such that $n$, $n-1$, $n-2$, $n/2$, $(n-1)/2$, $(n-2)/2$ are not primes.

2 Some of short paths

We know every permutation $\alpha\in A_{n}$ is either a cycle or factorization (which is a product of disjoint cycles). We also know that taking an appropriate power shows every permutation in $A_{n}$ is adjacent to a permutation of prime order. We determine some of short paths between elements of prime order the proper power graph of the alternating group.

Lemma 2.1**.**

([4], Lemma $2.1$) Let $G$ be a finite group and $x,y\in G\backslash\{1\}$ such that $xy=yx$ and $gcd(o(x),o(y))=1$. Then $x\thicksim xy\thicksim y$. In particular, $d(x,y)=2$.

Due to the Lemma  2.1, the distance a permutation of prime order from any permutation of its fixed points having a different prime order is $2$. For each $\alpha\in A_{n}$, the support of $\alpha$ is defined by $S(\alpha)=\{i:\ \alpha(i)\neq i\}$ and show the complement of set $S(\alpha)$ by $S^{c}(\alpha)$.

Lemma 2.2**.**

*([4], Lemma 4.3.)

For $n\geq 10$, there is a path of length at most four in ${\mathcal{P}}^{*}(A_{n})$ between any two $3$-cycles.*

Lemma 2.3**.**

Let $\alpha,\beta\in A_{n}$ have respective prime orders $p$ and $q$ such that $\alpha=\alpha_{1}\cdots\alpha_{m}$ and $\beta=\beta_{1}\cdots\beta_{m^{\prime}}$. If $pq\leq 9$, then $d(\alpha,\beta)\leq 6$.

Proof.

Assume that $k=n-|S(\alpha)|$, $k^{\prime}=n-|S(\beta)|$. We consider three cases as follows:

Case 1: Let $p=q=3$.

• $(1.1)$

If $k,k^{\prime}\geq 4$, then there are $j_{1},j_{2},j_{3},j_{4}\notin S(\alpha)$ and $x=(j_{1},j_{2})(j_{3},j_{4})$. Since $\alpha$ and $x$ commute and $gcd(o(\alpha),o(x))=1$, Lemma  2.1 implies that $d(\alpha,x)=2$. If $S(x)\cap S(\beta)=\emptyset$, by Lemma  2.1, $d(x,\beta)=2$. Therefore $d(\alpha,\beta)\leq 4$. If $S(x)\cap S(\beta)\neq\emptyset$ and $n-|S(\beta)\cup S(x)|\geq 5$, then for distinct $i_{1},i_{2},i_{3},i_{4},i_{5}\notin S(\beta)\cup S(x)$, there is a $5$-cycle $y=(i_{1},i_{2},i_{3},i_{4},i_{5})$. Since $y$ and $x$ commute and $gcd(o(y),o(x))=1$, Lemma  2.1 implies that $d(y,x)=2$. Likewise, $d(y,\beta)=2$. Therefore $d(\alpha,\beta)\leq 6$. Otherwise, $m^{\prime}>8$ and there are $\beta_{j_{6}},\beta_{j_{7}},\beta_{j_{8}}$ and $\beta_{j_{9}}$ such that $S(x)\subseteq S(\beta_{j_{6}}\cdots\beta_{j_{9}})\cup S^{c}(\beta)$. Say $\beta_{j_{1}}=(z_{11},z_{12},z_{13}),\beta_{j_{2}}=(z_{21},z_{22},z_{23}),\cdots$, $\beta_{j_{5}}=(z_{51},z_{52},z_{53})$, and set

[TABLE]

Then $\beta\sim\sigma\beta_{j_{6}}^{l}\cdots\beta_{j_{m^{\prime}}}^{l}\sim\sigma^{3}$, in which $({\beta_{j}^{l}})^{5}=\beta_{j}$. By Lemma  2.1, $d(\sigma^{3},x)=2$. Therefore $d(\alpha,\beta)\leq 6$.

• $(1.2)$

If $k\geq 4$ and $k^{\prime}<4$, then $m^{\prime}>16$. Since $k\geq 4$, then for distinct $j_{1},j_{2},j_{3},j_{4}\notin S(\alpha)$, there is $x=(j_{1},j_{2})(j_{3},j_{4})$ with $d(\alpha,x)=2$. Since $m^{\prime}>16$, as above there is a $15$-cycle $\sigma$ such that $S(x)\cap S(\sigma)=\emptyset$ and $d(\beta,\sigma^{3})=2$. By Lemma  2.1, $d(\sigma^{3},x)=2$. Therefore $d(\alpha,\beta)\leq 6$.

• $(1.3)$

If $k^{\prime}\geq 4$ and $k<4$, then the proof is similar to $(1.2)$.

• $(1.4)$

If $k<4$ and $k^{\prime}<4$, then $m^{\prime}>16$ and $m>16$. Let $\alpha_{1}=(r_{11},r_{12},r_{13})$, $\alpha_{2}=(r_{21},r_{22},r_{23})$, $\alpha_{3}=(r_{31},r_{32},r_{33})$, $\alpha_{4}=(r_{41},r_{42},r_{43})$, and set $\tau=(r_{11},r_{21},r_{12}$, $r_{22},r_{13},r_{23})(r_{31},r_{41}$, $r_{32},r_{42},r_{33},r_{43})$. Then $\alpha\sim\tau\alpha_{5}^{l}\cdots\alpha_{{m}}^{l}\sim\tau^{3}$, in which $({\alpha_{i}^{l}})^{3}=\alpha_{i}$. There are $\beta_{j_{1}},\cdots,\beta_{j_{t-1}}$ and $\beta_{j_{t}}$ such that $S(\tau)\subseteq S(\beta_{j_{1}}\cdots\beta_{j_{t}})\cup S^{c}(\beta)$. Since $t\leq 12$ and $m^{\prime}>16$, there is $\sigma$ in $A_{n}$ such that $d(\beta,\sigma)=2$, $o(\sigma)=5$ and $S(\sigma)\subseteq S(\beta)-S(\beta_{j_{1}}\cdots\beta_{j_{t}})$. By Lemma  2.1, $d(\sigma,\tau^{3})=2$. Therefore $d(\alpha,\beta)\leq 6$.

Case 2: Let $p=q=2$, the proof is similar to case $1$.

Case 3: Let $p=2$ and $q=3$. Assume that $k\geq 3$ and $k^{\prime}\geq 4$. If $S(\alpha)\cap S(\beta)=\emptyset$, then $d(\alpha,\beta)=2$. Suppose $S(\alpha)\cap S(\beta)\neq\emptyset$. For $n-|S(\alpha)\cup S(\beta)|\geq 5$, there is a $5$-cycle $y$ such that $d(\alpha,y)=d(\beta,y)=2$. Thus $d(\alpha,\beta)\leq 4$.

For $n-|S(\alpha)\cup S(\beta)|<5$, since $k\geq 3$, for distinct $c_{1},c_{2},c_{3}\notin S(\alpha)$, the $3$-cycle $c=(c_{1},c_{2},c_{3})$ is at distance $2$ from $\alpha$. If $m^{\prime}\geq 7$, then there are $x$, $\beta_{j_{1}}$, $\beta_{j_{2}}$, $\beta_{j_{3}}$ and $\beta_{j_{4}}$ such that $S(c)\cap S(\beta_{j_{1}}\beta_{j_{2}}\beta_{j_{3}}\beta_{j_{4}})=\emptyset$, $d(x,\beta)=2$ and $S(x)=S(\beta_{j_{1}}\cdots\beta_{j_{4}})$, where $o(x)=2$. By Lemma  2.1, $d(x,c)=2$. Therefore $d(\alpha,\beta)\leq 6$.

If $m^{\prime}<7$, then $n-|S(\beta)\cup S(c)|\geq 4$. Hence for distinct $j_{1},j_{2},j_{3},j_{4}\notin S(\beta)\cup S(c)$, there is $x=(j_{1},j_{2})(j_{3},j_{4})$. By Lemma  2.1, $d(c,x)=d(x,\beta)=2$. Therefore $d(\alpha,\beta)\leq 6$. The remaining cases are similar.

∎

Lemma 2.4**.**

Let $\alpha,\beta\in A_{n}$ have respective prime orders $p$ and $q$ such that $\alpha=\alpha_{1}\cdots\alpha_{m}$ and $\beta=\beta_{1}\cdots\beta_{m^{\prime}}$, where $p\leq q$. If $m^{\prime}=3$, $n-|S(\beta)|<3$ and $n-|S(\alpha)|<3$, then $d(\alpha,\beta)\leq 10$. Otherwise, $d(\alpha,\beta)\leq 8$.

Proof.

Assume that $k=n-|S(\alpha)|$ and $k^{\prime}=n-|S(\beta)|$. When $k<3$, since ${\mathcal{P}}^{*}(A_{n})$ is connected and $o(\alpha)$ is a prime, $m\geq 3$. Likewise, when $k^{\prime}<3$, $m^{\prime}\geq 3$.

For $k^{\prime}<3$ and $m^{\prime}=3$, let $\beta_{1}=(z_{11},\cdots,z_{1q})$, $\beta_{2}=(z_{21},\cdots,z_{2q})$, $\beta_{3}=(z_{31},\cdots,z_{3q})$, and set $y=(z_{11},z_{21},z_{31},z_{12},z_{22},z_{32},\cdots$, $z_{1q},z_{2q},z_{3q})$. Then $\beta\thicksim y\thicksim y^{q}$ and let $y^{q}=y_{1}\cdots y_{q}$ be a product of $3$-cycles.

If $k<3$, we have $m\geq 3$. As above there is a $3p$-cycle $x$ such that $\alpha\thicksim x\thicksim x^{p}$ and let $x^{p}=x_{1}\cdots x_{p}$ be a product of $3$-cycles. By Lemma  2.3, $d(x^{p},y^{q})\leq 6$, thus $d(\alpha,\beta)\leq 10$.

Let $k\geq 3$ and $p\neq 3$. There is a $3$-cycle $c$ with $d(\alpha,c)=2$. Since $q\geq 13$, there are $\tau$, $y_{i_{1}}$, $y_{i_{2}}$, $y_{i_{3}}$ and $y_{i_{4}}$ such that $S(c)\cap S(y_{i_{1}}y_{i_{2}}y_{i_{3}}y_{i_{4}})=\emptyset$, $d(y^{q},\tau)=2$ and $S(\tau)=S(y_{i_{1}}y_{i_{2}}y_{i_{3}}y_{i_{4}})$, where $o(\tau)=2$. By Lemma  2.1, $d(c,\tau)=2$. Consequently $d(y^{q},c)\leq 4$ and $d(\alpha,\beta)\leq 8$.

Let $k\geq 3$ and $p=3$. Say $\alpha_{1}=(r_{11},r_{12},r_{13})$, $\alpha_{2}=(r_{21},r_{22},r_{23})$, and set $\sigma=(r_{11},r_{21},r_{12},r_{22},r_{13},r_{23})$. Then for distinct $u,v\notin S(\alpha)$, $\alpha\sim(u,v)\sigma{\alpha_{3}}^{l}\cdots{\alpha_{m}}^{l}\sim(u,v)\sigma^{3}$, in which $({{\alpha_{j}}^{l}})^{2}={\alpha_{j}}$. Since $q\geq 13$, there are $\tau$, $y_{i_{1}}$, $\cdots$, $y_{i_{4}}$ and $y_{i_{5}}$ such that $S((u,v)\sigma)\cap S(y_{i_{1}}\cdots y_{i_{5}})=\emptyset$, $d(y^{q},\tau)=2$ and $S(\tau)=S(y_{i_{1}}\cdots y_{i_{5}})$, where $o(\tau)=5$. By Lemma  2.1, $d((u,v)\sigma^{3},\tau)=2$. Consequently $d(y^{q},(u,v)\sigma^{3})\leq 4$ and $d(\alpha,\beta)\leq 8$.

For the rest of the proof, we consider four cases as follows:

Case $(1)$: Let $p=q$ and $p\geq 5$.

We know if $k<3$, then for $p<11$, $m\geq 7$, for $p=11$, $m\geq 5$ and for $p\geq 13$, $m\geq 4$.

• $(1.1)$

Let $k\geq 3$ and $k^{\prime}\geq 3$. Then there are cycles $c$ and $c^{\prime}$ of length $3$ in $A_{n}$ such that $d(\alpha,c)=d(\beta,c^{\prime})=2$. By Lemma  2.2, $d(c,c^{\prime})\leq 4$, thus $d(\alpha,\beta)\leq 8$.

• $(1.2)$

Let $k<3$, $k^{\prime}<3$. Then $m^{\prime}>3$. Assume that $\alpha^{\prime}=\alpha_{4}\cdots\alpha_{m}$, $\gamma_{1}=\beta_{4}\beta_{5}\cdots\beta_{m^{\prime}}$, $\gamma_{2}=\beta_{1}\beta_{5}\cdots\beta_{m^{\prime}}$, $\gamma_{3}=\beta_{2}\beta_{5}\cdots\beta_{m^{\prime}}$ and $\gamma_{4}=\beta_{3}\beta_{5}\cdots\beta_{m^{\prime}}$. Since $|(S(\alpha^{\prime})\cap S(\gamma_{1}))\cup(S(\alpha^{\prime})\cap S(\gamma_{2}))\cup(S(\alpha^{\prime})\cap S(\gamma_{3}))\cup(S(\alpha^{\prime})\cap S(\gamma_{4}))|\geq(m-3)p\geq 13$, there is $\gamma_{j}$ such that $|(S(\alpha^{\prime})\cap S(\gamma_{j}))|\geq 4$. Hence there are $\tau$ and $\sigma$ in $A_{n}$ such that $d(\alpha,\tau^{p})=d(\beta,\sigma^{q})=2$, $S(\tau)=S(\alpha_{1}\alpha_{2}\alpha_{3})$, $S(\sigma)=S(\beta)-S(\gamma_{j})$, $o(\tau)=3p$ and $o(\sigma)=3q$. Since $n-|S(\tau)\cup S(\sigma)|\geq 4$, for distinct $j_{1},j_{2},j_{3},j_{4}\notin S(\tau)\cup S(\sigma)$, there is $x=(j_{1},j_{2})(j_{3},j_{4})$. By Lemma  2.1, $d(x,\tau^{p})=d(x,\sigma^{q})=2$. Consequently $d(\alpha,\beta)\leq 8$.

• $(1.3)$

Let $k<3$ and $k^{\prime}\geq 3$. Since $k^{\prime}\geq 3$, there is a $3$-cycle $c$ with $d(\beta,c)=2$. Since $k<3$, $m\geq 3$. As proved above there is a $3p$-cycle $\tau$ such that $d(\alpha,\tau^{p})=2$ and $S(c)\cap S(\tau)\geq 1$. Hence $n-|S(\tau)\cup S(c)|\geq 4$ and there is $x$ in $A_{n}$ such that $d(x,\tau^{p})=d(x,c)=2$ and $o(x)=2$. Consequently $d(\tau^{p},c)\leq 4$ and $d(\alpha,\beta)\leq 8$.

• $(1.4)$

Let $k\geq 3$ and $k^{\prime}<3$. The proof is similar to $(1.3)$.

The proof of the following cases is similar to the previous case.

Case $(2)$: Let $p\neq q$, $p\geq 5$ and $q\geq 5$.

Case $(3)$: Let $q\geq 5$ and $p=3$.

Case $(4)$: Let $q\geq 5$ and $p=2$. ∎

3 Diameter bounds

In this section, we consider diameter bounds. Next, we derive a criterion for the upper bound on $diam({\mathcal{P}}^{*}(A_{n}))$.

Corollary 3.1**.**

If ${\mathcal{P}}^{*}(A_{n})$ is connected, then $diam({\mathcal{P}}^{*}(A_{n}))\leq 11$.

Proof.

Let $x,y\in A_{n}$. If $x$ and $y$ are elements of prime orders, then $d(x,y)\leq 10$, by Lemmas  2.3,  2.4. Suppose $o(y)$ is not prime and $q$ is the least prime factor of $o(y)$, thus $y\thicksim y^{o(y)/q}$ and let $y^{o(y)/q}=y_{1}\cdots y_{m^{\prime}}$ be the factorization of $y^{o(y)/q}$ into cycles. Hence $m^{\prime}=3$, $n-|S(y^{o(y)/q})|\geq 3$ or $m^{\prime}\neq 3$. If $o(x)$ is not prime, then $x\thicksim x^{o(x)/p}$ and let $x^{o(x)/p}=x_{1}\cdots x_{m}$ be the factorization of $x^{o(x)/p}$ into cycles, where $p$ is the least prime factor of $o(x)$. Hence $m=3$, $n-|S(x^{o(x)/p})|\geq 3$ or $m\neq 3$. By Lemmas  2.3 and  2.4, $d(x^{o(x)/p},y^{o(y)/q})\leq 8$. Therefore $d(x,y)\leq 10$. If $o(x)$ be prime, then $d(x,y^{o(y)/q})\leq 10$, by Lemmas  2.4 and  2.3. Therefore $d(x,y)\leq 11$. Consequently $diam({\mathcal{P}}^{*}(A_{n}))\leq 11$. ∎

We need several results, to prove a lower bound of the diameter.

Lemma 3.2**.**

([9], Lemma 3.1.) Let $G$ be a finite group and ${\mathcal{P}}^{*}(G)$ is connected. If $x$ and $y$ are elements of prime orders and $\langle x\rangle\neq\langle y\rangle$, then $d(x,y)=2t$ and there are elements $x_{0}=x,x_{1},\cdots,x_{2t}=y$ such that $o(x_{2i})$ is prime for $i\in\{0,1,\cdots,t\}$, $o(x_{2i+1})=o(x_{2i})o(x_{2i+2})$ for $i\in\{0,1,\cdots,t-1\}$, and $x_{i}$ is adjacent to $x_{i+1}$ for $i\in\{0,1,\cdots,2t-1\}$, where $t$ is a positive integer.

Theorem 3.3**.**

(Chebyshev’s theorem ([3], p. 124)) If $n>1$, then there is always at least one prime $p$, between $n$ and $2n$.

Lemma 3.4**.**

*([7], Theorem 2.)

Let $\alpha=\alpha_{1}\cdots\alpha_{m}$ such that $\alpha_{1},\cdots,\alpha_{m}$ are cycles of length $t$. Then $\langle\alpha_{i}\rangle\cong\mathbb{Z}_{t}$, $(1\leq i\leq m)$ and $C_{S_{mt}}(\alpha)\cong(\mathbb{Z}_{t})^{m}\rtimes S_{m}$, where $C_{S_{mt}}(\alpha)$ is the centralizer of $\alpha$ in $S_{mt}$.*

Theorem 3.5**.**

$diam({\mathcal{P}}^{*}(A_{n}))\geq 6$.

Proof.

By Chebyshev’s theorem, there is a prime $p$, such that $[n/2]<p<n$. Suppose $x=(1,\cdots,p)$ and $y=(n,n-1,\cdots,n-p+1)$. By Lemma  3.2, $d(x,y)=2t$, where $t$ is a positive integer. Since $o(x)=o(y)=p$, $d(x,y)>2$. If $t=2$, then there are $z,z_{1}$ and $z_{2}$ in $S_{n}$ such that $x\thicksim z_{2}\thicksim z\thicksim z_{1}\thicksim y$, $o(z_{1})=o(z_{2})=pq$ and $o(z)=q$, where $q$ is a prime. So $z\in C_{S_{n}}(x)\cap C_{S_{n}}(y)$. Since $C_{S_{n}}(x)\cong\langle x\rangle\times S_{\{p+1,\cdots,n\}}$, $C_{S_{n}}(y)\cong\langle y\rangle\times S_{\{1,\cdots,n-p\}}$ and $q\nmid p$, we have $z\in S_{\{1,\cdots,n-p\}}\cap S_{\{p+1,\cdots,n\}}=1$, which is a contradiction. Therefore $d(x,y)\geq 6$. The proof is completed. ∎

Lemma 3.6**.**

Let $p^{\prime}$ is the maximum prime factor of the $n(n-1)(n-2)$ and $[\frac{n-2}{p^{\prime}}]\geq 3p^{\prime}+2$. Then $diam({\mathcal{P}}^{*}(A_{n}))\leq 8$.

Proof.

Let $x,y\in A_{n}$. Suppose $o(x)$ and $o(y)$ are not primes, and let $p$ and $q$ are at least prime factors of $o(x)$ and $o(y)$, respectively. Then $x\thicksim x^{o(x)/p}$ and $y\thicksim y^{o(y)/q}$. Now we let $x^{o(x)/p}=\alpha=\alpha_{1}\cdots\alpha_{m}$ and $y^{o(y)/q}=\beta=\beta_{1}\cdots\beta_{m^{\prime}}$. Assume that $k=n-|S(\alpha)|$, $k^{\prime}=n-|S(\beta)|$ and $p\leq q$. If $pq\leq 9$, then $d(\alpha,\beta)\leq 6$, by Lemma  2.3. Therefore $d(x,y)\leq 8$.

For $pq>9$, we consider four cases as follows:

Case 1: Let $p\neq q$, $p\geq 5$ and $q\geq 5$.

• $(1.1)$

If $k\geq 3$, $k^{\prime}\geq 3$ and $n-|S(\alpha)\cup S(\beta)|\geq 3$, there is the $3$-cycle $c$ at distance $2$ from both $\alpha$ and $\beta$, where $S(c)\subset S^{c}(\alpha)\cap S^{c}(\beta)$. Therefore $d(\alpha,\beta)\leq 4$.

• $(1.2)$

If $k\geq 3$, $k^{\prime}\geq 3$ and $n-|S(\alpha)\cup S(\beta)|<3$, then $mp+m^{\prime}q>p^{\prime}(3p^{\prime}+2)$, thus $m+m^{\prime}>3p^{\prime}+2\geq 17$, accordingly $m\geq 7$ or $m^{\prime}\geq 7$. Let $m^{\prime}\geq 7$. Since $k\geq 3$, there is the $3$-cycle $c$ with $d(\alpha,c)=2$. Since $m^{\prime}\geq 7$, there are $x$, $\beta_{j_{1}}$, $\beta_{j_{2}}$, $\beta_{j_{3}}$ and $\beta_{j_{4}}$ such that $S(c)\cap S(\beta_{j_{1}}\cdots\beta_{j_{4}})=\emptyset$, $d(x,\beta)=2$ and $S(x)=S(\beta_{j_{1}}\cdots\beta_{j_{4}})$, where $o(x)=2$. By Lemma  2.1, $d(c,x)=2$. Thus $d(\alpha,\beta)\leq 6$.

• $(1.3)$

If $k\geq 3$ and $k^{\prime}<3$, then $m^{\prime}>7$. Since $k\geq 3$, there is a $3$-cycle $c$ with $d(\alpha,c)=2$. Since $m^{\prime}\geq 7$, there is $y$ in $A_{n}$ such that $d(\beta,y)=2$ and $S(y)\cap S(c)=\emptyset$, where $o(y)=2$. By Lemma  2.1, $d(c,y)=2$. Therefore $d(\alpha,\beta)\leq 6$.

• $(1.4)$

If $k<3$ and $k^{\prime}\geq 3$, then the proof is similar to $(1.3)$.

• $(1.5)$

If $k<3$ and $k^{\prime}<3$, then there are $\alpha_{i_{1}}$, $\alpha_{i_{2}}$, $\alpha_{i_{3}}$, $\beta_{j_{1}}$ and $\beta_{j_{2}}$ such that $S(\alpha_{i_{1}})\cap S(\beta_{j_{1}})\neq\emptyset$, $|S(\alpha_{i_{2}})\cap S(\beta_{j_{1}}\beta_{j_{2}})|\geq 2$ and $S(\alpha_{i_{3}})\cap S(\beta_{j_{2}})\neq\emptyset$. Also, there is $f$ in $A_{n}$ such that $d(f,\alpha)=2$ and $S(f)=S(\alpha_{i_{1}}\alpha_{i_{2}}\alpha_{i_{3}})$, where $o(f)=3$. Hence $S(f)\subset S(\beta_{j_{1}}\cdots\beta_{j_{t}})\cup S^{c}(\beta)$ and $t\leq 3p-2$. Since $m^{\prime}\geq 3p+2$, $m^{\prime}-t\geq 4$, thus there is $g$ in $A_{n}$ such that $d(\beta,g)=2$, $S(g)=S(\beta_{j_{t+1}}\cdots\beta_{j_{t+4}})$, where $o(g)=2$. Since $S(f)\cap S(g)=\emptyset$, $d(f,g)=2$, by Lemma  2.1. Therefore $d(\alpha,\beta)\leq 6$. Consequently $d(x,y)\leq 8$.

The proof of the following cases is similar to the previous case.

Case 2: Let $p=q$ and $p\geq 5$.

Case 3: Let $q\geq 5$ and $p=3$.

Case 4: Let $q\geq 5$ and $p=2$.

Suppose $o(x)$ and $o(y)$ are prime. As proved above $d(x,y)\leq 6$. Let $o(x)$ is a prime and $o(y)$ is not a prime. Then $y\sim y^{o(y)/q}$ and let $y^{o(y)/q}=\beta=\beta_{1}\cdots\beta_{m^{\prime}}$, where $q$ is the least prime factor $o(y)$. As proved above $d(x,\beta)\leq 6$. Consequently $d(x,y)\leq 8$. The proof is completed. ∎

Example 3.7**.**

For $n=$ $2025,2432,3250,5292,7106,7569,9802$, $10621$, $10880$, $10881$, $11286$, $11440$, $11662$, $13312$, $13456$, $13690$ and $14337$, $diam({\mathcal{P}}^{*}(A_{n}))\leq 8$.

Theorem 3.8**.**

Let ${\mathcal{P}}^{*}(A_{n})$ is connected and $p$ is the maximum prime factor of the $n(n-1)(n-2)$. Then

• $(1)$

If $[\frac{n-2}{p}]\geq 3p+2$, then $6\leq diam({\mathcal{P}}^{*}(A_{n}))\leq 8$.

• $(2)$

Otherwise, $6\leq diam({\mathcal{P}}^{*}(A_{n}))\leq 11$.

Proof.

By Theorem  3.5, Corollary  3.1, and Lemma  3.6, the proof is completed.

∎

**Acknowledgements

**We would like to show our gratitude to the referee for their careful review of our manuscript and some helpful suggestions.