Vertex connectivity of the power graph of a finite cyclic group

TL;DR

This paper investigates the vertex connectivity of the power graph of finite cyclic groups, providing exact values for certain cases and bounds for others, enhancing understanding of the graph's structural properties.

Contribution

It determines the exact vertex connectivity of power graphs for cyclic groups with multiple prime factors under specific conditions, and offers bounds in more general cases.

Findings

Exact vertex connectivity for r=1 case.

Exact values for certain composite orders with multiple primes.

Upper bounds for general cases, sharp in many instances.

Abstract

Let $n=p_1^{n_1}p_2^{n_2}\ldots p_r^{n_r}$, where $r,n_1,\ldots, n_r$ are positive integers and $p_1,p_2,\ldots,p_r$ are distinct prime numbers with $p_1<p_2<\cdots <p_r$. For the cyclic group $C_n$ of order $n$, let $\mathcal{P}(C_n)$ be the power graph of $C_n$ and $\kappa(\mathcal{P}(C_n))$ be the vertex connectivity of $\mathcal{P}(C_n)$. It is known that $\kappa(\mathcal{P}(C_n))=p_1^{n_1} -1$ if $r=1$. For $r\geq 2$, we determine the exact value of $\kappa(\mathcal{P}(C_n))$ when $2\phi(p_1\ldots p_{r-1})\geq p_1\ldots p_{r-1}$, and give an upper bound for $\kappa(\mathcal{P}(C_n))$ when $2\phi(p_1\ldots p_{r-1}) < p_1\ldots p_{r-1}$, which is sharp for many values of $n$ but equality need not hold always.

Equations

$$r(x)=\bigcup _{y\in M(G)\setminus ⟨x⟩}(⟨x⟩\cap ⟨y⟩).$$

$$\kappa (\mathcal{P}(G))\le \min \{|r(x)|:x\in M(G)\}.$$

$$n=p_1^{n_1}{p}_2^{n_2}\dots p_r^{n_r},$$

$$p_1<p_2<\cdots <p_r.$$

$$\kappa (\mathcal{P}(C_n))\ge |A(C_n)|=\varphi (n)+1,$$

$$\kappa (\mathcal{P}(C_n))\le \varphi (n)+p_1^{n_1-1}{p}_2^{n_2-1}.$$

$$\kappa (\mathcal{P}(C_n))\le \varphi (n)+p_1{p}_2-\varphi (p_1{p}_2).$$

$$\kappa (\mathcal{P}(C_n))=\varphi (n)+p_1^{n_1-1}\dots p_{r-1}^{n_{r-1}-1}{p}_r^{n_r-1}\left[p_1{p}_2\dots p_{r-1}-\varphi (p_1{p}_2\dots p_{r-1})\right].$$

$$\kappa (\mathcal{P}(C_n))\le \varphi (n)+p_1^{n_1-1}\dots p_{r-1}^{n_{r-1}-1}\left[p_1\dots p_{r-1}+\varphi (p_1\dots p_{r-1})(p_r^{n_r-1}-2)\right].$$

$$\kappa (\mathcal{P}(C_n))=\varphi (n)+2^{n_1-1}{p}_2^{n_2-1}.$$

$$\kappa (\mathcal{P}(C_n))=\varphi (n)+p_1^{n_1-1}\dots p_{r-1}^{n_{r-1}-1}{p}_r^{n_r-1}\left[p_1{p}_2\dots p_{r-1}-\varphi (p_1{p}_2\dots p_{r-1})\right].$$

$$\kappa (\mathcal{P}(C_n))=\varphi (n)+2^{n_1-1}{p}_2^{n_2-1}\left[(p_2-1)p_3^{n_3-1}+2\right].$$

$$\varphi (n)=p_1^{n_1-1}(p_1-1)\dots p_r^{n_r-1}(p_r-1)=p_1^{n_1-1}\dots p_r^{n_r-1}\varphi (p_1{p}_2\dots p_r).$$

$$\varphi \left(\frac{n}{p_i}\right)\ge \varphi \left(\frac{n}{p_r^{n_r}}\right)p_r^{n_r-1},$$

$$\varphi \left(\frac{n}{p_i}\right)$$

$$>p_1^{n_1-1}\dots p_{i-1}^{n_{i-1}-1}{p}_{i+1}^{n_{i+1}-1}\dots p_r^{n_r-1}\varphi (p_1\dots p_i\dots p_{r-1})=\varphi \left(\frac{n}{p_r^{n_r}}\right)p_r^{n_r-1}.$$

$$\varphi \left(\frac{n}{p_i}\right)$$

$$\ge p_1^{n_1-1}\dots p_i^{n_i-1}\dots p_{r-1}^{n_{r-1}-1}{p}_r^{n_r-1}\varphi (p_1{p}_2\dots p_{r-1})=\varphi \left(\frac{n}{p_r^{n_r}}\right)p_r^{n_r-1}.$$

$$\varphi \left(p_1{p}_2\dots p_t\right)-p_1{p}_2\dots p_t+\underset{k=1}{\sum ^t}\frac{p_1{p}_2\dots p_t}{p_k}$$

$$m_{i_1,i_2,\dots ,i_k}=\frac{n}{p_{i_1}{p}_{i_2}\dots p_{i_k}}.$$

$${\alpha }_k=p_1^{n_1}{p}_2^{n_2}\dots p_{r-1}^{n_{r-1}}{p}_r^k.$$

$${\beta }_{k,i_1,i_2,\dots ,i_l}=\frac{{\alpha }_k}{p_{i_1}\dots p_{i_l}}=\frac{p_1^{n_1}\dots p_{r-1}^{n_{r-1}}{p}_r^k}{p_{i_1}\dots p_{i_l}}.$$

$$Z(r,k)=E_{{\alpha }_{k+1}}\cup \dots \cup E_{{\alpha }_{n_r-1}}\cup E_n\cup S_{{\beta }_{k,1}}\cup \dots \cup S_{{\beta }_{k,r-1}}.$$

$$|Z(r,k)|=\varphi (n)+{\beta }_{0,1,\dots ,r-1}\left[p_r^{n_r-1}\varphi (p_1\dots p_{r-1})+p_r^k\left[p_1\dots p_{r-1}-2\varphi (p_1\dots p_{r-1})\right]\right].$$

$$|Z(r,k)|=\left|E_n\right|+\underset{i=k+1}{\bigcup ^{n_r-1}}\left|E_{{\alpha }_i}\right|+\left|\underset{i=1}{\bigcup ^{r-1}}{S}_{{\beta }_{k,i}}\right|.$$

$$\underset{i=k+1}{\bigcup ^{n_r-1}}\left|E_{{\alpha }_i}\right|$$

$$=p_1^{n_1-1}{p}_2^{n_2-1}\dots p_{r-1}^{n_{r-1}-1}\varphi (p_1{p}_2\dots p_{r-1})(p_r^{n_r-1}-p_r^k).$$

$$\left|\underset{i=1}{\bigcup ^{r-1}}{S}_{{\beta }_{k,i}}\right|$$

$$=\sum _i{\beta }_{k,i}-\sum _{i<j}{\beta }_{k,i,j}+\dots +(-1{)}^{r-2}{\beta }_{k,1,2,\dots ,r-1}$$

$$=p_1^{n_1-1}\dots p_{r-1}^{n_{r-1}-1}{p}_r^k\left[\underset{i=1}{\sum ^{r-1}}\frac{p_1\dots p_{r-1}}{p_i}-\sum _{i<j}\frac{p_1\dots p_{r-1}}{p_i{p}_j}+\dots +(-1{)}^{r-2}\right]$$

$$=p_1^{n_1-1}\dots p_{r-1}^{n_{r-1}-1}{p}_r^k\left[p_1{p}_2\dots p_{r-1}-\varphi (p_1{p}_2\dots p_{r-1})\right].$$

$$\kappa (\mathcal{P}(C_n))\le \varphi (n)+p_1^{n_1-1}\dots p_{r-1}^{n_{r-1}-1}\left[p_r^{n_r-1}\varphi (p_1\dots p_{r-1})+p_r^k\left[p_1\dots p_{r-1}-2\varphi (p_1\dots p_{r-1})\right]\right].$$

$$\kappa (\mathcal{P}(C_n))\le \varphi (n)+p_1^{n_1-1}\dots p_{r-1}^{n_{r-1}-1}\left[p_1\dots p_{r-1}+\varphi (p_1\dots p_{r-1})(p_r^{n_r-1}-2)\right],$$

Full text

Key words: Power graph, Vertex connectivity, Cyclic group, Euler’s totient function

AMS subject classification. 05C25, 05C40, 20K99

1 Introduction

The notion of directed power graph of a group was introduced by Kelarev et al. in [9], which was further extended to semigroups in [10, 11]. Chakrabarty et al. defined the notion of undirected power graph of a semigroup, in particular, of a group in [3]. Since then many researchers have investigated both directed and undirected power graphs of groups from different view points. More on these graphs can be found in the survey paper [1] and the references therein.

Let $G$ be a finite group. The power graph $\mathcal{P}(G)$ of $G$ is the simple undirected graph with vertex set $G$, in which two distinct vertices are adjacent if and only if one of them can be obtained as a power of the other. Since $G$ is finite, the identity element is adjacent to all other vertices and so $\mathcal{P}(G)$ is connected. The vertex connectivity of $\mathcal{P}(G)$, denoted by $\kappa(\mathcal{P}(G))$, is the minimum number of vertices which need to be removed from the vertex set $G$ so that the induced subgraph of $\mathcal{P}(G)$ on the remaining vertices is disconnected or has only one vertex. The latter case arises only when $\mathcal{P}(G)$ is a complete graph.

Note that $\kappa(\mathcal{P}(G))=|G|-1$ if and only if $\mathcal{P}(G)$ is a complete graph. Chakrabarty et al. proved in [3, Theorem 2.12] that $\mathcal{P}(G)$ is a complete graph if and only if $G$ is a cyclic group of prime power order. We have $\kappa(\mathcal{P}(G))=1$ if and only if $|G|=2$ or the subgraph $\mathcal{P}(G^{\ast})$ of $\mathcal{P}(G)$ is disconnected, where $G^{\ast}=G\setminus\{1\}$. This is equivalent to saying that $\kappa(\mathcal{P}(G))>1$ if and only if $|G|\geq 3$ and $\mathcal{P}(G^{\ast})$ is connected. Recently, some authors have studied the connectedness of $\mathcal{P}(G^{\ast})$, see [8] and [13] for related works in this direction.

Let $A(G)$ be the subset of $G$ consisting of those vertices which are adjacent to all other vertices of $\mathcal{P}(G)$. The identity element of $G$ is in $A(G)$. The groups $G$ for which $|A(G)|>1$ were described in [2, Proposition 4].

Theorem 1.1**.**

[2]** Let $G$ be a finite group for which $|A(G)|>1$. Then one of the following holds:

1. (1)

$G$* is a cyclic group of prime power order and $A(G)=G$;* 2. (2)

$G$* is a cyclic group of non-prime-power order and $A(G)$ consists of the identity element and the generators of $G$;* 3. (3)

$G$* is a generalized quaternion $2$-group and $A(G)$ contains the identity element and the unique involution in $G$.*

Determining $\kappa(\mathcal{P}(G))$ for an arbitrary finite group $G$ is an interesting problem in the study of power graphs. When $\mathcal{P}(G)$ is not a complete graph, in order to find $\kappa(\mathcal{P}(G))$, first of all one needs to remove the vertices in $A(G)$ from $G$. This gives $\kappa(\mathcal{P}(G))\geq|A(G)|$. If $G$ is not a cyclic group, then the following upper bound for $\kappa(\mathcal{P}(G))$ was obtained in [12, Theorem 10].

Theorem 1.2**.**

[12]** Let $G$ be a non-cyclic finite group. Let $M(G)$ be the subset of $G$ consisting of elements $x$ for which $\langle x\rangle$ is a maximal cyclic subgroup of $G$. For $x\in M(G)$, define

[TABLE]

Then

[TABLE]

To our knowledge, $\kappa(\mathcal{P}(G))$ is not determined even when $G$ is an arbitrary finite cyclic group. In this paper, we consider $G=C_{n}$, a finite cyclic group of order $n$ and study the vertex connectivity $\kappa(\mathcal{P}(C_{n}))$ of $\mathcal{P}(C_{n})$. We note that the graph $\mathcal{P}(C_{n})$ has the maximum number of edges among all the power graphs of finite groups of order $n$. This property of $\mathcal{P}(C_{n})$ was conjectured by Mirzargar et al. in [12] and proved by Curtin and Pourgholi in [6]. So it is natural to expect that $\kappa(\mathcal{P}(C_{n}))$ would be large in general.

If $n=1$ or if $n$ is divisible by only one prime number, then $\mathcal{P}\left(C_{n}\right)$ is a complete graph and so $\kappa\left(\mathcal{P}\left(C_{n}\right)\right)=n-1$. Therefore, for the rest of the paper, we assume that $n$ is divisible by at least two distinct prime numbers. We write the prime power decomposition of $n$ as

[TABLE]

where $r\geq 2$, $n_{1},\ldots,n_{r}$ are positive integers and $p_{1},\ldots,p_{r}$ are distinct prime numbers with

[TABLE]

Since $r\geq 2$, the set $A(C_{n})$ consists of the identity element and the generators of $C_{n}$ (see Theorem 1.1). So

[TABLE]

where $\phi:\mathbb{N}\longrightarrow\mathbb{N}$ is the Euler’s totient function. Note that the number of generators of $C_{n}$ is $\phi(n)$. If $n=p_{1}p_{2}$, then equality holds in (1) by [4, Theorem 3(ii)]. We shall show that the converse is also true (see Lemma 2.5). When $n=p_{1}^{n_{1}}p_{2}^{n_{2}}$, it was proved in [5, Theorem 2.7] that

[TABLE]

If $n=p_{1}p_{2}p_{3}$, then [5, Theorem 2.9] gives that

[TABLE]

Here we determine the exact value of $\kappa(\mathcal{P}(C_{n}))$ when $2\phi(p_{1}\ldots p_{r-1})\geq p_{1}\ldots p_{r-1}$ and give an upper bound for $\kappa(\mathcal{P}(C_{n}))$ when $2\phi(p_{1}\ldots p_{r-1})<p_{1}\ldots p_{r-1}$.

1.1 Main Results

For a given subset $X$ of $C_{n}$, we define $\overline{X}=C_{n}\setminus X$ and denote by $\mathcal{P}(\overline{X})$ the induced subgraph of $\mathcal{P}(C_{n})$ with vertex set $\overline{X}$. We prove the following in Sections 3 and 4.

Theorem 1.3**.**

Let $n=p_{1}^{n_{1}}\ldots p_{r}^{n_{r}}$, where $r\geq 2$, $n_{1},\ldots,n_{r}$ are positive integers and $p_{1},\ldots,p_{r}$ are distinct prime numbers with $p_{1}<p_{2}<\cdots<p_{r}$. Then the following hold:

1. (i)

If $2\phi(p_{1}\ldots p_{r-1})>p_{1}\ldots p_{r-1}$, then

[TABLE]

Further, there is only one subset $X$ of $C_{n}$ with $|X|=\kappa(\mathcal{P}(C_{n}))$ such that $\mathcal{P}(\overline{X})$ is disconnected. 2. (ii)

If $2\phi(p_{1}\ldots p_{r-1})<p_{1}\ldots p_{r-1}$, then

[TABLE] 3. (iii)

If $2\phi(p_{1}\ldots p_{r-1})=p_{1}\ldots p_{r-1}$, then $r=2$, $p_{1}=2$ (so that $n=2^{n_{1}}p_{2}^{n_{2}}$) and

[TABLE]

Moreover, there are $n_{2}$ subsets $X$ of $C_{n}$ with $|X|=\kappa(\mathcal{P}(C_{n}))$ for which $\mathcal{P}(\overline{X})$ is disconnected.

Note that the values in the right hand side of the (in)equalities in Theorem 1.3 are equal if $n_{r}=1$. As a consequence of Theorem 1.3(i) and (iii), we obtain the following when the smallest prime divisor of $n$ is at least the number of distinct prime divisors of $n$.

Corollary 1.4**.**

If $p_{1}\geq r$, then

[TABLE]

In order to show that the bound in Theorem 1.3(ii) is sharp for many values of $n$, we prove the following in Section 5 when $r=3$.

Theorem 1.5**.**

Let $n=p_{1}^{n_{1}}p_{2}^{n_{2}}p_{3}^{n_{3}}$, where $n_{1},n_{2},n_{3}$ are positive integers and $p_{1},p_{2},p_{3}$ are distinct prime numbers with $p_{1}<p_{2}<p_{3}$. If $2\phi(p_{1}p_{2})<p_{1}p_{2}$, then $p_{1}=2$ and

[TABLE]

Further, there is only one subset $X$ of $C_{n}$ with $|X|=\kappa(\mathcal{P}(C_{n}))$ such that $\mathcal{P}(\overline{X})$ is disconnected.

We further note that, for certain values of $n$, equality may not hold in the bound obtained in Theorem 1.3(ii), see Example 3.4 below. So it would be interesting to find $\kappa(\mathcal{P}(C_{n}))$ when $2\phi(p_{1}\ldots p_{r-1})<p_{1}\ldots p_{r-1}$.

2 Preliminaries

Recall that $\phi$ is a multiplicative function, that is, $\phi(ab)=\phi(a)\phi(b)$ for any two positive integers $a,b$ which are relatively prime. So

[TABLE]

Lemma 2.1**.**

For $1\leq i\leq r-1$, we have

[TABLE]

where the inequality is strict except when $r=2$, $p_{1}=2$, $p_{2}=3$ and $n_{1}\geq 2$.

Proof.

Let $1\leq i\leq r-1$. We have $\phi(p_{r})=p_{r}-1\geq p_{i}>p_{i}-1=\phi(p_{i})$. If $n_{i}=1$, then

[TABLE]

If $n_{i}\geq 2$, then

[TABLE]

The last part of the lemma can be seen easily. Note that $\phi(p_{r})>p_{i}$ for $r\geq 3$. ∎

Lemma 2.2**.**

We have $\phi\left(p_{1}p_{2}\ldots p_{t}\right)-p_{1}p_{2}\ldots p_{t}+\underset{k=1}{\overset{t}{\sum}}\frac{p_{1}p_{2}\ldots p_{t}}{p_{k}}\geq 0$ for all $t\geq 1$. Further, equality holds if and only if $t=1$.

Proof.

Clearly, equality holds if $t=1$. So let $t\geq 2$. We shall prove by induction on $t$ that strict inequality holds. Since $\phi\left(p_{1}p_{2}\right)-p_{1}p_{2}+p_{1}+p_{2}=1>0$, the statement is true for $t=2$. Assume that $t\geq 3$. Then

[TABLE]

In the above, the last inequality follows using the induction hypothesis. ∎

For $x\in C_{n}$, we denote by $o(x)$ the order of $x$. If two elements $x,y$ are adjacent in $\mathcal{P}(C_{n})$, then $o(x)\mid o(y)$ or $o(y)\mid o(x)$ according as $x\in\langle y\rangle$ or $y\in\langle x\rangle$. The converse statement is also true (which does not hold for an arbitrary finite group), which follows from the property that $C_{n}$ has a unique subgroup of order $d$ for every positive divisor $d$ of $n$. We shall use this fact frequently without mention. For $x\in C_{n}$, let $N(x)$ be the neighborhood of $x$ in $\mathcal{P}(C_{n})$, that is, the set of all elements of $C_{n}$ which are adjacent to $x$. If $o(x)=o(y)$ for $x,y\in C_{n}$, then it is clear that $N(x)\cup\{x\}=N(y)\cup\{y\}$, also see [7, Lemma 3].

Let $X$ be a subset of $C_{n}$. For two disjoint nonempty subsets $A$ and $B$ of $\overline{X}$, we say that $A\cup B$ is a separation of $\mathcal{P}(\overline{X})$ if $\overline{X}=A\cup B$ and there is no edge containing vertices from both $A$ and $B$. Thus $\mathcal{P}(\overline{X})$ is disconnected if and only if there exists a separation of it. For a positive divisor $d$ of $n$, we define the following two sets:

1. $E_{d}$ = the set of all elements of $C_{n}$ whose order is $d$,
2. $S_{d}$ = the set of all elements of $C_{n}$ whose order divides $d$.

Then $S_{d}$ is a cyclic subgroup of $C_{n}$ of order $d$ and $E_{d}$ is precisely the set of generators of $S_{d}$. So $\left|S_{d}\right|=d$ and $\left|E_{d}\right|=\phi(d).$ The following result is very useful throughout the paper.

Lemma 2.3**.**

Let $X$ be a subset of $C_{n}$ of minimum possible size with the property that $\mathcal{P}(\overline{X})$ is disconnected. Then either $E_{d}\subseteq X$ or $E_{d}\cap X$ is empty for each divisor $d$ of $n$.

Proof.

Suppose that $E_{d}\cap X\neq\emptyset$. We show that $E_{d}\subseteq X$. Fix a separation $A\cup B$ of $\mathcal{P}(\overline{X})$. Let $x\in E_{d}\cap X$. The minimality of $|X|$ implies that the subgraph $\mathcal{P}(\overline{X\setminus\{x\}})$ of $\mathcal{P}(C_{n})$ is connected. So there exist $a\in A$ and $b\in B$ such that $x$ is adjacent to both $a$ and $b$.

Suppose that there exists an element $y\in E_{d}$ which is not in $X$. Then $y\neq x$. Since $o(x)=o(y)=d$, we have $N(x)\cup\{x\}=N(y)\cup\{y\}$. If $y\in A$ (respectively, $y\in B$), then the fact that $x$ is adjacent to $b$ (respectively, to $a$) implies $y$ is adjacent to $b$ (respectively, to $a$). This contradicts that $A\cup B$ is a separation of $\mathcal{P}(\overline{X})$. ∎

Remark 2.4**.**

Under the hypothesis of Lemma 2.3, it follows that there are three possibilities for the set $E_{d}$, where $d$ is a divisor of $n$: either $E_{d}\subseteq X$, $E_{d}\subseteq A$ or $E_{d}\subseteq B$, where $A\cup B$ is any separation of $\mathcal{P}(\overline{X})$.

We complete this section with the following lemma. For a given subset $\{i_{1},i_{2},\ldots,i_{k}\}$ of $\{1,2,\ldots,r\}$, we define the integer $m_{i_{1},i_{2},\ldots,i_{k}}$ by

[TABLE]

Lemma 2.5**.**

$\kappa(\mathcal{P}(C_{n}))=\phi(n)+1$* if and only if $r=2$ and $n=p_{1}p_{2}$.*

Proof.

If $n=p_{1}p_{2}$, then $\kappa(\mathcal{P}(C_{n}))=\phi(n)+1$ by [4, Theorem 3(ii)]. For the converse part, let $X=E_{1}\cup E_{n}$. It is enough to show that $\mathcal{P}(\overline{X})$ is connected whenever $r\geq 3$, or $r=2$ and one of $n_{1},n_{2}$ is at least $2$.

Let $x$ and $y$ be two distinct elements of $\overline{X}$. Then $x\in S_{m_{j}}$ and $y\in S_{m_{k}}$ for some $j,k\in\{1,2,\ldots,r\}$. So $x$ (respectively, $y$) is adjacent with the elements of $E_{m_{j}}\setminus\{x\}$ (respectively, $E_{m_{k}}\setminus\{y\}$). If $j=k$, then $x$ and $y$ are connected through the elements of $E_{m_{j}}$. Assume that $j\neq k$. Since $r\geq 3$, or $r=2$ and one of $n_{1},n_{2}$ is at least $2$, the set $E_{m_{j,k}}$ is non-empty. Then the elements of both $E_{m_{j}}$ and $E_{m_{k}}$ are adjacent with the elements of $E_{m_{j,k}}$. It follows that $x$ and $y$ are connected by a path. ∎

3 Upper bounds and Proof of Theorem 1.3(ii) and (iii)

We shall prove the bounds in Theorem 1.3 by identifying suitable subsets $X$ of $C_{n}$ of required size such that $\mathcal{P}(\overline{X})$ is disconnected. Let $0\leq k\leq n_{r}-1$. Define the following integers:

[TABLE]

So $\alpha_{k}=\alpha_{0}p_{r}^{k}$. For given $k$ and subset $\{i_{1},i_{2},\ldots,i_{l}\}$ of $\{1,2,\ldots,r-1\}$, define the integer $\beta_{k,i_{1},i_{2},\ldots,i_{l}}$ by

[TABLE]

Set

[TABLE]

Proposition 3.1**.**

The subgraph $\mathcal{P}(\overline{Z(r,k)})$ of $\mathcal{P}(C_{n})$ is disconnected.

Proof.

For $x\in\overline{Z(r,k)}$, observe that the order $o(x)$ of $x$ is one of the following two types:

1. (I)

$o(x)=p_{1}^{n_{1}}p_{2}^{n_{2}}\ldots p_{r-1}^{n_{r-1}}p_{r}^{s}$, where $0\leq s\leq k$; 2. (II)

$o(x)=p_{1}^{l_{1}}\ldots p_{r-1}^{l_{r-1}}p_{r}^{t}$, where $k+1\leq t\leq n_{r}$, $0\leq l_{i}\leq n_{i}$ for $i\in\{1,\ldots,r-1\}$ and $(l_{1},\ldots,l_{r-1})\neq(n_{1},\ldots,n_{r-1})$.

Let $A$ (respectively, $B$) be the subset of $\overline{Z(r,k)}$ consisting of all the elements whose order is of type (I) (respectively, of type (II)). Then $A,B$ are nonempty sets and $\overline{Z(r,k)}=A\cup B$ is a disjoint union. Since $t>s$, no element of $B$ can be obtained as an integral power of any element of $A$. Again, since $(l_{1},\ldots,l_{r-1})\neq(n_{1},\ldots,n_{r-1})$, no element of $A$ can be obtained as an integral power of any element of $B$. It follows that $A\cup B$ is a separation of $\mathcal{P}(\overline{Z(r,k)})$. ∎

Proposition 3.2**.**

The number of elements in $Z(r,k)$ is given by:

[TABLE]

Proof.

The sets $E_{\alpha_{k+1}},\ldots,E_{\alpha_{n_{r}-1}},E_{n}$ are pairwise disjoint and each of them is disjoint from $\underset{i=1}{\overset{r-1}{\bigcup}}S_{\beta_{k,i}}$. So

[TABLE]

We have

[TABLE]

and

[TABLE]

Now the lemma follows from the above, as $\beta_{0,1,\ldots,r-1}=p_{1}^{n_{1}-1}\ldots p_{r-1}^{n_{r-1}-1}$. ∎

In the next section, we shall have occasions to calculate the cardinality of the union of certain subgroups, which will be similar to that of calculating $\left|\underset{i=1}{\overset{r-1}{\bigcup}}S_{\beta_{k,i}}\right|$ as in the above. As a consequence of Propositions 3.1 and 3.2, we have the following.

Corollary 3.3**.**

The vertex connectivity $\kappa(\mathcal{P}(C_{n}))$ of $\mathcal{P}(C_{n}))$ satisfies the following:

[TABLE]

Proof of Theorem 1.3(ii).

If $2\phi(p_{1}\ldots p_{r-1})<p_{1}\ldots p_{r-1}$, then the minimum of $|Z(r,k)|$ occurs when $k=0$. This gives

[TABLE]

thus proving Theorem 1.3(ii). ∎

The following example shows that equality may not hold in (2).

Example 3.4**.**

Consider $n=2310=2\times 3\times 5\times 7\times 11$. The value in the right hand side of (2) is $\phi(n)+162$. Now set

[TABLE]

Taking $A=E_{30}$ and $B=C_{n}\setminus(X\cup A)$, it can be seen that $A\cup B$ is a separation of $\mathcal{P}(\overline{X})$ and so $\mathcal{P}(\overline{X})$ is disconnected. Calculating $|X|$, we get

[TABLE]

So equality may not hold in (2).

If $2\phi(p_{1}\ldots p_{r-1})>p_{1}\ldots p_{r-1}$, then the minimum of $|Z(r,k)|$ occurs when $k=n_{r}-1$ and this gives

[TABLE]

In the next section, we shall show that equality holds in (3) which will prove Theorem 1.3(i). Observe that the bounds (2) and (3) coincide if $n_{r}=1$, or if $r=2$ and $p_{1}=2$.

Proof of Theorem 1.3(iii).

Note that $2\phi(p_{1}\ldots p_{r-1})=p_{1}\ldots p_{r-1}$ if and only if $r=2$ and $p_{1}=2$. So $n=2^{n_{1}}p_{2}^{n_{2}}$ in this case. For $r=2$ and $0\leq k\leq n_{2}-1$, we have

[TABLE]

and that $|Z(2,k)|=\phi(n)+2^{n_{1}-1}p_{2}^{n_{2}-1}$ is independent of $k$. Thus

[TABLE]

Now, let $X$ be a subset of $C_{n}$ of minimum possible size such that $\mathcal{P}(\overline{X})$ is disconnected. In order to prove Theorem 1.3(iii), it is enough to show that $X=Z(2,t)$ for some $0\leq t\leq n_{2}-1$. Write $T=X\setminus(E_{1}\cup E_{n})$. Since $X$ contains $E_{1}\cup E_{n}$ and $|X|\leq\phi(n)+2^{n_{1}-1}p_{2}^{n_{2}-1}$, we get

[TABLE]

We claim that the set $E_{2^{n_{1}-1}p_{2}^{n_{2}}}$ is disjoint from $X$. Otherwise, $E_{2^{n_{1}-1}p_{2}^{n_{2}}}\subseteq T$ by Lemma 2.3 and so $\left|E_{2^{n_{1}-1}p_{2}^{n_{2}}}\right|\leq|T|$. On the other hand, using Lemma 2.1, we get

[TABLE]

a contradiction.

Let $A\cup B$ be a separation of $\mathcal{P}(\overline{X})$. We may assume that $E_{2^{n_{1}-1}p_{2}^{n_{2}}}$ is contained in $B$. Then, for $x\in A$, we must have

[TABLE]

for some $j$ with $0\leq j\leq n_{2}-1$. Let $t\in\{0,1,\ldots,n_{2}-1\}$ be the largest integer for which $A$ has an element of order $2^{n_{1}}p_{2}^{t}$. Then, using Lemma 2.3 together with the fact that $A\cup B$ is a separation of $\mathcal{P}(\overline{X})$, the following hold:

1. (i)

$E_{2^{n_{1}}p_{2}^{t}}\subseteq A$, 2. (ii)

$E_{2^{n_{1}}p_{2}^{t+1}},E_{2^{n_{1}}p_{2}^{t+2}},\ldots,E_{2^{n_{1}}p_{2}^{n_{2}-1}}$ are contained in $T$, 3. (iii)

the subgroup $S_{2^{n_{1}-1}p_{2}^{t}}$ is contained in $X$.

Thus $X$ contains $Z(2,t)$. Since $|X|\leq\phi(n)+2^{n_{1}-1}p_{2}^{n_{2}-1}=|Z(2,t)|$, it follows that $X=Z(2,t)$. This completes the proof. ∎

4 Proof of Theorem 1.3(i) and Corollary 1.4

Assume, throughout this section, that $2\phi(p_{1}\ldots p_{r-1})>p_{1}\ldots p_{r-1}$ (and so $p_{1}\geq 3$). Let $X$ be a subset of $C_{n}$ of minimum possible size with the property that $\mathcal{P}(\overline{X})$ is disconnected. By (3),

[TABLE]

Set $T=X\setminus E_{n}$. Since $E_{n}\subseteq X$, we have

[TABLE]

Proposition 4.1**.**

Each of the sets $E_{m_{i}}$, $1\leq i\leq r-1$, is disjoint from $X$.

Proof.

Suppose that $E_{m_{i}}\cap X\neq\emptyset$. Then $E_{m_{i}}\subseteq X$ by Lemma 2.3, in fact, $E_{m_{i}}\subseteq T$. So $|E_{m_{i}}|\leq|T|$. Since $1\leq i\leq r-1$, $|E_{m_{i}}|=\phi\left(\frac{n}{p_{i}}\right)>\phi\left(\frac{n}{p_{r}^{n_{r}}}\right)p_{r}^{n_{r}-1}$ by Lemma 2.1 and so

[TABLE]

Then the inequalities (4) and (5) together imply

[TABLE]

Since $2\phi(p_{1}p_{2}\ldots p_{r-1})>p_{1}p_{2}\ldots p_{r-1}$, it follows that $|E_{m_{i}}|>|T|$, a contradiction. ∎

We shall prove later that the set $E_{m_{r}}$ is also disjoint from $X$. However, the argument used in the proof of Proposition 4.1 can not be applied (when $n_{r}\geq 2$) to prove this statement.

Fix a separation $A\cup B$ of $\mathcal{P}(\overline{X})$. Proposition 4.1 implies that each $E_{m_{i}}$, $1\leq i\leq r-1$, is contained either in $A$ or in $B$.

Proposition 4.2**.**

Suppose that $r\geq 3$. If $E_{m_{i}}\subseteq A$ and $E_{m_{j}}\subseteq B$ for some $1\leq i\neq j\leq r-1$, then $E_{m_{r}}$ is disjoint from $X$.

Proof.

Since $A\cup B$ is a separation of $\mathcal{P}(\overline{X})$, the subgroup $S_{m_{i,j}}$ of $C_{n}$ is contained in $T$. Suppose that $E_{m_{r}}\cap X\neq\emptyset$. Then $E_{m_{r}}\subseteq T$ by Lemma 2.3. So $|S_{m_{i,j}}|+|E_{m_{r}}|=|S_{m_{i,j}}\cup E_{m_{r}}|\leq|T|$. If $n_{r}\geq 2$, then

[TABLE]

and so

[TABLE]

If $n_{r}=1$, then $|S_{m_{i,j}}|+|E_{m_{r}}|=\frac{n}{p_{i}p_{j}}+p_{1}^{n_{1}-1}\ldots p_{r-1}^{n_{r-1}-1}\phi(p_{1}p_{2}\ldots p_{r-1})$ and so

[TABLE]

In both cases, it follows that $|S_{m_{i,j}}|+|E_{m_{r}}|>|T|$, a contradiction. ∎

Proposition 4.3**.**

All the sets $E_{m_{i}}$, $1\leq i\leq r-1$, are contained either in $A$ or in $B$.

Proof.

Clearly, this holds for $r=2$. Assume that $r\geq 3$. Suppose that some of the sets $E_{m_{i}}$, $1\leq i\leq r-1$, are contained in $A$ and some are in $B$. Then $E_{m_{r}}$ is disjoint from $X$ by Proposition 4.2. Without loss, we may assume that $E_{m_{r}}$ is contained in $A$. Let

[TABLE]

Then $1\leq a\leq r-2$ by our assumption. We shall get a contradiction by showing that $a\notin\{1,2,\ldots,r-2\}$.

Claim-1: $a\neq r-2$. Suppose that $a=r-2$. Let $E_{m_{i_{1}}},E_{m_{i_{2}}},\ldots,E_{m_{i_{a}}}$ be the sets contained in $A$ and $E_{m_{i_{a+1}}}$ be contained in $B$, where $\{i_{1},\ldots,i_{a},i_{a+1}\}=\{1,2,\ldots,r-1\}$. Since $E_{m_{r}}$ is contained in $A$, the subgroups

[TABLE]

are contained in $T$. This follows as $A\cup B$ is a separation of $\mathcal{P}(\overline{X})$. Let $Q$ be the union of these $r-1$ subgroups. Then $|Q|\leq|T|$. We shall get a contradiction by showing that $|Q|>|T|$.

The subscript $i_{a+1}$ is common to all the above $r-1$ subgroups. Applying a similar calculation as in the proof of Proposition 3.2, we get

[TABLE]

Then

[TABLE]

The last inequality holds as $1\leq i_{a+1}\leq r-1$. So $|Q|>|T|$, a contradiction. This proves Claim-1.

Claim-2: $a\notin\{1,2,\ldots,r-3\}$. Suppose that $1\leq a\leq r-3$ (we must have $r\geq 4$ as $a\geq 1$). Set $b=r-1-a$. Then $b\geq 2$. Let $E_{m_{i_{1}}},\ldots,E_{m_{i_{a}}}$ be the sets contained in $A$ and $E_{m_{i_{a+1}}},E_{m_{i_{a+2}}},\ldots,E_{m_{i_{a+b}}}$ be contained in $B$, where

[TABLE]

So the following subgroups

[TABLE]

are contained in $T$. Let $1\leq s\leq a$ and $1\leq t\leq b$. Note that $E_{m_{i_{s},i_{a+t}}}$ is the set of generators of the subgroup $S_{m_{i_{s},i_{a+t}}}$ and so is contained in $T$. Define the set

[TABLE]

Since $R$ is contained in $T$, we have $|R|\leq|T|$. We shall get a contradiction by showing that $|R|>|T|$.

The subscript $r$ is common to all the $b$ subgroups contained in $R$. Applying a similar calculation as in the proof of Proposition 3.2, we have

[TABLE]

We next calculate a lower bound for $\left|E_{m_{i_{s},i_{a+t}}}\right|$. Applying Lemma 2.1,

[TABLE]

It can easily be seen (irrespective of $n_{i_{s}}=1$ or $n_{i_{s}}\geq 2$) that

[TABLE]

Using the inequalities (6) and (7), we get

[TABLE]

So

[TABLE]

Observe that the right hand side of (8) is independent of $t$. Since $b\geq 2$ and the sets $E_{m_{i_{s},i_{a+t}}}$ are pairwise disjoint, we get

[TABLE]

The hypothesis that $2\phi(p_{1}\ldots p_{r-1})>p_{1}\ldots p_{r-1}$ implies $2\phi(p_{j_{1}}\ldots p_{j_{l}})>p_{j_{1}}\ldots p_{j_{l}}$ for any subset $\{j_{1},\ldots,j_{l}\}$ of $\{1,2,\ldots,r-1\}$. So the last inequality holds in the above. Note that each of the sets $E_{m_{i_{s},i_{a+t}}}$ is disjoint from $\underset{j=1}{\overset{b}{\bigcup}}S_{m_{r,i_{a+j}}}$. So

[TABLE]

Then

[TABLE]

The last inequality holds by Lemma 2.2. It follows that $\left|R\right|>\left|T\right|$, a contradiction. This proves Claim-2. ∎

The following proves Theorem 1.3(i). Recall the integers $\alpha_{k}$ and $\beta_{k,i_{1},i_{2},\ldots,i_{l}}$ defined in Section 3 for $0\leq k\leq n_{r}-1$ and subsets $\{i_{1},i_{2},\ldots,i_{l}\}$ of $\{1,2,\ldots,r-1\}$.

Proposition 4.4**.**

The set $E_{m_{r}}$ is disjoint from $X$. As a consequence, $X=Z(r,n_{r}-1)$ and

[TABLE]

Proof.

By Proposition 4.3, we may assume that all the sets $E_{m_{i}}$, $1\leq i\leq r-1$, are contained in $B$. We show that $E_{m_{r}}$ is contained in $A$.

Note that the order of an element in $A$ is of the form $\alpha_{j}=p_{1}^{n_{1}}p_{2}^{n_{2}}\ldots p_{r-1}^{n_{r-1}}p_{r}^{j}$ for some $j$ with $0\leq j\leq n_{r}-1$. This follows, since the elements of $A$ are not adjacent with the elements of $E_{m_{i}}$, $1\leq i\leq r-1$, in $B$. Let $t\in\{0,1,\ldots,n_{r}-1\}$ be the largest integer for which $A$ has an element of order $\alpha_{t}$. Then Lemma 2.3 implies that $E_{\alpha_{t}}\subseteq A$. We claim that $t=n_{r}-1$.

If $n_{r}=1$, then there is nothing to prove. So consider $n_{r}\geq 2$. Suppose that $t<n_{r}-1$. Since $A\cup B$ is a separation of $\mathcal{P}(\overline{X})$, the sets $E_{\alpha_{k}}$ ($t+1\leq k\leq n_{r}-1$) and the subgroups $S_{\beta_{t,l}}$ ($1\leq l\leq r-1$) are contained in $T$. Set

[TABLE]

Then $|P|+|Q|=|P\cup Q|\leq|T|$. We now calculate $|P|$ and $|Q|$. Since the sets $E_{\alpha_{k}}$ are pairwise disjoint, we have

[TABLE]

Applying a similar calculation as in the proof of Proposition 3.2, we get

[TABLE]

Then

[TABLE]

The last inequality holds, since $0\leq t<n_{r}-1$ and $2\phi(p_{1}\ldots p_{r-1})>p_{1}\ldots p_{r-1}$. It follows that $|P|+|Q|>|T|$, a contradiction. Hence $t=n_{r}-1$ and $E_{m_{r}}$ is contained in $A$.

We now show that $X=Z(r,n_{r}-1)$. Since $E_{m_{r}}$ is contained in $A$ and $E_{m_{i}}$, $1\leq i\leq r-1$, are contained in $B$, the subgroups

[TABLE]

of $C_{n}$ must be contained in $X$. Since $E_{n}\subseteq X$, it follows that $X$ contains $Z(r,n_{r}-1)$. Then minimality of $|X|$ implies that $X=Z(r,n_{r}-1)$ and so

[TABLE]

This completes the proof. ∎

As a consequence of Theorem 1.3(i) and (iii), we now prove Corollary 1.4.

Proof of Corollary1.4.

Since $p_{1}\geq r\geq 2$, we get

[TABLE]

where the inequality is strict except when $r=2$ and $p_{1}=2$. Thus $2\phi(p_{1}\ldots p_{r-1})\geq p_{1}\ldots p_{r-1}$, with equality if and only if $(r,p_{1})=(2,2)$. Then the corollary follows from Theorem 1.3(i) and (iii). ∎

5 Proof of Theorem 1.5

Here $r=3$. Since $2\phi(p_{1}p_{2})<p_{1}p_{2}$, it follows from the proof Corollary 1.4 that $p_{1}<3=r$. So $p_{1}=2$ and $n=2^{n_{1}}p_{2}^{n_{2}}p_{3}^{n_{3}}$. Since $2<p_{2}<p_{3}$, we have $p_{2}\geq 3$ and $p_{3}\geq 5$. Let $X$ be a subset of $C_{n}$ of minimum size such that $\mathcal{P}(\overline{X})$ is disconnected. Then, using the bound (2), we have $|X|\leq\phi(n)+2^{n_{1}-1}p_{2}^{n_{2}-1}\left[p_{3}^{n_{3}-1}\phi(p_{2})+2\right].$ Setting $\Gamma=X\setminus E_{n}$, we get

[TABLE]

Proposition 5.1**.**

$E_{m_{1}}$* is disjoint from $X$.*

Proof.

Otherwise, $E_{m_{1}}\subseteq\Gamma$ by Lemma 2.3. Since the identity element of $C_{n}$ is in $\Gamma$ but not in $E_{m_{1}}$, we get $|E_{m_{1}}|<|\Gamma|$. On the other hand, using (9), we have

[TABLE]

Since $p_{2}\geq 3$ and $p_{3}\geq 5$, it follows that $|E_{m_{1}}|-|\Gamma|\geq 0$, a contradiction. ∎

Fix a separation $A\cup B$ of $\mathcal{P}(\overline{X})$. By the above proposition, $E_{m_{1}}$ is contained either in $A$ or in $B$. Without loss of generality, we may assume that $E_{m_{1}}\subseteq A$.

Proposition 5.2**.**

At least one of $E_{m_{2}}$ and $E_{m_{3}}$ is disjoint from $X$.

Proof.

Otherwise, both $E_{m_{2}}$ and $E_{m_{3}}$ are contained in $\Gamma$ by Lemma 2.3 and so $|E_{m_{2}}|+|E_{m_{3}}|\leq|\Gamma|$. Set $\theta=|E_{m_{2}}|+|E_{m_{3}}|-|\Gamma|$. Using (9), the following can be verified:

[TABLE]

Since $p_{2}\geq 3$ and $p_{3}\geq 5$, it follows that $\theta>0$ in all the four cases. This gives $|E_{m_{2}}|+|E_{m_{3}}|>|\Gamma|$, a contradiction. ∎

Proposition 5.3**.**

$E_{m_{2}}$* is disjoint from $X$.*

Proof.

Otherwise, $E_{m_{2}}$ is contained in $\Gamma$ by Lemma 2.3. Then, by Proposition 5.2, $E_{m_{3}}$ is disjoint from $X$ and so is contained either in $A$ or in $B$.

Case 1: $E_{m_{3}}\subseteq B$. Since $E_{m_{1}}\subseteq A$, the subgroup $S_{m_{1,3}}$ must be contained in $\Gamma$ and so $|E_{m_{2}}|+|S_{m_{1,3}}|\leq|\Gamma|$. We calculate $|E_{m_{2}}|+|S_{m_{1,3}}|-|\Gamma|$ using (9). If $n_{2}=1$, then

[TABLE]

If $n_{2}\geq 2$, then

[TABLE]

Thus $|E_{m_{2}}|+|S_{m_{1,3}}|>|\Gamma|$, a contradiction.

Case 2: $E_{m_{3}}\subseteq A$. Since $E_{m_{2}}\subseteq X$ and $A\cup B$ is a separation of $\mathcal{P}(\overline{X})$, the order of any element of $B$ is of the form $2^{n_{1}}p_{2}^{k}p_{3}^{n_{3}},$ where $0\leq k\leq n_{2}-2$ (this is possible only when $n_{2}\geq 2$). Let $t\in\{0,1,\ldots,n_{2}-2\}$ be the largest integer for which $B$ has an element of order $2^{n_{1}}p_{2}^{t}p_{3}^{n_{3}}$. Then the sets

[TABLE]

and the two subgroups

[TABLE]

are contained in $\Gamma$. Set $P=\underset{k=t+1}{\overset{n_{2}-1}{\bigcup}}E_{2^{n_{1}}p_{2}^{k}p_{3}^{n_{3}}}$ and $Q=S_{2^{n_{1}}p_{2}^{t}p_{3}^{n_{3}-1}}\cup S_{2^{n_{1}-1}p_{2}^{t}p_{3}^{n_{3}}}$ . Then $|P|+|Q|=|P\cup Q|\leq|\Gamma|$. We have $|Q|=2^{n_{1}-1}p_{2}^{t}p_{3}^{n_{3}-1}(p_{3}+1)$ and

[TABLE]

So

[TABLE]

Then $|P|+|Q|-|\Gamma|\geq 2^{n_{1}-1}p_{2}^{n_{2}-1}[p_{3}^{n_{3}-1}(p_{3}-p_{2})-2]+2^{n_{1}}p_{2}^{t}p_{3}^{n_{3}-1}>0$ for any $t$. This gives $|P|+|Q|>|\Gamma|$, a contradiction. ∎

Proposition 5.4**.**

$E_{m_{2}}$* is contained in $A$.*

Proof.

Proposition 5.3 implies that $E_{m_{2}}$ is contained either in $A$ or in $B$. Suppose that $E_{m_{2}}\subseteq B$. Since $E_{m_{1}}\subseteq A$, the subgroup $S_{m_{1,2}}$ is contained in $\Gamma$. Consider the set $E_{m_{3}}$, which would be contained either in $A,B$ or $\Gamma$. We show that none of these possibilities occurs. If $E_{m_{3}}\subseteq\Gamma$, then

[TABLE]

This gives $|E_{m_{3}}|+|S_{m_{1,2}}|>|\Gamma|$, a contradiction. If $E_{m_{3}}\subseteq A$, then the subgroup $S_{m_{2,3}}$ is contained in $\Gamma$. Since

[TABLE]

we get $|S_{m_{1,2}}\cup S_{m_{2,3}}|>|\Gamma|$, a contradiction. Finally, assume that $E_{m_{3}}\subseteq B$. Then the subgroup $S_{m_{1,3}}$ is contained in $\Gamma$. In this case, we get

[TABLE]

giving $|S_{m_{1,2}}\cup S_{m_{1,3}}|>|\Gamma|$, a contradiction. ∎

Proposition 5.5**.**

The order of any element of $B$ is $2^{n_{1}}p_{2}^{n_{2}}$.

Proof.

We have $E_{m_{1}}\subseteq A$ by our assumption and $E_{m_{2}}\subseteq A$ by Proposition 5.4. Since $A\cup B$ is a separation of $\mathcal{P}(\overline{X})$, the order of any element of $B$ is of the form $2^{n_{1}}p_{2}^{n_{2}}p_{3}^{k},$ where $0\leq k\leq n_{3}-1$. Let $t\in\{0,1,\ldots,n_{3}-1\}$ be the largest integer for which $B$ has an element of order $2^{n_{1}}p_{2}^{n_{2}}p_{3}^{t}$ (and so $E_{2^{n_{1}}p_{2}^{n_{2}}p_{3}^{t}}\subseteq B$). Then the sets

[TABLE]

and the two subgroups

[TABLE]

are contained in $\Gamma$. Set $P_{1}=\underset{k=t+1}{\overset{n_{3}-1}{\bigcup}}E_{2^{n_{1}}p_{2}^{n_{2}}p_{3}^{k}}$ and $Q_{1}=S_{2^{n_{1}}p_{2}^{n_{2}-1}p_{3}^{t}}\bigcup S_{2^{n_{1}-1}p_{2}^{n_{2}}p_{3}^{t}}$. Then $|P_{1}|+|Q_{1}|=|P_{1}\cup Q_{1}|\leq|\Gamma|$. On the other hand, we have

[TABLE]

So $|P_{1}|+|Q_{1}|=2^{n_{1}-1}p_{2}^{n_{2}-1}p_{3}^{n_{3}-1}(p_{2}-1)+2^{n_{1}}p_{2}^{n_{2}-1}p_{3}^{t}.$ Then, using (9), we get

[TABLE]

If $t\geq 1$, then it would follow that $|P_{1}|+|Q_{1}|>|\Gamma|$ which is not possible. So $t=0$ and every element in $B$ is of order $2^{n_{1}}p_{2}^{n_{2}}$. ∎

Now, Proposition 5.5 together with the facts that $E_{m_{1}}$, $E_{m_{2}}$ are contained in $A$ imply the sets $E_{2^{n_{1}}p_{2}^{n_{2}}p_{3}^{k}}$ with $k\in\{1,2,\ldots,n_{3}-1\}$ and the two subgroups $S_{2^{n_{1}}p_{2}^{n_{2}-1}},S_{2^{n_{1}-1}p_{2}^{n_{2}}}$ are contained in $\Gamma$. Also $E_{n}\subseteq X$. Since $Z(3,0)$ is precisely the union of these sets, it follows that $X$ contains $Z(3,0)$. Then, by the minimality of $|X|$, we get $X=Z(3,0)$ and hence

[TABLE]

This completes the proof of Theorem 1.5.