# On *p*-regular *G*-conjugacy class sizes

- Xianhe Zhao
^{1}Email author, - Ruifang Chen
^{2}and - Xu Geng
^{1}

**2014**:34

https://doi.org/10.1186/1029-242X-2014-34

© Zhao et al.; licensee Springer. 2014

**Received: **4 October 2013

**Accepted: **7 January 2014

**Published: **24 January 2014

## Abstract

Let *N* be a normal subgroup of a *p*-solvable group *G*. The purpose of this paper is to investigate some properties of *N* under the condition that the two longest sizes of the non-central *p*-regular *G*-conjugacy classes of *N* are coprime. Some known results are generalized.

**MSC:**20E45.

## Keywords

*G*-conjugacy class sizes

*p*-regular elements

## 1 Introduction

All groups considered in this paper are finite. Let *G* be a group, and *x* an element of *G*. We denote by ${x}^{G}$ the conjugacy class of *G* containing *x* and by $|{x}^{G}|$ the size of ${x}^{G}$.

The relationship between the *p*-regular conjugacy class sizes and the structure of a group *G* has been studied by many authors, see, for example, [1–5]. Let *N* be a normal subgroup of a group *G*. Clearly *N* is the union of some conjugacy classes of *G*. So, it is interesting to decide the structure of *N* by some arithmetical properties of the *G*-conjugacy class contained in *N*, for example, [3, 6, 7]. Particularly, in [3], we decided the structure of *N* when *N* possesses two *G*-conjugate class sizes. In this paper, the case considered is that *N* has more than two *G*-conjugate class sizes.

In a recent paper [1], the authors studied the structure of *G* under the condition that the largest two *p*-regular conjugacy class sizes (say, *m* and *n*) of *G* are coprime, where $m>n$ and $p\nmid n$. Notice that, when $G=N$, the condition *n* dividing $|N/Z(N)|$ is of course true, so our aim is, by eliminating the assumption $p\nmid n$, to investigate the properties of *N* under the corresponding condition. More precisely, we prove the following.

**Theorem A**

*Let*

*N*

*be a normal subgroup of a*

*p*-

*solvable group*

*G*.

*If*$m=|{b}^{G}|>|{a}^{G}|=n$

*are the two longest sizes of the non*-

*central*

*p*-

*regular*

*G*-

*conjugacy classes of*

*N*

*with*$(m,n)=1$

*and*

*n*

*dividing*$|N/Z(N)|$,

*where*$a,b\in N$,

*then either a*

*p*-

*complement of*$N/Z(N)$

*is a prime power order group or*

- (i)
${|N|}_{{p}^{\prime}}={|{a}^{N}|}_{{p}^{\prime}}{|{b}^{N}|}_{{p}^{\prime}}{|Z(N)|}_{{p}^{\prime}}$;

- (ii)
$|{x}^{G}|=m$

*for any non*-*central**p*-*regular element*$x\in {C}_{N}(b)$.*Furthermore*, ${C}_{N}{(b)}_{{p}^{\prime}}$*is abelian*; - (iii)
*if**d**and**t**are two non*-*central**p*-*regular elements of**N**such that*$|{t}^{G}|\ne m=|{d}^{G}|$,*then*${({C}_{N}(t)\cap {C}_{N}(d))}_{{p}^{\prime}}=Z{(N)}_{{p}^{\prime}}\le Z(G)$*and*${n}_{{p}^{\prime}}$*divides*$|{t}^{N}|$.

Based on this, in Section 4, we give our improvement and generalization of [1] by considering the case that *p* does not divide *n*.

Let *π* be a set of some primes; we use ${x}_{\pi}$ and ${x}_{{\pi}^{\prime}}$ for the *π*-component and the ${\pi}^{\prime}$-component of *x*, respectively. Moreover, ${G}_{\pi}$ denotes a Hall *π*-subgroup of *G*, ${G}_{{\pi}^{\prime}}$ a Hall ${\pi}^{\prime}$-subgroup of *G*, ${n}_{\pi}$ the *π*-part of *n* whenever *n* is a positive integer. Apart from these, we call an element *x* non-central if $x\notin Z(G)$, where $Z(G)$ is the center of *G*. Following [8] a group *G* is said to be quasi-Frobenius if $G/Z(G)$ is a Frobenius group and then the inverse images of the kernel and a complement of $G/Z(G)$ are called the kernel and complement of *G*.

## 2 Preliminaries

We first list some lemmas which are useful in the proof of our main result.

**Lemma 2.1** [[9], Lemma 1.1]

*Let*

*N*

*be a normal subgroup of a group*

*G*

*and*

*x*

*an element of G*.

*Then*:

- (a)
$|{x}^{N}|$

*divides*$|{x}^{G}|$; - (b)
$|{(Nx)}^{G/N}|$

*divides*$|{x}^{G}|$.

**Lemma 2.2**

*Let*

*N*

*be a*

*p*-

*solvable normal subgroup of a group*

*G*

*and*$B={b}^{G}$, $C={c}^{G}$

*with*$(|B|,|C|)=1$,

*where*

*b*,

*c*

*are two*

*p*-

*regular elements of*

*N*.

*Then*:

- (a)
$G={C}_{G}(b){C}_{G}(c)$.

- (b)
$BC=CB$

*is a**p*-*regular**G*-*conjugacy class of**N**and*$|BC|$*divides*$|B||C|$.

*Proof* Set $G=N$ in Lemma 1 of [4], the proof is finished. □

**Lemma 2.3**

*Let*

*N*

*be a*

*p*-

*solvable normal subgroup of a group*

*G*

*and*${B}_{0}$

*be a non*-

*central*

*p*-

*regular*

*G*-

*conjugacy class of*

*N*

*with the largest size*.

*Then the following properties hold*:

- (a)
*Let**C**be a**p*-*regular**G*-*conjugacy class of**N**with*$(|{B}_{0}|,|C|)=1$,*then*$|\u3008{C}^{-1}C\u3009|$*divides*$|{B}_{0}|$. - (b)
*Let*$n,m=|{B}_{0}|$*be two largest**p*-*regular**G*-*conjugacy class sizes of**N**with*$(m,n)=1$*and**D**be a**p*-*regular**G*-*conjugacy class of**N**with*$|D|>1$.*If*$(|D|,n)=1$,*then*$|D|=m$.

*Proof*(a) By Lemma 2.2(b), $C{B}_{0}$ is a

*p*-regular

*G*-conjugacy class of

*N*. Clearly, $|C{B}_{0}|\ge |{B}_{0}|$, so the hypotheses of the lemma imply that $|C{B}_{0}|=|{B}_{0}|$, from which it follows that ${C}^{-1}C{B}_{0}={B}_{0}$, and hence $\u3008{C}^{-1}C\u3009{B}_{0}={B}_{0}$, consequently $|\u3008{C}^{-1}C\u3009|$ divides $|{B}_{0}|$.

- (b)
Suppose that

*A*is a*p*-regular*G*-conjugacy class and $|A|=n$. Lemma 2.2(b) implies that*DA*is a*p*-regular*G*-conjugacy class. Also $|DA|\ge |A|$, so $|DA|=n\text{or}m$. If $|DA|=n$, then ${D}^{-1}DA$ is a*p*-regular*G*-conjugacy class, and hence ${D}^{-1}DA=A$, which implies that $\u3008{D}^{-1}D\u3009A=A$. It follows that $|\u3008{D}^{-1}D\u3009|$ divides $|A|$. On the other hand, $\u3008{D}^{-1}D\u3009\subseteq \u3008{A}^{-1}A\u3009$, so $|\u3008{D}^{-1}D\u3009|$ divides $|\u3008{A}^{-1}A\u3009|$. By (a), we find that $|\u3008{A}^{-1}A\u3009|$ divides $|{B}_{0}|$, from which it follows that $|\u3008{D}^{-1}D\u3009|$ divides $|{B}_{0}|$, a contradiction. Consequently $|DA|=m$, equivalently, $|{B}_{0}|$ divides $|A||D|$, it follows that $|D|=|{B}_{0}|$ by the hypotheses of the lemma, as wanted. □

**Lemma 2.4**

*Suppose that*

*N*

*is a*

*p*-

*solvable normal subgroup of a group*

*G*,

*Let*${B}_{0}$

*be a non*-

*central*

*p*-

*regular*

*G*-

*conjugacy class of*

*N*

*with the largest size*.

*Write*

*Then* ${M}_{{p}^{\prime}}$ *is abelian*, *furthermore*, *if* ${(Z(G)\cap N)}_{{p}^{\prime}}<{M}_{{p}^{\prime}}$, *then* $\pi ({M}_{{p}^{\prime}}/{(Z(G)\cap N)}_{{p}^{\prime}})\subseteq \pi ({B}_{0})$.

*Proof*Write

By the definition of *M* and *K*, we have $K=[M,G]$. Let $d\in D$, where *D* is a *p*-regular *G*-conjugacy class of *N* with $(|D|,|{B}_{0}|)=1$. Applying Lemma 2.3(a), we have $\pi (K)\subseteq \pi ({B}_{0})$, which implies that $(|K|,|D|)=1$, hence $|{d}^{K}|=1$. It shows that $K={C}_{K}(d)$, so $K\le Z(M)$. Notice that $M/K\le Z(G/K)$, we find that *M* is nilpotent, hence $M=P\times {M}_{{p}^{\prime}}$, where $P\in {Syl}_{p}(M)$. Obviously, ${(Z(G)\cap N)}_{{p}^{\prime}}\le {M}_{{p}^{\prime}}$. If ${(Z(G)\cap N)}_{{p}^{\prime}}<{M}_{{p}^{\prime}}$, let $r\in \pi ({M}_{{p}^{\prime}}/{(Z(G)\cap N)}_{{p}^{\prime}})$, $R\in {Syl}_{r}(M)$, then $R\u22b4G$. Also $1\ne [R,G]\le [M,G]=K$, we have $r\in \pi (K)\subseteq \pi ({B}_{0})$, which implies that $\pi ({M}_{{p}^{\prime}}/{(Z(G)\cap N)}_{{p}^{\prime}})\subseteq \pi ({B}_{0})$. Suppose that *D* is a generating class of *M* and $d\in D$, then $|{d}^{R}|$ divides $|R|$, and $|{d}^{R}|$ divides $|D|$. The fact that $(|R|,|D|)=1$ implies that $R={C}_{R}(d)$, so $R\le Z(M)$ by the nilpotence of *M*, which shows that ${M}_{{p}^{\prime}}$ is abelian. □

## 3 Proof of Theorem A

In this section we are equipped to prove the main result.

*Proof of Theorem A* Suppose that $N/Z(N)$ is not a prime power order group. We will complete the proof by the following steps:

**Step 1** *We may assume that* ${N}_{r}\nleqq Z(G)$ *for every* ${p}^{\prime}$-*prime factor* *r* *of* $|N|$.

Otherwise, there exists a ${p}^{\prime}$-prime factor *r* of $|N|$ such that ${N}_{r}\le Z(G)$, then $N={N}_{{r}^{\prime}}\times {N}_{r}$. Obviously, ${N}_{{r}^{\prime}}$ satisfies the condition of the theorem. Application of the induction hypothesis to $|N|$ shows that the conclusion of the theorem holds, and hence we may assume that ${N}_{r}\nleqq Z(G)$ for every ${p}^{\prime}$-prime factor *r* of $|N|$.

**Step 2** *If the* *p*-*regular element* $x\in Z({C}_{G}(b))\cap N$, *then either* $x\in Z(G)$, *or* ${C}_{G}(x)={C}_{G}(b)$.

Obviously, ${C}_{G}(b)\le {C}_{G}(x)$, which implies that $|{x}^{G}|$ divides $|{b}^{G}|$, it follows that $(|{x}^{G}|,n)=1$. If $x\notin Z(G)$, Lemma 2.3 shows that $|{x}^{G}|=m$, so ${C}_{G}(x)={C}_{G}(b)$.

**Step 3** *We may assume that* *b* *is a prime power order* *q*-*element* ($q\ne p$).

Let *q* be a prime factor of $o(b)$, ${b}_{q}$ be the *q*-component and ${C}_{G}({b}_{q})\ne G$. Notice that ${C}_{G}(b)={C}_{G}({b}_{q}{b}_{{q}^{\prime}})={C}_{G}({b}_{q})\cap {C}_{G}({b}_{{q}^{\prime}})\subseteq {C}_{G}({b}_{q})$, applying Step 2, we have ${C}_{G}({b}_{q})={C}_{G}(b)$, and this completes the proof by replacing *b* with ${b}_{q}$.

**Step 4** ${C}_{N}(b)={P}_{b}{Q}_{b}\times L$, *where* ${P}_{b}$ *is a Sylow* *p*-*subgroup of* ${C}_{N}(b)$, ${Q}_{b}$ *is a Sylow* *q*-*subgroup of* ${C}_{N}(b)$, *L* *is a* ${\{p,q\}}^{\prime}$-*Hall subgroup of* ${C}_{N}(b)$ *with* $L\le Z({C}_{G}(b))$. *Particularly*, *if* $L\nleqq Z(G)$, *then* ${C}_{N}{(b)}_{{p}^{\prime}}\le Z({C}_{G}(b))$.

Let $x\in {C}_{N}(b)$ be a ${\{p,q\}}^{\prime}$-element. Notice that ${C}_{G}(bx)={C}_{G}(b)\cap {C}_{G}(x)\le {C}_{G}(b)$, we find that $|{b}^{G}|$ divides $|{(bx)}^{G}|$, so the maximality of $|{b}^{G}|$ implies that $|{b}^{G}|=|{(bx)}^{G}|$, and hence ${C}_{G}(bx)={C}_{G}(b)\le {C}_{G}(x)$, from which it follows that $x\in Z({C}_{G}(b))$. Consequently, ${C}_{N}(b)={P}_{b}{Q}_{b}\times L$, where ${P}_{b}$ is a Sylow *p*-subgroup of ${C}_{N}(b)$, ${Q}_{b}$ is a Sylow *q*-subgroup of ${C}_{N}(b)$, *L* is a ${\{p,q\}}^{\prime}$-Hall subgroup of ${C}_{N}(b)$ with $L\le Z({C}_{G}(b))$.

Particularly, if $L\nleqq Z(G)$, let $y\in L$ be a non-central prime power order *r*-element, then Step 2 implies that $|{y}^{G}|=m$, and hence ${C}_{G}(y)={C}_{G}(b)$. By the above argument, we have ${C}_{N}(y)={P}_{y}{R}_{y}\times {L}_{y}$, where ${P}_{y}$ is a Sylow *p*-subgroup of ${C}_{N}(b)$, ${R}_{y}$ is a Sylow *r*-subgroup of ${C}_{N}(b)$, ${L}_{y}$ is a ${\{p,r\}}^{\prime}$-Hall subgroup of ${C}_{N}(b)$ with ${L}_{y}\le Z({C}_{G}(b))$. Clearly, ${C}_{N}(b)={P}_{b}\times L{L}_{y}$, so ${C}_{N}{(b)}_{{p}^{\prime}}=L{L}_{y}\le Z({C}_{G}(b))$.

**Step 5** $q\nmid m$.

*q*-subgroup of ${C}_{N}(b)$, which implies that $b\in {C}_{N}(a)\cap {C}_{N}(b)$, therefore $a\in {C}_{N}(b)$. We distinguish two cases according to the structure of ${C}_{N}(b)$.

- (1)
If $L\nleqq Z(G)$, Step 4 implies that ${C}_{N}{(b)}_{{p}^{\prime}}\le Z({C}_{G}(b))$. By the above $a\in {C}_{N}(b)$, we have ${C}_{G}(b)\le {C}_{G}(a)$, which leads to $|{a}^{G}|$ dividing $|{b}^{G}|$, a contradiction.

- (2)
If $L\le Z(G)$, then we may assume that

*a*is a*q*-element since $a\in {C}_{N}(b)={P}_{b}{Q}_{b}\times L$. For every ${\{p,q\}}^{\prime}$-element $x\in {C}_{N}(a)$, we have ${C}_{G}(ax)={C}_{G}(a)\cap {C}_{G}(x)\le {C}_{G}(a)$, the hypothesis of the theorem shows that ${C}_{G}(ax)={C}_{G}(a)\le {C}_{G}(x)$, from which it follows that $x\in Z({C}_{G}(a))$. Notice that $b\in {C}_{N}(a)$, so $x\in {C}_{N}(b)$, which implies that $x\in L$, consequently a*p*-complement of $N/Z(N)$ is a prime power order group, a contradiction.

**Step 6** *We may assume that* *a* *is a* ${\{p,q\}}^{\prime}$-*element*.

Let $a={a}_{q}{a}_{{q}^{\prime}}$, where ${a}_{q}$, ${a}_{{q}^{\prime}}$ are the *q*-component and ${q}^{\prime}$-component of *a*, respectively. Notice that ${C}_{G}(a)={C}_{G}({a}_{q})\cap {C}_{G}({a}_{q}^{\prime})\subseteq {C}_{G}({a}_{q})$; we have ${a}_{q}\in M$, where *M* is ever defined in Lemma 2.4. If ${a}_{q}\notin Z(G)$, then, by Lemma 2.4, $q\in \pi ({M}_{{p}^{\prime}}/{(Z(G)\cap N)}_{{p}^{\prime}})\subseteq \pi (m)$, in contradiction to Step 5.

**Step 7** ${C}_{N}{(a)}_{q}\le Z{(G)}_{q}$.

Suppose that there exists a non-central *q*-element $y\in {C}_{N}(a)$, then ${C}_{G}(ay)={C}_{G}(a)\cap {C}_{G}(y)\subseteq {C}_{G}(a)$, so we have ${C}_{G}(ay)={C}_{G}(a)$ by the hypothesis of the theorem. It follows that $ay\in M$, which implies that $y\in M$ because of $a\in M$. So $q\in \pi ({M}_{{p}^{\prime}}/{(Z(G)\cap N)}_{{p}^{\prime}})\subseteq \pi (m)$, in contradiction to Step 5. Hence, ${C}_{N}{(a)}_{q}\le Z{(G)}_{q}$, as required.

**Step 8**

(8.1) ${({C}_{N}(a)\cap {C}_{N}(b))}_{{p}^{\prime}}=Z{(N)}_{{p}^{\prime}}\le Z(G)$.

(8.2) ${|N|}_{{p}^{\prime}}={|{a}^{N}|}_{{p}^{\prime}}{|{b}^{N}|}_{{p}^{\prime}}{|Z(N)|}_{{p}^{\prime}}$.

(8.1) Our immediate object is to show that ${({C}_{N}(a)\cap {C}_{N}(b))}_{{p}^{\prime}}\le Z(G)$. Otherwise, then there exists a non-central *p*-regular element $y\in {C}_{N}(a)\cap {C}_{N}(b)$. In view of Step 4, if $L\le Z(G)$, then we may assume that *y* is a *q*-element. So the *q*-element *y* lies in ${C}_{N}(a)$, in contradiction to Step 7. So another possibility $L\nleqq Z(G)$ must prevail, which implies that $y\in {C}_{N}{(b)}_{{p}^{\prime}}$. Keeping in mind that ${C}_{N}{(b)}_{{p}^{\prime}}\le Z({C}_{G}(b))$, we have ${C}_{G}(y)={C}_{G}(b)$ by the hypotheses, from which it follows that $a\in {C}_{N}(y)={C}_{N}(b)$. Applying ${C}_{N}{(b)}_{{p}^{\prime}}\le Z({C}_{G}(b))$ once again, we have ${C}_{G}(b)\le {C}_{G}(a)$, a contradiction, as wanted. Consequently, ${({C}_{N}(a)\cap {C}_{N}(b))}_{{p}^{\prime}}\le {(Z(G)\cap N)}_{{p}^{\prime}}$. Notice that ${(Z(G)\cap N)}_{{p}^{\prime}}\le Z{(N)}_{{p}^{\prime}}$ and $Z{(N)}_{{p}^{\prime}}\le {({C}_{N}(a)\cap {C}_{N}(b))}_{{p}^{\prime}}$, so ${({C}_{N}(a)\cap {C}_{N}(b))}_{{p}^{\prime}}=Z{(N)}_{{p}^{\prime}}$, as required.

(8.2) Next, the conclusion ${|N|}_{{p}^{\prime}}={|{a}^{N}|}_{{p}^{\prime}}{|{b}^{N}|}_{{p}^{\prime}}{|Z(N)|}_{{p}^{\prime}}$ is to be dealt with. Obviously, $(|{a}^{N}|,|{b}^{N}|)=1$ by Lemma 2.1 in terms of $(|{a}^{G}|,|{b}^{G}|)=1$, which implies that $N={C}_{N}(a){C}_{N}(b)$. This leads to $|N|=|{C}_{N}(a)||{C}_{N}(b)|/|{C}_{N}(a)\cap {C}_{N}(b)|$, and hence $|N|=|{a}^{N}||{b}^{N}||{C}_{N}(a)\cap {C}_{N}(b)|$. Notice that ${({C}_{N}(a)\cap {C}_{N}(b))}_{{p}^{\prime}}=Z{(N)}_{{p}^{\prime}}$, and we have ${|N|}_{{p}^{\prime}}={|{a}^{N}|}_{{p}^{\prime}}{|{b}^{N}|}_{{p}^{\prime}}{|Z(N)|}_{{p}^{\prime}}$, as wanted.

**Step 9** *It followed that* $n=|{a}^{N}|{p}^{\alpha}$, *where* $\alpha \ge 0$. *And hence*, *if* $L\le Z(G)$, $|{a}^{G}|$ *is at most a* $\{p,q\}$-*number*.

Notice that $|N|=|{a}^{N}||{b}^{N}||{C}_{N}(a)\cap {C}_{N}(b)|$ and ${({C}_{N}(a)\cap {C}_{N}(b))}_{{p}^{\prime}}=Z{(N)}_{{p}^{\prime}}$, and by the hypothesis that *n* divides $|N/Z(N)|$, we have $n=|{a}^{N}|{p}^{\alpha}$, where $\alpha \ge 0$. On the other hand, $|{a}^{N}|=|{C}_{N}(b)|/|{C}_{N}(a)\cap {C}_{N}(b)|$. By what has already been proved, we find that, if $L\le Z(G)$, then $|{a}^{N}|$ is at most a $\{p,q\}$-number, and so is $|{a}^{G}|$.

**Step 10** $|{x}^{G}|=m$ *for any non*-*central* *p*-*regular element* $x\in {C}_{N}(b)$, *and therefore* ${C}_{N}{(b)}_{{p}^{\prime}}$ *is abelian*.

- (1)
If $L\nleqq Z(G)$, then ${C}_{N}{(b)}_{{p}^{\prime}}\le Z({C}_{G}(b))$ by Step 4. Obviously, ${C}_{N}{(b)}_{{p}^{\prime}}$ is abelian. In addition, for any non-central

*p*-regular element $x\in {C}_{N}(b)$, we find that $|{x}^{G}|$ divides $|{b}^{G}|$, application of Lemma 2.3 yields $|{x}^{G}|=m$. - (2)
If $L\le Z(G)$, by Step 9, we know $|{a}^{G}|$ is at most a $\{p,q\}$-number. Also, $x\in {C}_{N}(b)={P}_{b}{Q}_{b}\times L$ is a non-central

*p*-regular element; we may assume that $x\in {Q}_{b}\setminus Z(G)$ if necessary by a suitable conjugate.

*S*, and by coprime action properties, we have

If $|{g}^{G}|=m$, then $|{x}^{G}|=m$ by Lemma 2.3 since $|{x}^{G}|$ divides $|{g}^{G}|$.

If $|{g}^{G}|=n$, notice that $|{x}^{G}|$ divides $|{g}^{G}|$, then $x\in {M}_{q}$. However, ${M}_{q}\le {C}_{N}{(a)}_{q}\le Z{(G)}_{q}$, a contradiction.

*G*is a

*p*-solvable group, let

*T*be a Hall $\{p,q\}$-subgroup of

*G*,

*ST*is a subgroup of

*G*since

*S*is a normal subgroup of

*G*. Notice that

*S*is abelian, $S\cap T=1$, and $S\le {C}_{G}(w)$ we have

This implies that $|D|:{|{C}_{G}(a)|}_{\{p,q\}}>|S:{C}_{S}(x)|=|U|$.

*D*normalizes

*U*. Also,

a contradiction.

So the argument on the above three cases forces $|{x}^{G}|=m$.

Next, we show that ${({C}_{N}(b))}_{{p}^{\prime}}$ is abelian when $L\le Z(G)$. Notice that ${C}_{N}(b)={P}_{b}{Q}_{b}\times L$, it is enough to show that ${Q}_{b}$ is abelian. For any non-central element $x\in {Q}_{b}$, we have ${C}_{S}(x)\le Z{(G)}_{{p}^{\prime}}$. Otherwise, by the above, we have $|{x}^{G}|=m$. Replacing *b* with *x*, we have ${({C}_{N}(a)\cap {C}_{N}(b))}_{{p}^{\prime}}\nleqq Z(G)$, which contradicts Step 8. Therefore ${Q}_{b}/{Q}_{b}\cap Z(G)$ acts on the group $S/S\cap Z(G)$ fixed-point-freely, which implies that ${Q}_{b}/{Q}_{b}\cap Z(G)$ is cyclic or a generalized quaternion. So, if ${Q}_{b}$ is not abelian, then $q=2$ and ${Q}_{b}/{Q}_{b}\cap Z(G)$ is a generalized quaternion group. Now $b\in Z({Q}_{b})$ but $b\notin Z(G)$, so there exists an element $y\in {Q}_{b}$ and $y\notin Z({Q}_{b})$ such that $b={y}^{2}c$ where $c\in Z(G)\cap {Q}_{b}$. So ${C}_{G}(y)\le {C}_{G}(b)$, which indicates that $y\in Z({C}_{G}(b))$, and of course we have $y\in Z({Q}_{b})$, a contradiction. Therefore ${Q}_{b}$ is abelian, as required.

**Step 11** *If* *d* *and* *t* *are two non*-*central* *p*-*regular elements of* *N* *such that* $|{t}^{G}|\ne m=|{d}^{G}|$, *then*:

(11.1) ${({C}_{N}(t)\cap {C}_{N}(d))}_{{p}^{\prime}}=Z{(N)}_{{p}^{\prime}}\le Z(G)$.

(11.2) ${|{a}^{N}|}_{{p}^{\prime}}$ *divides* $|{t}^{N}|$, *and hence* ${n}_{{p}^{\prime}}$ *divides* $|{t}^{N}|$.

*p*-regular element $y\in {C}_{N}(t)\cap {C}_{N}(d)$, and we will distinguish two cases as for the structure of ${C}_{N}(d)$.

- (1)
If $L\nleqq Z(G)$, then ${C}_{N}{(d)}_{{p}^{\prime}}\le Z({C}_{G}(d))$ by Step 4, which implies that $y\in Z({C}_{G}(d))$. However,

*y*is non-central, the hypotheses of the theorem shows that ${C}_{G}(y)={C}_{G}(d)$, from which it follows that $t\in {C}_{N}{(d)}_{{p}^{\prime}}$. Notice that ${C}_{N}{(d)}_{{p}^{\prime}}\le Z({C}_{G}(d))$, we have ${C}_{G}(t)={C}_{G}(d)$ by the hypotheses of the theorem, but this contradicts the fact that $|{t}^{G}|\ne m=|{d}^{G}|$. Therefore ${({C}_{N}(t)\cap {C}_{N}(d))}_{{p}^{\prime}}\le Z(G)$. - (2)
Suppose that $L\le Z(G)$. Our immediate object is to show that

*t*can be assumed to be a ${q}^{\prime}$-element. In fact, let ${t}_{q}$ be the*q*-component of*t*. If ${t}_{q}\notin Z(G)$, Step 10 implies that $|{{t}_{q}}^{G}|=m$. Notice that ${C}_{G}(t)\le {C}_{G}({t}_{q})$, we have $|{t}^{G}|=m$ by Lemma 2.3, against the fact that $|{t}^{G}|\ne m=|{d}^{G}|$. It follows that ${t}_{q}\in Z(G)$, so ${C}_{G}(t)={C}_{G}({t}_{{q}^{\prime}})$, where ${t}_{{q}^{\prime}}$ is the ${q}^{\prime}$-component of*t*. Thus, without loss of generality, we may assume that*t*is a ${q}^{\prime}$-element.

*y*is a

*q*-element. Keeping in mind that $q\nmid m$, application of Step 10 once again, we have $|{y}^{G}|=m$. Moreover,

which implies that $|{(ty)}^{G}|=m$ by the maximality of *m*. So ${C}_{G}(ty)={C}_{G}(y)\le {C}_{G}(t)$, which shows that $|{t}^{G}|=m$ by Lemma 2.3(b), a contradiction. Thus, ${({C}_{N}(d)\cap {C}_{N}(t))}_{{p}^{\prime}}\le Z(G)$.

Next, in a similar manner as in Step (8.1), the equality in (11.1) follows.

Clearly, ${({C}_{N}(b)/Z(N))}_{{p}^{\prime}}$ acts as a group on the set $\{{t}^{N}\}$ through the above action. Obviously, ${t}^{N}\cap {C}_{N}(b)=\mathrm{\varnothing}$ and ${({C}_{N}(t)\cap {C}_{N}(b))}_{{p}^{\prime}}=Z{(N)}_{{p}^{\prime}}$, this shows that the group ${({C}_{N}(b)/Z(N))}_{{p}^{\prime}}$ acts on the set $\{{t}^{N}\}$ fixed-point-freely. Therefore ${|{C}_{N}(b)/Z(N)|}_{{p}^{\prime}}$ divides $|{t}^{N}|$. Notice that ${|{C}_{N}(b)/Z(N)|}_{{p}^{\prime}}={|{a}^{N}|}_{{p}^{\prime}}$, so we find that ${|{a}^{N}|}_{{p}^{\prime}}$ divides $|{t}^{N}|$. By Step 9, ${n}_{{p}^{\prime}}$ divides $|{t}^{N}|$, which is fairly straightforward. □

**Corollary 1** *Suppose that* *G* *is a group*. *Let* $|{b}^{G}|=m>n=|{a}^{G}|$ *be the two longest sizes of the non*-*central conjugacy classes of* *G*. *If* $(m,n)=1$, *then* *G* *is solvable and the conjugacy class size of the element in* *G* *is exactly* 1, *n* *or* *m*.

*Proof* Let *p* be a prime and *p* be not a prime factor of $|G|$, then *G* is *p*-solvable. Obviously, *n* divides $|G/Z(G)|$. So, in Theorem A, by taking $N=G$, we have $|{x}^{G}|={n}_{{p}^{\prime}}=n$ if $|{x}^{G}|\ne m$, from which it follows that $x\in {M}_{{p}^{\prime}}$. On the other hand, $|{y}^{G}|=n$ for any non-central element $y\in {C}_{G}(a)$. In fact, if $|{y}^{G}|=m$, we have $y\in {C}_{G}(a)\cap {C}_{G}(y)$, in contradiction to (iii) in Theorem A. It follows that ${C}_{G}(a)={M}_{{p}^{\prime}}$ is abelian. By Theorem A, ${C}_{G}(b)$ is abelian, and the solvability of *G* is obtained. □

## 4 Application of Theorem A

Based on Theorem A, we consider the case that *p* does not divide *n*.

**Theorem B**

*Let*

*N*

*be a normal subgroup of a*

*p*-

*solvable group*

*G*.

*If*$m=|{b}^{G}|>|{a}^{G}|=n$

*are the two longest sizes of the non*-

*central*

*p*-

*regular*

*G*-

*conjugacy classes of*

*N*

*with*$(m,n)=1$

*and*

*n*

*dividing*$|N/Z(N)|$,

*where*$a,b\in N$

*and*$p\nmid n$,

*then either a*

*p*-

*complement of*$N/Z(N)$

*is a prime power order group or*

- (i)
*the**p*-*regular**G*-*conjugacy class sizes of**N**are exactly*1,*n**and**m*. - (ii)
*Let**x**be a non*-*central**p*-*regular element of**N*.*If*$x\in {C}_{N}(a)$,*then*$|{x}^{G}|=n$;*if*$x\in {C}_{N}(b)$,*then*$|{x}^{G}|=m$. - (iii)
*A**p*-*complement of**N**is a solvable quasi*-*Frobenius group with abelian kernel and complement*. - (iv)
*The conjugacy class sizes of a**p*-*complement of**N**are exactly*1, $|{a}^{N}|$,*and*${|{b}^{N}|}_{{p}^{\prime}}$.

*Proof*Suppose that a

*p*-complement of $N/Z(N)$ is not a prime power order subgroup, and

*t*,

*d*are two non-central

*p*-regular elements of

*N*with $|{t}^{G}|\ne m=|{d}^{G}|$.

- (i)
Now, we show the

*G*-conjugacy class size of the*p*-regular element of*N*is 1,*n*or*m*. Obviously, $|{t}^{G}|=n$ is to be dealt with. Since*n*is a ${p}^{\prime}$-number, by Theorem A, we find that*n*divides $|{t}^{G}|$, forcing $|{t}^{G}|=n$, as wanted. - (ii)
Let $x\in {C}_{N}(a)$ be a non-central

*p*-regular element. If $|{x}^{G}|=m$, then $x\in {({C}_{N}(a)\cap {C}_{N}(x))}_{{p}^{\prime}}$, in contradiction to (iii) in Theorem A. So $|{x}^{G}|=n$. By Theorem A once again, the conclusion (ii) is obtained. - (iii)
Let ${C}_{N}{(a)}_{{p}^{\prime}}{C}_{N}{(b)}_{{p}^{\prime}}=H$, then $H={M}_{{p}^{\prime}}{C}_{N}{(b)}_{{p}^{\prime}}$ is a

*p*-complement of*N*. Now, ${C}_{N}{(a)}_{{p}^{\prime}}$ ($={M}_{{p}^{\prime}}$) and ${C}_{N}{(b)}_{{p}^{\prime}}$ are abelian, we find that*H*is solvable. Taking into account ${({C}_{N}(a)\cap {C}_{N}(b))}_{{p}^{\prime}}=Z{(N)}_{{p}^{\prime}}$ by Theorem A, we find that $Z(H)=Z{(N)}_{{p}^{\prime}}$. For convenience, we employ ‘_’ to work in the factor group modulo $Z(H)$. Notice that $|\overline{{M}_{{p}^{\prime}}}|=|\overline{{C}_{N}{(a)}_{{p}^{\prime}}}|$ divides $|{b}^{N}|$ and $|\overline{{C}_{N}{(b)}_{{p}^{\prime}}}|$ divides $|{a}^{N}|$, so we have $(|\overline{{C}_{N}{(b)}_{{p}^{\prime}}}|,|\overline{{M}_{{p}^{\prime}}}|)=1$.

so ${C}_{N}(x)\cap {C}_{N}(y)$ contains a non-central *p*-regular element.

*p*-complement of

*N*is a solvable quasi-Frobenius group with the abelian kernel ${M}_{{p}^{\prime}}$ and complement ${C}_{N}{(b)}_{{p}^{\prime}}$.

- (iv)Let $x\in H$ be a non-central element. We have $|{x}^{G}|=n$ or
*m*by (i). If $|{x}^{G}|=n$, then$\left|{x}^{H}\right|=|H:{C}_{H}(x)|=|{C}_{N}{(b)}_{{p}^{\prime}}/Z{(N)}_{{p}^{\prime}}|=|{C}_{N}(b)/{C}_{N}(a)\cap {C}_{N}(b)|=\left|{a}^{N}\right|.$

*p*-complement of

*N*. In view of the

*p*-solvability of

*N*, replacing

*H*if necessary by a suitable conjugate, we may assume that ${C}_{N}{(x)}_{{p}^{\prime}}\le H$. Notice that ${C}_{N}{(x)}_{{p}^{\prime}}\cap {M}_{{p}^{\prime}}\le {({C}_{N}(x)\cap {C}_{N}(a))}_{{p}^{\prime}}=Z{(N)}_{{p}^{\prime}}$, we have $|{C}_{N}{(x)}_{{p}^{\prime}}|=|{C}_{N}{(b)}_{{p}^{\prime}}|$. Also,

*H*, we find that $\overline{{C}_{N}{(x)}_{{p}^{\prime}}}$ and $\overline{{C}_{N}{(b)}_{{p}^{\prime}}}$ are conjugate in $\overline{H}$. Consequently, we may assume that there exists an element $g\in N$ such that ${x}^{g}\in {C}_{N}(b)$, so

So the conjugacy class sizes of *H* are 1, $|{a}^{N}|$, and ${|{b}^{N}|}_{{p}^{\prime}}$. □

**Corollary 2** [[1], Theorem A]

*Suppose that*

*G*

*is a*

*p*-

*solvable group*.

*Let*$m>n>1$

*be the two longest sizes of the non*-

*central*

*p*-

*regular conjugacy classes of*

*G*.

*Suppose that*$(m,n)=1$

*and*

*p*

*is not a prime divisor of*

*n*.

*Then*

*G*

*is solvable and*

- (a)
*the**p*-*regular conjugacy class sizes of**G**are exactly*1,*n*,*and**m*; - (b)
*a**p*-*complement of**G**is a quasi*-*Frobenius group with abelian kernel and complement*.*Furthermore*,*its conjugacy class sizes are exactly*1,*n*,*and*${m}_{{p}^{\prime}}$.

*Proof* Obviously, *n* divides $|G/Z(G)|$. So, in Theorem B, by taking $N=G$, the proof of this corollary is finished. □

## Declarations

### Acknowledgements

We wish to thank the referees for careful reading and valuable comments for the origin draft. This work is supported by the National Natural Science Foundation of China (10771172, 11271301), NSFC (U1204101) and Major project of Henan Education Department (13B110085).

## Authors’ Affiliations

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