On p-regular G-conjugacy class sizes

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.

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 $|〈C^{-1}C〉|$ 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 $〈C^{-1}C〉B_0=B_0$, consequently $|〈C^{-1}C〉|$ 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 $〈D^{-1}D〉A=A$. It follows that $|〈D^{-1}D〉|$ divides $|A|$. On the other hand, $〈D^{-1}D〉\subseteq 〈A^{-1}A〉$, so $|〈D^{-1}D〉|$ divides $|〈A^{-1}A〉|$. By (a), we find that $|〈A^{-1}A〉|$ divides $|B_0|$, from which it follows that $|〈D^{-1}D〉|$ 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

$$M=〈D|D\mathit{\text{ is a }}p\mathit{\text{-regular }}G\mathit{\text{-conjugacy class of }}N\mathit{\text{ with }}(|D|,|B_0|)=1〉.$$

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

$$K=〈D^{-1}D|D\text{ is a }p\text{-regular }G\text{-conjugacy class of }N\text{ with }(|D|,|B_0|)=1〉.$$

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 $R⊴G$. 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 $q\nmid m$.

If $q|m$, then $q\nmid n$, and of course we have $q\nmid |a^N|$ by Lemma 2.1. Notice that $|a^N|=|N:C_N(a)|=|C_N(b):C_N(a)\cap C_N(b)|$, hence $C_N(a)\cap C_N(b)$ contains a Sylow 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.

By the structure of $C_N(b)$, we distinguish two cases:

• (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.

On the other hand, we know $M_{p^{\prime }}$ is abelian and $Z{(G)}_q\cap N\le M_{p^{\prime }}\le C_N(a)$, by Step 7, we can write $M_{p^{\prime }}=S\times (Z{(G)}_q\cap N)$ where $q\nmid |S|$. Notice that $〈x〉$ acts coprimely on the abelian subgroup S, and by coprime action properties, we have

$$S=[S,〈x〉]\times C_S(x).$$

Denote by $U=[S,〈x〉]$. As $a\in S$, we can write $a=uw$ with $u\in U$, $w\in C_S(x)$. Consider the element $g=wx$; we have

$$C_G(g)=C_G(w)\cap C_G(x)\le C_G(x).$$

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.

So we are left with only one alternative: $|g^G|<n$. Keeping in mind that 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

$$\begin{array}{rcl}|ST:C_{ST}(g)|& \le & \left|g^G\right|<n=\left|a^G\right|\\ =& |G:C_G(a)|\\ =& |G:C_G(a){|}_{\{p,q\}}\\ =& |T|:|C_G(a){|}_{\{p,q\}}.\end{array}$$

(3.1)

Moreover, since $S⊴G$, S is abelian, $S\cap T=1$, and $S\le C_G(w)$ we have

$$\begin{array}{rcl}{C}_{ST}(g)& =& C_{ST}(w)\cap C_{ST}(x)\\ =& S{C}_T(w)\cap C_S(x)C_T(x)\\ =& C_S(x)[S{C}_T(w)\cap C_T(x)]\\ =& C_S(x)[C_T(w)\cap C_T(x)].\end{array}$$

(3.2)

We denote $D=C_T(w)\cap C_T(x)$. Combining (3.1) and (3.2), we have

$$\frac{|T|}{{|C_G(a)|}_{\{p,q\}}}>\frac{|S||T|}{|C_S(x)||D|}.$$

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

On the other hand, as $D\le C_G(x)$ and $U=[S,〈x〉]$, then D normalizes U. Also,

$$C_D(u)=C_G(u)\cap D=C_T(u)\cap C_T(w)\cap C_T(x)\le C_T(a)\cap C_T(x),$$

so

$$|C_D(u)|\le |C_T(a)\cap C_T(x)|\le |C_G(a){|}_{\{p,q\}}.$$

Therefore

$$\left|u^D\right|=|D:C_D(u)|=|D|:|C_D(u)|\ge |D|:|C_G(a){|}_{\{p,q\}}>|U|,$$

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 $|t^N|$.

(11.1) Firstly, it will be established that ${(C_N(t)\cap C_N(d))}_{p^{\prime }}\le Z(G)$. Otherwise, there exists a non-central 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.

Also, in this case, we may assume that y is a q-element. Keeping in mind that $q\nmid m$, application of Step 10 once again, we have $|y^G|=m$. Moreover,

$$C_G(ty)=C_G(t)\cap C_G(y)\subseteq C_G(y),$$

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.

(11.2) Consider the quotient group ${(C_N(b)/Z(N))}_{p^{\prime }}$ and the set $\{t^N\}$. For any $\overline{x}\in {(C_N(b)/Z(N))}_{p^{\prime }}$ and $y\in \{t^N\}$, without loss of generality, we may assume that $\overline{x}=xZ(N)$ where $x\in C_N(b)$, and we set

$$y^{\overline{x}}=y^x.$$

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 $(|\overline{C_N{(b)}_{p^{\prime }}}|,|\overline{M_{p^{\prime }}}|)=1$.

Now, $\overline{M_{p^{\prime }}}$ is an abelian normal subgroup of $\overline{H}$. To prove that $\overline{H}$ is a Frobenius group, we are left to show $C_{\overline{M_{p^{\prime }}}}(\overline{x})=1$ for any non-central element $x\in C_N{(b)}_{p^{\prime }}$. Otherwise, there exists a non-central element $y\in M_{p^{\prime }}$ such that $1\ne \overline{y}\in C_{\overline{M_{p^{\prime }}}}(\overline{x})$, then ${(\overline{xy})}^{o(\overline{x})}={(\overline{x}\overline{y})}^{o(\overline{x})}={\overline{y}}^{o(\overline{x})}\ne 1$ since $(o(\overline{x}),o(\overline{y}))=1$. Hence

$${\overline{xy}}^{o(\overline{x})}={\overline{y}}^{o(\overline{x})}\in \overline{C_N(x)}\cap \overline{C_N(xy)}=\overline{C_N(x)\cap C_N(xy)}=\overline{C_N(x)\cap C_N(y)},$$

so $C_N(x)\cap C_N(y)$ contains a non-central p-regular element.

Notice that $|x^G|=m$ and $|y^G|=n$; by Theorem A, we have ${(C_N(x)\cap C_N(y))}_{p^{\prime }}\le Z(G)$, in contradiction to the previous paragraph. So a 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

  $$\left|x^H\right|=|H:C_H(x)|=|C_N{(b)}_{p^{\prime }}/Z{(N)}_{p^{\prime }}|=|C_N(b)/C_N(a)\cap C_N(b)|=\left|a^N\right|.$$

If $|x^G|=m$, then $C_N{(x)}_{p^{\prime }}{M}_{p^{\prime }}$ is also a 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,

$$C_N{(b)}_{p^{\prime }}/Z{(N)}_{p^{\prime }}=C_N{(b)}_{p^{\prime }}/Z(H)=\overline{C_N{(b)}_{p^{\prime }}},$$

keeping in mind that $(|\overline{C_N{(b)}_{p^{\prime }}}|,|\overline{M_{p^{\prime }}}|)=1$, by the solvability of 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

$$\begin{array}{rcl}\left|x^H\right|& =& \left|{x^g}^H\right|=|H:C_H\left(x^g\right)|=|C_N{(a)}_{p^{\prime }}/Z{(N)}_{p^{\prime }}|\\ =& |C_N(a)/C_N(a)\cap C_N(b){|}_{p^{\prime }}\\ =& |b^N{|}_{p^{\prime }}.\end{array}$$

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. □