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On pregular Gconjugacy class sizes
Journal of Inequalities and Applications volume 2014, Article number: 34 (2014)
Abstract
Let N be a normal subgroup of a psolvable group G. The purpose of this paper is to investigate some properties of N under the condition that the two longest sizes of the noncentral pregular Gconjugacy 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 pregular 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 Gconjugacy class contained in N, for example, [3, 6, 7]. Particularly, in [3], we decided the structure of N when N possesses two Gconjugate class sizes. In this paper, the case considered is that N has more than two Gconjugate class sizes.
In a recent paper [1], the authors studied the structure of G under the condition that the largest two pregular 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 psolvable group G. If $m={b}^{G}>{a}^{G}=n$ are the two longest sizes of the noncentral pregular Gconjugacy classes of N with $(m,n)=1$ and n dividing $N/Z(N)$, where $a,b\in N$, then either a pcomplement 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 noncentral pregular element $x\in {C}_{N}(b)$. Furthermore, ${C}_{N}{(b)}_{{p}^{\prime}}$ is abelian;

(iii)
if d and t are two noncentral pregular 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 πcomponent and the ${\pi}^{\prime}$component of x, respectively. Moreover, ${G}_{\pi}$ denotes a Hall πsubgroup of G, ${G}_{{\pi}^{\prime}}$ a Hall ${\pi}^{\prime}$subgroup of G, ${n}_{\pi}$ the πpart of n whenever n is a positive integer. Apart from these, we call an element x noncentral if $x\notin Z(G)$, where $Z(G)$ is the center of G. Following [8] a group G is said to be quasiFrobenius 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 psolvable normal subgroup of a group G and $B={b}^{G}$, $C={c}^{G}$ with $(B,C)=1$, where b, c are two pregular elements of N. Then:

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

(b)
$BC=CB$ is a pregular Gconjugacy class of N and $BC$ divides $BC$.
Proof Set $G=N$ in Lemma 1 of [4], the proof is finished. □
Lemma 2.3 Let N be a psolvable normal subgroup of a group G and ${B}_{0}$ be a noncentral pregular Gconjugacy class of N with the largest size. Then the following properties hold:

(a)
Let C be a pregular Gconjugacy 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 pregular Gconjugacy class sizes of N with $(m,n)=1$ and D be a pregular Gconjugacy 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 pregular Gconjugacy 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 pregular Gconjugacy class and $A=n$. Lemma 2.2(b) implies that DA is a pregular Gconjugacy class. Also $DA\ge A$, so $DA=n\text{or}m$. If $DA=n$, then ${D}^{1}DA$ is a pregular Gconjugacy 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 $AD$, it follows that $D={B}_{0}$ by the hypotheses of the lemma, as wanted. □
Lemma 2.4 Suppose that N is a psolvable normal subgroup of a group G, Let ${B}_{0}$ be a noncentral pregular Gconjugacy 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 pregular Gconjugacy 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 pregular 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 qelement ($q\ne p$).
Let q be a prime factor of $o(b)$, ${b}_{q}$ be the qcomponent 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 psubgroup of ${C}_{N}(b)$, ${Q}_{b}$ is a Sylow qsubgroup 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 psubgroup of ${C}_{N}(b)$, ${Q}_{b}$ is a Sylow qsubgroup 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 noncentral prime power order relement, 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 psubgroup of ${C}_{N}(b)$, ${R}_{y}$ is a Sylow rsubgroup 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 $qm$, 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 qsubgroup 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 qelement 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 pcomplement 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 qcomponent 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 noncentral qelement $y\in {C}_{N}(a)$, 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 noncentral pregular 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 qelement. So the qelement 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 noncentral pregular 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 noncentral pregular 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 noncentral pregular 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 $\u3008x\u3009$ acts coprimely on the abelian subgroup S, and by coprime action properties, we have
Denote by $U=[S,\u3008x\u3009]$. 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
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 psolvable 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
Moreover, since $S\u22b4G$, S is abelian, $S\cap T=1$, and $S\le {C}_{G}(w)$ we have
We denote $D={C}_{T}(w)\cap {C}_{T}(x)$. Combining (3.1) and (3.2), we have
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 $U=[S,\u3008x\u3009]$, then D normalizes U. Also,
so
Therefore
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 noncentral 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 $S/S\cap Z(G)$ fixedpointfreely, 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 noncentral pregular 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 noncentral pregular 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 noncentral, 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 qcomponent 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 qelement. 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.
(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
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 $\{{t}^{N}\}$ fixedpointfreely. 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 noncentral 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 psolvable. 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 noncentral 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 psolvable group G. If $m={b}^{G}>{a}^{G}=n$ are the two longest sizes of the noncentral pregular Gconjugacy 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 pcomplement of $N/Z(N)$ is a prime power order group or

(i)
the pregular Gconjugacy class sizes of N are exactly 1, n and m.

(ii)
Let x be a noncentral pregular 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 pcomplement of N is a solvable quasiFrobenius group with abelian kernel and complement.

(iv)
The conjugacy class sizes of a pcomplement of N are exactly 1, ${a}^{N}$, and ${{b}^{N}}_{{p}^{\prime}}$.
Proof Suppose that a pcomplement of $N/Z(N)$ is not a prime power order subgroup, and t, d are two noncentral pregular elements of N with ${t}^{G}\ne m={d}^{G}$.

(i)
Now, we show the Gconjugacy class size of the pregular 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 noncentral pregular 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 pcomplement 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 noncentral element $x\in {C}_{N}{(b)}_{{p}^{\prime}}$. Otherwise, there exists a noncentral 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
so ${C}_{N}(x)\cap {C}_{N}(y)$ contains a noncentral pregular 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 pcomplement of N is a solvable quasiFrobenius group with the abelian kernel ${M}_{{p}^{\prime}}$ and complement ${C}_{N}{(b)}_{{p}^{\prime}}$.

(iv)
Let $x\in H$ be a noncentral 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 pcomplement of N. In view of the psolvability 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,
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
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 psolvable group. Let $m>n>1$ be the two longest sizes of the noncentral pregular 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 pregular conjugacy class sizes of G are exactly 1, n, and m;

(b)
a pcomplement of G is a quasiFrobenius 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. □
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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).
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Zhao, X., Chen, R. & Geng, X. On pregular Gconjugacy class sizes. J Inequal Appl 2014, 34 (2014). https://doi.org/10.1186/1029242X201434
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Keywords
 normal subgroups
 Gconjugacy class sizes
 pregular elements