On a Finite Group with Restriction on Set of Conjugacy Classes Size

Abstract

The greatest power of a prime p dividing the natural number n will be denoted by $n_p$. For a set of primes $\pi$ and a natural number n we will denote $n_{\pi }=\prod _{p\in \pi }n_p$. Let G be a finite group with trivial center, and $p,q>5$ be distinct prime divisors of |G|. We prove that if for every nonunity conjugacy classes size $\alpha$, it is true that $\alpha _{\{p,q\}}\in \{p^n,q^m,p^nq^m\}$, where n and m depend only on p and q, then $|G|_{\{p,q\}}=p^nq^m$, and $C_G(g)\cap C_G(h)=1$ for every p-element g and every q-element h.

1 Introduction

There has been considerable work over the years into the relation between the structure of a finite group and the cardinality of its conjugacy classes. One of the first results in this direction, due to Burnside, is that if a finite group has a conjugacy class with prime-power cardinality, the group is not simple. Many papers, for example [1] and [2], studied the structure of groups in which the orders of some elements, and their conjugacy classes are known. This paper studies the structure of groups with restrictions on the set of conjugacy classes size.

In this paper, all groups are finite. The number of elements of a set $\pi$ is denoted by $|\pi |$. Denote the set of prime divisors of positive integer n by $\pi (n)$, and the set $\pi (|G|)$ for a group G by $\pi (G)$.

Let G be a group and take $a\in G$. We denote by $a^G$ the conjugacy class of G containing a, put $N(G)=\{|x^G|, x \in G\} {\setminus }\{1\}$. Denote by $|G||_p$ the number $p^n$ such that N(G) contains $\alpha$ multiple of $p^n$ and avoids the multiple of $p^{n+1}$. For $\pi \subseteq \pi (G)$ put $|G||_{\pi }=\prod _{p\in \pi }|G||_p$. For brevity, write |G|| to mean $|G||_{\pi (G)}$. Observe that $|G||_p$ divides $|G|_p$ for each $p\in \pi (G)$. However, $|G||_p$ can be less than $|G|_p$.

Definition

Let p and q be distinct numbers. Say that a group G satisfies the condition $\{p,q\}^*$ and write $G\in \{p, q\}^*$ if $\alpha _{\{p, q\}}\in \{|G||_p, |G||_q, |G||_{\{p, q\}}\}$ for every $\alpha \in N(G)$.

Camina [3] proved that a group G with $\{p,q\}^*$-property is nilpotent if $N(G)=\{1,p^n,q^m,p^nq^m\}$. Beltran and Felipe [4] extended Camina’s theorem in the following way: Let G be a finite soluble group whose conjugacy class sizes are $\{1,n,m,nm \}$, where n and m are coprime positive integers; then G is nilpotent and the integers n and m are prime-power numbers; in particular $G\in \{n,m\}^*$. Kong and Guo [5] investigated groups such that the set of conjugacy class sizes of biprimary elements is precisely ${1,p^{\alpha },m,p^{\alpha }m}$, where $p^{\alpha }$ is a prime power, $(p,m)=1$, and there is a p-element whose conjugacy class size is $p^{\alpha }$. They proved that in this case group is nilpotent and $m=q^{\beta }$ for some prime number $q\ne p$.

In the general case, a group with the $\{p,q\}^*$-property does not be nilpotent. For example, let $G\simeq L_n(k)$. Then $G\in \{p,q\}^*$ where p is a primitive prime divisor of $k^n-1$ and q is a primitive prime divisor of $k^{n-1}-1$.

In this paper, we inspect the groups with $\{p,q\}^*$-properties and with trivial center.

Theorem

If $G\in \{p,q\}^*$ is a group with trivial center, where $p,q\in \pi (G)$ and $p>q>5$, then $|G|_{\{p,q\}}=|G||_{\{p, q\}}$.

Corollary

In the hypotheses of the theorem, $C_G(g)\cap C_G(h)=1$ for every p-element g and every q-element h.

2 Definitions and Preliminary Results

Lemma 1

([6, Lemma 1.4]) For a finite group G, take $K\unlhd G$ and put ${\overline{G}}= G/K$. Take $x\in G$ and ${\overline{x}}=xK\in G/K$. The following claims hold:

• (i) $|x^K|$ and $|{\overline{x}}^{{\overline{G}}}|$ divide $|x^G|$.

• (ii) For neighboring members L and M of a composition series of G, with $L<M$, take $x\in M$ and the image ${\widetilde{x}}=xL$ of x. Then $|{\widetilde{x}}^S|$ divides $|x^G|$, where $S=M/L$.

• (iii) If $y\in G$ with $xy=yx$ and $(|x|,|y|)=1$ then $C_G(xy)=C_G(x)\cap C_G(y)$.

• (iv) If $(|x|, |K|) = 1$, then $C_{{\overline{G}}}({\overline{x}}) = C_G(x)K/K$.

• (v) $\overline{C_G(x)}\le C_{{\overline{G}}}({\overline{x}})$.

Lemma 2

[7, Theorem V.8.7] Let G be a Frobenius group with kernel A and complement B. Then the following statements are true:

• (i) |B| divides $|A|-1$;

• (ii) A is nilpotent, and if the order of B is even, then A is abelian;

• (iii) the Sylow p-subgroups of the group B are cyclic for odd p and cyclic or general quaternion groups for $p=2$.

• (iv) Every subgroup of B of order pq, where p and q are primes, is cyclic.

Lemma 3

[3, Theorem 2] Let G be a finite group whose conjugacy classes have either 1, $p^a$, $q^b$ or $p^a q^b$ elements, where p and q are primes and a and b are integers. If all of the values actually occur, then G is nilpotent.

Lemma 4

Take $g\in G$. If each conjugacy class of G contains an element h such that $g\in C_G(h)$, then $g\in Z(G)$.

Proof

The assertion follows from the fact that a finite group generated by any set of representatives of conjugacy classes. $\square$

The prime graph GK(G) of G is defined as follows. The vertex set is $\pi (G)$ and two distinct primes $r, s\in \pi (G)$ considered as vertices of the graph are adjacent if and only if there exists element $g\in G$ such that $|g|=rs$. Denote by $\pi _i(G)$ the set of vertices of the ith prime graph component of G. If G has even order, then we always assume that $2\in \pi _1(G)$.

Lemma 5

[8, Theorem A] If a finite group G has disconnected prime graph, then one of the following conditions holds:

• (a) $s(G)=2$ and G is a Frobenius or 2-Frobenius group;

• (b) there is a nonabelian simple group S such that $S\le G = G/F(G)\le Aut(S)$, where F(G) is the maximal normal nilpotent subgroup of G; moreover, F(G) and G / S are $\pi _1(G)$-subgroups, $s(S)\ge s(G)$, and for every i with $2\le i\le s(G)$ there is j with $2\le j\le s(S)$ such that $\pi _i(G)=\pi _j (S)$.

Lemma 6

[1, Theorem 1] Let G be a finite group, and let p and q be different primes. Then some Sylow p-subgroup of G commutes with some Sylow q-subgroup of G if and only if the class sizes of the q-elements of G are not divisible by p and the class sizes of the p-elements of G are not divisible by q.

Lemma 7

[2] Let G be a finite group and p a prime, $p\not \in \{3, 5\}$. Then G has abelian Sylow p-subgroups if and only if $|x^G|_p=1$ for all p-elements x of G.

Given a set $\pi$ of primes, a finite group is said to have property $D_{\pi }$ whenever it includes a Hall $\pi$-subgroup and all its Hall $\pi$-subgroups are conjugate. For brevity, we write $G\in D_{\pi }$ when a group G has property $D_{\pi }$.

Lemma 8

[9] Let G be a group and $\pi$ be a set of prime numbers. If G has a nilpotent Hall $\pi$-subgroup then $G\in D_{\pi }$.

Lemma 9

[10, Corollary 6.7] Suppose that G is a group and $\pi$ is a set of primes. Then $G\in D_{\pi }$ if and only if all composition factor of G have property $D_{\pi }$.

Lemma 10

[11, Theorem 3.3.2] Let G be a $p'$-group of automorphisms of an abelian p-group V and suppose $V_1$ is a G-invariant direct factor of V. Then $V=V_1\times V_2$, where $V_2$ is also G-invariant.

Lemma 11

[11, Theorem 5.2.3] Let A be a $\pi (G)'$-group of automorphisms of an abelian group G. Then $G=C_G(A)\times [G,A]$.

Lemma 12

Let $S\le G \le Aut(S)$, where S is a nonabelian simple group such that GK(G) has two connected components, $|\pi _2(G)|=1$ and for some $5<q\in \pi _1(G)$ the Sylow q-subgroups is abelian. Then there exists $g\in G$ such that $\pi (|g|)\subseteq \pi _1(G)$ and $|g^G|_q>1$.

Proof

Suppose that S is an alternating group. It is easy to show that G contains a 2-element g such that $\pi (|C_G(g)|)\subseteq \{2,3\}$. Consequently, in this case the lemma is hold. Using [8, 12, 13], we can infer that the lemma is true if S is isomorphic to one of the sporadic simple groups. Let S be a group of Lie type over a field of characteristic t. It follows from [8] and [12] that $t\in \pi _1(G)$. [14, Proposition 5.1.2] shows that S contains a regular unipotent element h, in particular, $\pi (|C_S(h)|)=\{t\}$. If $t=q$ then $|C_G(x)|_t<|G|_t$ for every $q'$-element x. Since $|\pi _1(G)|>1$, we can take $g\in G$ with $\pi (|g|)\subseteq \pi _1(G){\setminus }\{t\}$. Otherwise $g=h$. $\square$

Lemma 13

Let $G\in \{p,q\}^*$, N be a normal subgroup of G such that $|N|_{\{p,q\}}=|G|_{\{p,q\}}$. Then $N\in \{p,q\}^*$.

Proof

Since N includes Sylow p- and q-subgroups of G, we get $Syl_p(C_G(g))=Syl_p(C_N(g))$, $Syl_q(C_G(g))=Syl_q(C_N(g))$. Therefor, $Ind_G(g)_{\{p,q\}}=Ind_N(g)_{\{p,q\}}$.

$\square$

3 Proof of the Theorem and the Corollary

Assume that the theorem is false and that $|G|_p>|G||_p$. Then the centralizer of each element of G contains an element of order p. Let $P\in Syl_p(G)$, $Q\in Syl_q(G)$.

Lemma 14

$|G|_q>|G||_q$.

Proof

If $a\in Z(Q)$ then $|a^G|_q=1$ and hence $|a^G|_p=|G||_p$. Let $g\in C_G(a)$ be a p-element. If $|g^G|_p=1$ then $|g^G|_q=|G||_q$ and, consequently, the lemma is proved. We have $|g^G|_p=|G||_p$. Consequently $|(ag)^G|_p=|g^G|_p$. From Lemma 1 we get $C_G(ag)=C_G(a)\cap C_G(g)$ and hence $C_G(a)$ contains some Sylow p-subgroup of $C_G(g)$. Therefore, $C_G(a)$ includes $Z({\overline{P}})$ for some ${\overline{P}}\in Syl_p(G)$. Thus $C_G(a)$ contains a p-element $g'$ such that $|g'^G| _p = 1$ and the lemma is proved. $\square$

Lemma 14 shows that $|C_G(a)|_q>1$ for every $a\in G$. In particular, the centralizer of every element contains an element of order q.

Lemma 15

$\pi (G)\ne \{p,q\}$.

Proof

Assume that G is a $\{p, q\}$-group. We have $N(G)\subseteq \{|G||_p, |G||_q, |G||\}$. Suppose that $N(G)=\{|G||_p\}$ or $N(G)=\{|G||_p, |G||\}$. Take $g\in Z(P)$. We have $|g^G|_p=1$. Consequently, $|g^G|=1$ and $g\in Z(G)$, a contradiction. Assume that $N(G)=\{|G||_p, |G||_q\}$. Take $g\in Z(P)$. Lemma 14 shows that $C_G(g)$ contains a q-element h. We have $C_G(hg)=C_G(h)\cap C_G(g)$; in particular, $|(gh)^G|$ is a multiple of both $|g^G|$ and $|h^G|$. Therefore, $C_G(h)=C_G(g)$. Now $C_G(h)$ contains an element $h'$ such that $|h'^G|_q=1$. Thus, $|(gh')^G|_{\{p,q\}}=|G||_p|G||_q$; a contradiction. We have $N(G)=\{|G||_p, |G||_q, |G||\}$. It follows from Lemma 3 that G is nilpotent. Hence, $Z(G)>1$; a contradiction.$\square$

Lemma 16

For every $g\in P$ we have $|g^G|_{q}=|G||_q$.

Proof

Assume that there exists $g\in P$ such that $|g^G|_q=1$. Let us show that in the centralizer of each element there is a subgroup conjugate to Z(P). Take $x\in G$ such that $C_G(x)$ does not include the center of any Sylow p-subgroup of G.

Assume that $x=ab=ba$, where a is a $p'$-element and b is a nontrivial p-element. Suppose that $Z(P)<C_G(a)$. If $|a^G|_p=1$ then $b^{C_G (a)}$ contains an element c with $Z(P)<C_G(ac)$. Hence, $|a^G|_p=|G||_p$ and $|a^G|_p=|(ab)^G|_p$. Therefore, x is contained in some Sylow p-subgroup of $C_G(a)$, and hence $C_G(x)$ includes the center of a Sylow p-subgroup. Thus, we may assume that x is a $p'$-element.

Suppose that x is a $\{p,q\}'$-element. If $|x^G|_p= 1$ then clearly there exists ${\overline{P}}\in Syl_p(G)$ with $Z({\overline{P}})<C_G(x)$, a contradiction. Hence, $|x^G|_p=|G||_p$. Take $h\in C_G(x)$ a p-element. If $|h^G|_p=|G||_p$ then $|(xh)^G|_p=|h^G|_p$. Hence, $C_G(x)$ includes a Sylow p-subgroup S of $C_G(h)$. The group S contains the center of some Sylow p-subgroup of G, a contradiction. Therefore, $|h^G|_p=1$ for every p-element h of $C_G(x)$. As we noted above, the centralizer of each element contains an element of order q. Take $y\in C_G (x)$ a q-element. Assume that $|y^G|_p=|G||_p$. Then $|(xy)^G|_p=|y^G|_p$. Hence, $C_G(x)$ includes some Sylow p-subgroup of $C_G(y)$. Since $C_G(g)$ includes a Sylow q-subgroup of G, we may assume that $g \in C_G(y)$. Hence in $C_G(x)$ there is a p-element $g'$ with $|g'^G|_p = |G||_p$; a contradiction. Thus, we may assume that $|y^G|_p=1$ for every q-element y of $C_G(x)$. We have $|(xy)^G|_q=|y^G|_q=|G||_q$. Hence, $C_G(x)$ includes some Sylow q-subgroup of $C_G(y)$. Every q-subgroup of $C_G(y)$ contains an element z of the center of some Sylow q-subgroup of G. Therefore, $C_G(x)$ contains $z'$ such that $|z'^G|_q=1$, a contradiction.

Therefore, $x=ab=ba$ where a is a $\{p,q\}'$-element and b is a q-element. If $|a^G|_p=|G||_p$ or $|b^G|_p = |G||_p$ then the preceding arguments imply claim. Thus, $|a^G|_p=|b^G|_p=1$. Hence $|a^G|_q=|G||_q$. Since $C_G(a)$ includes some Sylow p-subgroup of G, we may assume that $g\in C_G(a)$. Since $|(ag)^G|_q=|a^G|_q$, it follows that $C_G(ag)$ includes some Sylow q-subgroup of $C_G(a)$. Hence $C_G(a)$ contains h with $x\in C_G(g^h)$. We have $|(xg^h)^G|_p=|g^G|_p$ and hence $Z(P)<C_G(x^{h^{- 1}})$, a contradiction.

We get that centralizer of every element of G includes center of some Sylow p-subgroup. From Lemma 4, it follows that Z(G) includes a center of Sylow p-subgroup. In particular, Z(G) is nontrivial, a contradiction.$\square$

Lemma 17

For every $g\in Q$ we have $|g^G|_{p}=|G||_p$.

Proof

Similar to the proof of Lemma 16. $\square$

Lemma 18

We have $C_G(Q)=Z(Q)$ and $C_G(P)=Z(P)$.

Proof

Assume that there is a $\{p, q\}'$-element $a\in C_G(P)$. Take $c\in C_G(a)$ a q-element. Assume that $|c^G|_q=|G||_q$. We have $|(ac)^G|_q=|c^G|_q$ and, consequently, $C_G (a)$ contains a q-element d centralizing some Sylow q-subgroup of G. Therefore, $C_G(a)$ contains a q-element b such that $|b^G|_q=1$. To infer that $b\in Z(G)$, it suffices to show that the centralizer of each element contains an element conjugate to b.

Assume that there exists $g\in G$ such that $C_G(g)\cap b^G=\varnothing$. Let $g=xyz=xzy=zxy=zyx\in G$, where x is a q-element, y is a p-element and z is a $\{p, q \}'$-element.

Assume that $g=x$. Since b lies in the center of a Sylow q-subgroup of G, $C_G(g)$ contains an element conjugate to b; a contradiction.

Assume that $g=xy$ with nontrivial y. Since $C_G(a)$ includes P we may assume that $y\in C_G(a)$. Lemma 16 yields $|a^G|_q=|(ay)^G|_q=|y^G|_q$. Consequently, the Sylow q-subgroup of $|C_G(y)|$ is conjugate to a Sylow q-subgroup of $C_G(a)$. Since $|g^G|_q=|y^G|_q$, the Sylow q-subgroup of $C_G(g)$ is conjugate to a Sylow q-subgroup of $C_G(y)$. Hence, $C_G(g)$ contains an element conjugate to b, a contradiction.

Since the centralizer of any element of G contains both p- and q-elements, we may assume that x and y are not trivial and $g=xyz$. Note that $|g^G|_q\ge |z^G|_q$. Therefore, $C_G(g)$ contains some Sylow q-subgroup of $C_G(xy)$; in particular, $C_G(g)\cap b^G\ne \varnothing$. Therefore, Z(G) is nontrivial; a contradiction with the assumption of the theorem. Thus, $C_G(P)=Z(P)$. Similarly we show that $C_G(Q)=Z(Q)$. $\square$

Proposition 1

For every $g\in P$ we have $|g^G|_p=1$. For every $h\in Q$ we have $|h^G|_q=1$.

Proof. To prove the proposition, we need the following lemmas.

Lemma 19

Assume that there is a p-element $g\in P$ with $|g^G|_p=|G||_p$. If $H\in Syl_p (C_G (g))$, then $H<Z(C_G (g))$.

Proof

Since the centralizer of each element contains an element of order q, we obtain $\pi (C_G(g))\ne \{p\}$. Take a $p'$-element $x\in C_G (g)$. Then $|(gx)^G|_p=|g^G|_p$; therefore, $C_G(x)$ include some Sylow p-subgroup H of $C_G(g)$. It follows from Lemma 18, that $|x^G|_p=|G||_p$. Let $b\in H$. Then $|x^G|_p=|(xb)^G|_p$. Hence, $|b^{C_G (x)}|_p=1$. It follows from Lemma 7 that H is abelian. Thus, for each $h\in C_G(g)$, there exists $h'\in h^G$ such that $H<C_G(h')$. Lemma 4 implies that $H\le Z(C_G(g))$. $\square$

Lemma 20

Suppose that g is the same as in Lemma 19 and $D\in Syl_q(C_G (g))$. The centralizer of each element of G includes a subgroup conjugate to D.

Proof

Take $h\in P$. If $h\in Z(P)$ then Lemma 19 yields $h\in Z(C_G (g))$ and hence $D\in C_G(h)$. We may assume that $|h^G|_p=|G||_p$. Lemma 19 implies that $C_G(h)\le C_G(z)$, where $z \in Z(P)$. Since $|z^G|_q=|h^G|_q$ and $D^a<C_G(z)$ for some $a\in G$, it follows that $C_G(h)$ includes a subgroup conjugate to D. Thus, the centralizer of each p-element includes a subgroup conjugate to D. Take $x\in G$. Since the centralizer of each element contains a p-element, we may assume that $x=ab=ba$ where a is a $p'$-element and b is a nontrivial p-element. Lemma 16 yields $|x^G|_q=|b^G|_q$. Hence, $C_G(x)$ includes a Sylow q-subgroup of $C_G(b)$. Since a Sylow q-subgroup of $C_G(b)$ is conjugate to D, it follows that $C_G(x)$ includes a subgroup conjugate to D. $\square$

Lemmas 20 and 4 shows that $D\le Z(G)$. Since D is nontrivial, it follows that $Z(G)>1$, a contradiction. Thus, $|g^G|_p=1$ for every $g\in P$. For $h\in Q$ the proof is similar. $\square$

Note that from Lemma 7, it follows that the statement of Proposition 1 is equivalent to the fact that the Sylow p- and q-subgroups of G are abelian.

Proposition 2

$C_G(P)$ or $C_G(Q)$ contains a $\{p,q\}'$-element.

Proof

Suppose that $C_G(P)\le P$ and $C_G(Q)\le Q$. Lemma 15 shows that $\{p,q\}\ne \pi (G)$. Let $x\in G$ be a $\{p, q\}'$-element. Since $|x^G|_{\{p, q\}}=|G||_{\{p,q\}}$, we see that $|x^G|_{\{p,q\}}=|(xa)^G|_{\{p,q\}}$ for each $\{p,q\}$-element $a\in C_G(x)$. Hence, $|a^{C_G(x)}|_{\{p,q\}}=1$. From Lemmas 6 and 7, a Hall $\{p,q\}$-subgroup of $C_G(x)$ exists and is abelian. By Lemma 8 all Hall $\{p,q\}$-subgroups of $C_G(x)$ are conjugate in $C_G(x)$. Take ${\overline{P}}\in Syl_p(C_G (x))$ and ${\overline{Q}}\in Syl_q(C_G(x))$ with ${\overline{Q}}\le C_G({\overline{P}})$. Since P and Q are arbitrary we may assume that ${\overline{P}}<P$ and ${\overline{Q}}<Q$.

Lemma 21

Let $C=C_G({\overline{P}})$. Then C is solvable, $C/{\overline{P}}$ is a Frobenius group or a double-Frobenius group, p is a connected component of the graph $GK(C/{\overline{P}})$. In particular, one of the groups ${\overline{Q}}$ and $P/{\overline{P}}$ is cyclic.

Proof

Since P is an abelian group and $|P|/|{\overline{P}}|=|G||_p$, it follows that $p\in \pi (C/{\overline{P}})$. Note that p is a connected component of the prime graph of the group ${\overline{C}}=C/{\overline{P}}$. Otherwise, there exists a $p'$-element $y\in G$ such that $|C_G(y)|_p>|{\overline{P}}|$ and hence $|y^G|_p=1$, a contradiction with $C_G(P)\le P$.

Since $x\in C$, it follows that $\pi (C)\ne \{p, q\}$. Let $z\in C$ be a $\{p,q\}'$-element. The Hall $\{p,q\}$-subgroup of $C_G(z)$ is abelian. Lemma 8 implies that all Hall $\{p,q\}$-subgroups of $C_G(z)$ are conjugate. Hence, $C_G(z)$ includes a Hall $\{p,q\}$-subgroup X which is containing ${\overline{P}}$. Since X is abelian, it follows that $X<C$. Thus, each $\{p,q\}'$-element of C centralizes some Sylow q-subgroup of C. In particular, $GK({\overline{C}})$ has two connected components.

Assume that ${\overline{C}}$ is not solvable. Lemma 5 shows ${\overline{C}}\simeq K.S.A$, where K is a nilpotent $\pi _1({\overline{C}})$-subgroup, S is simple and $A\le Out(S)$ is a $\pi _1({\overline{C}})$-subgroup. Since $|\pi _1({\overline{C}})|>1$, we have $\{p\}=\pi _2({\overline{G}})$. Hence, p does not divide |K||A|. Assume that $q\in \pi (K)$. Since K is nilpotent, the Sylow q-subgroup R of K is unique and every Sylow q-subgroup of ${\overline{C}}$ includes R. Since the centralizer of each $p'$-element of C contains some Sylow q-subgroup of C, it follows that all $p'$-elements of ${\overline{C}}$ centralize R. Since S.A is generated by all its $p'$-elements, it follows that $R\le Z({\overline{C}})$, a contradiction with the property that $GK({\overline{C}})$ is disconnected.

Therefore, $|{\overline{Q}}|$ divides |S.A|. From Lemma 12, if follows that S.A contains a $p'$-element b such that $|b^{S.A}|_q>1$, a contradiction with the property that the centralizer of each $p'$-element of ${\overline{C}}$ includes a Sylow q-subgroup of ${\overline{C}}$. Thus, ${\overline{C}}$ is a solvable Frobenius group or a double-Frobenius group. By Lemma 2, the Sylow p-subgroup or q-subgroup of ${\overline{C}}$ is cyclic. $\square$

Lemma 22

Take a p-element h such that $C_G(h)$ contains a $\{p,q\}'$-element. Then there exists a subgroup $P_h$ of order $|G|_p/|G||_p$ such that $C_G(h)=C_G(P_h)$, in particular $P_h=Z(C_G (h))$.

Proof

By definition, $C_G(h)$ contains a $\{p,q\}'$-element x. Take $P_h\in Syl_p(C_G(x))$ with $h\in P_h$. We said above that the Hall $\{p,q\}$-subgroup of $C_G (x)$ is abelian. Obviously, $C_G(h)\ge C_G(P_h)$, and we verify the inverse inclusion. Take $Q_h\in Syl_q(C_G(x))$ with $Q_h\le C_G (P_h)$. Since $|(hy)^{G}|_q=|h^{G}|_q$ for each $\{p,q\}'$-element $y\in C_G (h)$, it follows that there exists $a\in y^{C_G(h)}$ with $Q_h<C_{C_G (h)}(a)$. We have $|(ab)^{C_G(h)}|_p=|b^{C_G(h)}|_p=|a^{C_G (h)}|_p$, where $b\in Q_h$. Hence, $C_{C_G (h)}(a)$ includes some Sylow p-subgroup of $C_{C_G (h)}(b)$. Therefore, $a^{C_G (h)}$ contains $a'$ such that $P_h Q_h<C_G(a')$. Hence, every $p'$-element of $C_G(h)$ centralizes some Sylow p-subgroup of $C_G(x)$. Since the Sylow p-subgroup of G are abelian, every p-element of $C_G(h)$ centralizes some subgroup conjugate to $P_h$ in $C_G(h)$. If $ba=ab=c\in C_G(h)$, $\pi (|a|)\subseteq \{p\}'$ and $\pi (b)=\{p\}$, then $|c^{C_G (h)}|_p=|a^{C_G (h)}|_p$. Hence, $C_G(c)$ includes a subgroup conjugate to $P_h$ in $C_G(h)$.

Hence, each conjugacy class of $C_G(h)$ contains an element of $C_G(P_h)$. Lemma 4 implies that $C_G(h)\le C_G(P_h)$. Thus, $P_h<Z(C_G (h))$.

Since $|C_G(y)|_p=|P_h|$ for every $p'$-element $y\in G$, it follows that $Z(C_G (h))=P_h$. $\square$

Lemma 23

If h is a q-element such that $C_G(h)$ contains a $\{p,q\}'$-element, then there exists a subgroup ${\widehat{Q}}_h$ of order $|G|_q/|G||_q$ with $C_G(h)=C_G({\widehat{Q}}_h)$; in particular, ${\widehat{Q}}_h=Z(C_G (h))$. Moreover $C_G(h)/{\widehat{Q}}_h$ is a Frobenius group or double-Frobenius group.

Proof

Similar to the proof of Lemma 22. $\square$

Lemma 24

Take a p-element $g\in G$ with $\pi (C_G(g))=\{p,q\}$. For each q-element $h\in C_G(g)$ we have $\pi (C_G(h))=\{p,q\}$. For each $a\in \langle g\rangle$ we have $\pi (C_G(a))=\{p,q\}$. Take a q-element $g'\in G$ with $\pi (C_G(g'))=\{p,q\}$. For each p-element $h'\in C_G(g')$ we have $\pi (C_G(h'))=\{p,q\}$. For each $b\in \langle g'\rangle$ we have $\pi (C_G(b))=\{p,q\}$.

Proof

Assume that there exists a $\{p,q\}'$-element $x\in C_G(h)$. Lemma 17 yields $|h^G|_p=|G||_p$. We have $|(xh)^G|_p=|h^G|_p$. Therefore, $x^G$ contains some $y\in C_G(g)$, a contradiction. The proof of the assertion about $g'$ is similar. We have $\langle g\rangle <C_G(h)$. Since $\pi (C_G(h))=\{p,q\}$, we obtain $\pi (C_G(a))=\{p,q\}$ for all $a\in \langle g\rangle$. $\square$

Let h be a p-element such that $\pi (C_G(h))\ne \{p,q\}$. Put $P_h=Z(C_G (h))$, $Q_h\in Syl_q(C_G(h))$. Lemma 22 shows $|P_h|=|G|_p/|G||_p$. For all $a\in P_h$ we have $P_h=P_a$. Therefore, $P_h$ and $P_v$ either coincide or intersect trivially, where v is the p-element whose centralizer contains the $\{p,q\}'$-element. Let $x \in C_G(h)$ be a $\{p,q\}'$-element. Therefore, $P_h$ is a Sylow p-subgroup of $C_G(x)$. From Lemma 21 infer that $C_G(h)/P_h$ is a solvable Frobenius group or double-Frobenius group. We have p is a connected component of the graph $GK(C_G(h)/P_h)$. The centralizer of each $p'$-element of $C_G(h)$ includes a group conjugate to $Q_h$. In particular, by Lemma 23 there exists ${\widehat{Q}}_{g'}$ for each $g'\in Q_h$ with ${\widehat{Q}}_{g'}=Z(C_G(g'))$. Since $Q_h<C_G(x)$, we conclude that ${\widehat{Q}}_{g'}=Q_h$. Since $P_h<C_G(Q_h)$, we find that $P_h\in Syl_p(C_G(g'))$.

Lemma 25

The group G contains a nonabelian composition factor whose order is a multiple of p or q.

Proof

Suppose that the orders of nonabelian composition factors of G are not divisible by p and q. Consider ${\widetilde{G}}=G/O_{\{p, q\}'}(G)$ and the natural homomorphism $\widetilde{\ }:G\rightarrow {\widetilde{G}}$. Suppose that ${\widetilde{P}}\lhd {\widetilde{G}}$. Let $x\in G$ be a p-element such that $C_G(x)$ contains a $\{p, q\}'$-element. Since $\widetilde{Q_x}<C_G(\widetilde{P_x})$, it follows that ${\widetilde{P}}.\widetilde{Q_x}/\widetilde{P_x}$ is a Frobenius group with the kernel ${\widetilde{P}}/\widetilde{P_x}$ and the complement $\widetilde{Q_x}\widetilde{P_x}/\widetilde{P_x}$. Lemma 2 shows that $Q_x$ is a cyclic group. Let $H\in Hall_{\{p,q\}}(C_{{\widetilde{G}}}(\widetilde{Q_x}))$. Note that $H/\widetilde{Q_x}$ is a Frobenius group and hence $Q/Q_x$ is a cyclic group. Take $b\in Q$ with $\langle bQ_x\rangle =Q/Q_x$ and assume that $b^z\in Q_x$ for some $z<|b|$. Lemma 24 implies that $C_G(b)$ contains a $\{p,q\}'$-element. Therefore, there exists ${\widehat{Q}}_b$. We have that for each $y\in Q_x$ there exists ${\widehat{Q}}_y$ and ${\widehat{Q}}_y=Q_x$. Therefore, $Q_x={\widehat{Q}}_b$, a contradiction. Thus, $Q=\langle a\rangle \times \langle b\rangle$, where a satisfies $\langle a\rangle =Q_x$.

Assume that $O_{\{p,q\}'}(G)$ is nontrivial. The centralizer of each element of $O_{\{p,q\}'}(G)$ contain a p-element. We may assume that $C=C_{O_{\{p,q\}'}(G)}(x)>1$. It follows from Lemma 22 that $C=C_{O_{\{p,q\}'}(G)}(P_x)$. The Sylow p-subgroup of $C_G(P_x)/P_x$ acts freely on $C/P_x$. Therefore, $P/P_x$ is cyclic. Hence, $P\simeq P_x.\langle {\overline{y}}\rangle$. Take the preimage $y\in P$ of ${\overline{y}}$. Assume that $y^m \in P_x$, where $m<|y|$. If $\pi (C_G(y))=\{p,q\}$ then we arrive at a contradiction with Lemma 24. Therefore, there exist $P_y$ and $P_y\ne P_y\cap P_x\ne 1$, a contradiction. Thus, $P=P_x\times \langle y \rangle$. We may assume that $y\in C_G(b)$. Thus, $C_{{\widetilde{G}}}({\widetilde{ab}})$ trivially intersects with ${\widetilde{P}}$. Since ${\widetilde{P}}$ is a normal subgroup of ${\widetilde{G}}$, it follows that $|(ab)^G|_p=|P|$; a contradiction. Thus, $O_{\{p,q\}'}(G)$ is trivial. Let ${\overline{G}}=G/P$ and the natural homomorphism $\overline{\ }:G\rightarrow {\overline{G}}$. From Lemma 13 follows that we can assume that ${\overline{G}}=O_{q'}({\overline{G}}).{\overline{Q}}$.

Observe that $C_G(Q_x)$ contains a $\{p,q\}'$-element. Lemma 23 implies that $C_G(Q_x)$ is solvable and q is a connected component of $GK(C_G(Q_x)/Q_x)$. Take ${\overline{H}}\in Hall_{p'}(C_G(Q_x)/Q_x)$. Since $Q=Q_x\times \langle b\rangle$, by the Schurr–Zassenhaus theorem $C_G(Q_x)$ contains a group H isomorphic to ${\overline{H}}$. The number q is the connected component of the graph GK(H). It follows from Lemma 5 that H is a Frobenius group or a double-Frobenius group. Take $N\in Hall_{q'}(H)$. Since $P_x$ is the unique Sylow p-subgroup of $C_G(Q_x)$ and each $q'$-element of $C_G(Q_x)$ centralizes some Sylow p-subgroup of $C_G(Q_x)$, we have $N\in C_G(P_x)$. Hence, N acts freely on $P/P_x$. Consequently, Sylow subgroups of N are cyclic or quaternion. Therefore, H is a Frobenius group. Since $C_{C_G(Q_x)}(P_x)\unlhd C_G(Q_x)$, we get $N\unlhd C_G (Q_x)$. Hence, N is the Frobenius kernel of H.

Therefore, $P.H/P_x$ is a double-Frobenius group. Let $R<G$ be the minimal preimage of $P.H/P_x$. Since N is cyclic, we infer that $R\cap P=[P,N]\times C_{R\cap P}(N)$. It follows from $C_{R\cap P}(N)<P_x$ that $C_{R\cap P}(N)$ is normal in R, and consequently $C_{R\cap P}(N)$ is trivial. Since $P_x<C_P(N)$, we obtain $R\cap P_x=1$. Therefore, $R\simeq P.H/P_x$ and $P=[P,N]\times P_x$.

Let $l\in P{\setminus } P_x$ be such that $C_G(l)\cap Q\in Syl_q(C_G(l))$. Since $P=[P, N]\times P_x$, we can uniquely represent l as vg, where $v\in [P,N]$ and $g\in P_x$. Assume that g is nontrivial. Take $w\in C_G(l)\cap Q$. Since $Q=Q_x\times \langle b\rangle$, we see that $w=cd$ with $c\in Q_x$ and $d\in \langle b\rangle$. We have $l=l^w=(vg)^{cd}=v^{cd}g^d$. The group $\langle b \rangle$ acts freely on $P_x$, and [P, N] is a normal subgroup of P.Q. Consequently, $d=1$. The $Q_x$ acts freely on [P, N]. Therefore, $c=1$; which is a contradiction. Consequently, if g is a p-element with $C_G(g)\cap Q \in Syl_q(C_G(g))$ then $g\in P_x\cup [P,N]$. In particular, for every p-element h, it follows that $h^G\cap P_x\ne \varnothing$ or $h^G\cap [P,N]\ne \varnothing$.

Assume that there exists $y\in Q$ such that $\pi (C_G(y))=\{p,q\}$. We have ${\overline{y}}$ acts freely on $O_{q'}({\overline{G}})$. Therefore, $Z(O_{q'}({\overline{G}}))>1$. Let $z\in Z(O_{q'}({\overline{G}}))>1$ such that $C_{{\overline{G}}}(z)\cap {\overline{Q}}\in Syl_q(C_{{\overline{G}}}(z))$. We can assume that $C_{{\overline{G}}}(z)\cap {\overline{Q}}=\overline{Q_x}$. Consequently $P_x$ is a Sylow p-subgroup of $C_G(z)$. Since $P_x\lhd P.O_{q'}({\overline{G}})$ and $G=P.O_{q'}({\overline{G}}).Q$, we obtain $P_x\lhd G$. Let $h\in C_P(b){\setminus } \{1\}$, $g\in P_x{\setminus } \{1\}$. We have $h\in [P,N]$. Since $P_x\lhd G$ we infer that $(hg)^G\cap (P_x\cup [P,N])=\varnothing$. Consequently $Ind_G(hg)_q=|Q|$, a contradiction.

Let $T=Soc({\overline{G}})$. Since $C_T(Q)=1$ we get that T is abelian. We can think that $C_T(a)>1$. Let $h\in C_T(a)$. We have $P_x<C_O(h)$. Consequently $P_x\lhd P.T.Q$. Therefore, [T, a] acts on $P/P_x$. If $[T,a]>1$ then [T, a] acts trivial on $P/P_x$. Therefore, Q contain y such that $[T,a]<C_{G}(y)$. Since $C_P(ya)$ is a subgroup of $P_x$ or [P, N] we get that $C_P(ay)=1$; a contradiction. We have $a\in T$, $T\le N$. Therefore, T is cyclic. Since $C_{{\overline{G}}}(a)\lhd {\overline{G}}$ and G include no normal subgroup which include Q we get $G=P.T.Q$. Therefore, $\pi (C_G(b))=\{p,q\}$, a contradiction. Thus, ${\widetilde{P}}$ is not a normal subgroup of ${\widetilde{G}}$.

Similarly, we can show that ${\widetilde{Q}}$ is not a normal subgroup of ${\widetilde{G}}$.

Put $F=Fit({\widetilde{G}})$ is a Fitting subgroup of ${\widetilde{G}}$. The definition of G and ${\widetilde{G}}$ implies that F is a $\{p,q\}$-group. Since p and q do not divide the orders of nonabelian composition factors, F is nontrivial. Assume that |F| is not divisible by q. Then F is a p-group. Since the Sylow p-subgroup of G is abelian, and any nonabelian composition factor of G is not divisible by p, we get ${\widetilde{P}}=F$, a contradiction.

Therefore, $p,q\in \pi (F)$. Since Q and P are abelian, we obtain that F contains a Hall $\{p,q\}$-subgroup of $C_G(g)$ for all $g\in F$. Consequently, $F={\widehat{P}}\times {\widehat{Q}}$, where ${\widehat{P}}<P$ with $|{\widehat{P}}|=|P|/|G||_p$ and ${\widehat{Q}}<Q$ with $|{\widehat{Q}}|=|Q|/|G||_q$. The ${\widetilde{P}}/{\widehat{P}}$ acts freely on ${\widehat{Q}}$. Hence, ${\widetilde{P}}/{\widehat{P}}$ is cyclic. Take $h\in {\widetilde{P}}$ with $\langle h{\widetilde{P}}\rangle ={\widetilde{P}}/{\widehat{P}}$ and a q-element $s\in C_{{\widetilde{G}}}(h)$. Since $|C_{{\widetilde{G}}}(s)|_p=|P|/|G||_p$, we obtain $|h|\le |P|/|G||_p$. Since $s\not \in {\widehat{Q}}$, we see that s acts freely on ${\widehat{P}}$. Hence, $|{\widetilde{G}}/{\widehat{P}}|_p\ge |P|/|G||_p$. Consequently, $|{\widetilde{G}}/{\widehat{P}}|_p=|P|/|G||_p$ and $\langle h\rangle$ intersect trivially with ${\widehat{P}}$. Similarly we can show that $|{\widetilde{G}}/{\widehat{Q}}|_q=|Q|/|G||_q$ and ${\widetilde{G}}$ contains an element f, where $|f|=|Q|/|G||_q$ such that f acts freely on ${\widetilde{P}}$. Thus, $|\langle {\widehat{P}},f\rangle |=|\langle {\widehat{Q}},h\rangle |=|G||_{\{p,q\}}$. We find that $\langle {\widehat{Q}},h\rangle$ is a Frobenius group with the kernel ${\widehat{Q}}$, $\langle {\widehat{P}},f\rangle$ is a Frobenius group with the kernel ${\widehat{P}}$. This contradicts the fact that the order of the kernel of the Frobenius group is greater than the order of the complement. Thus, G contains a nonabelian composition factor whose order is divisible by p or q.$\square$

Lemma 25 shows that G contains a nonabelian composition factor whose order is divisible by p or q. Assume that there exists a nonabelian composition factor of G whose order is divisible by p. Put ${\overline{G}}=G/O_{p'}(G)$. Assume that $O_{p'}$ is not solvable. Then it is easy to show that the centralizer of any p-element is not solvable, a contradiction. Put $C=Soc({\overline{G}})$. By definition, it follows that |C| is a multiple of p. Assume that C is not solvable. Then C is a simple group; otherwise, the centralizer of some p-element is not solvable. We have ${\overline{G}}/C\le Out(C)$. If ${\overline{G}}/C$ is divisible by p then ${\overline{G}}$ contains an p-element g that acts on C as an outer automorphism. Since p is greater than 5, we infer that g is a field or diagonal automorphism, and C is a group of Lie type. Therefore, $C_{{\overline{G}}}(g)$ is not solvable, a contradiction. Thus, $|C|_p=|G|_p$. Since N(C) contains a number $\alpha$ with $\alpha _p=|C|_p$, we arrive at a contradiction.

Now C is an elementary abelian p-group. Lemma 25 shows that ${\overline{G}}$ contains an nonabelian composition factor ${\widetilde{S}}$ whose order is a multiple of p. Let $S<{\overline{G}}$ be the minimal preimage of ${\widetilde{S}}$. Since ${\widetilde{S}}$ is a simple group, S is generated by the set of p-elements. Since the Sylow p-subgroup of ${\overline{G}}$ is abelian and every Sylow p-subgroup of ${\overline{G}}$ includes C, we get $S<C_{{\overline{G}}}(C)$; this contradicts the property that the centralizer of each p-element of ${\overline{G}}$ is solvable.$\square$

By Proposition 2, there exists a $\{p,q\}'$-element $a\in C_G (P)$ in contradiction with Lemma 18. Thus, $|G|_{p, q}=|G||_{p, q}$. The proof of our theorem is complete.

Assume that G contains a p-element x and a q-element y such that $C_G(x)\cap C_G(y)>1$. If $a\in C_G(x)\cap C_G(y)$ then$|a^G|_{\{p,q\}}\le |G|_{\{p,q\}}/(pq)$ and hence $|a^G|_{\{p,q\}}=1$, which is a contradiction. This justifies the corollary.