On p-regular G-conjugacy class sizes
© Zhao et al.; licensee Springer. 2014
Received: 4 October 2013
Accepted: 7 January 2014
Published: 24 January 2014
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.
All groups considered in this paper are finite. Let G be a group, and x an element of G. We denote by the conjugacy class of G containing x and by the size of .
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 , 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 , 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 and . Notice that, when , the condition n dividing is of course true, so our aim is, by eliminating the assumption , to investigate the properties of N under the corresponding condition. More precisely, we prove the following.
for any non-central p-regular element . Furthermore, is abelian;
if d and t are two non-central p-regular elements of N such that , then and divides .
Based on this, in Section 4, we give our improvement and generalization of  by considering the case that p does not divide n.
Let π be a set of some primes; we use and for the π-component and the -component of x, respectively. Moreover, denotes a Hall π-subgroup of G, a Hall -subgroup of G, the π-part of n whenever n is a positive integer. Apart from these, we call an element x non-central if , where is the center of G. Following  a group G is said to be quasi-Frobenius if is a Frobenius group and then the inverse images of the kernel and a complement of are called the kernel and complement of G.
We first list some lemmas which are useful in the proof of our main result.
Lemma 2.1 [, Lemma 1.1]
is a p-regular G-conjugacy class of N and divides .
Proof Set in Lemma 1 of , the proof is finished. □
Let C be a p-regular G-conjugacy class of N with , then divides .
Let be two largest p-regular G-conjugacy class sizes of N with and D be a p-regular G-conjugacy class of N with . If , then .
Suppose that A is a p-regular G-conjugacy class and . Lemma 2.2(b) implies that DA is a p-regular G-conjugacy class. Also , so . If , then is a p-regular G-conjugacy class, and hence , which implies that . It follows that divides . On the other hand, , so divides . By (a), we find that divides , from which it follows that divides , a contradiction. Consequently , equivalently, divides , it follows that by the hypotheses of the lemma, as wanted. □
Then is abelian, furthermore, if , then .
By the definition of M and K, we have . Let , where D is a p-regular G-conjugacy class of N with . Applying Lemma 2.3(a), we have , which implies that , hence . It shows that , so . Notice that , we find that M is nilpotent, hence , where . Obviously, . If , let , , then . Also , we have , which implies that . Suppose that D is a generating class of M and , then divides , and divides . The fact that implies that , so by the nilpotence of M, which shows that is abelian. □
3 Proof of Theorem A
In this section we are equipped to prove the main result.
Proof of Theorem A Suppose that is not a prime power order group. We will complete the proof by the following steps:
Step 1 We may assume that for every -prime factor r of .
Otherwise, there exists a -prime factor r of such that , then . Obviously, satisfies the condition of the theorem. Application of the induction hypothesis to shows that the conclusion of the theorem holds, and hence we may assume that for every -prime factor r of .
Step 2 If the p-regular element , then either , or .
Obviously, , which implies that divides , it follows that . If , Lemma 2.3 shows that , so .
Step 3 We may assume that b is a prime power order q-element ().
Let q be a prime factor of , be the q-component and . Notice that , applying Step 2, we have , and this completes the proof by replacing b with .
Step 4 , where is a Sylow p-subgroup of , is a Sylow q-subgroup of , L is a -Hall subgroup of with . Particularly, if , then .
Let be a -element. Notice that , we find that divides , so the maximality of implies that , and hence , from which it follows that . Consequently, , where is a Sylow p-subgroup of , is a Sylow q-subgroup of , L is a -Hall subgroup of with .
Particularly, if , let be a non-central prime power order r-element, then Step 2 implies that , and hence . By the above argument, we have , where is a Sylow p-subgroup of , is a Sylow r-subgroup of , is a -Hall subgroup of with . Clearly, , so .
Step 5 .
If , Step 4 implies that . By the above , we have , which leads to dividing , a contradiction.
If , then we may assume that a is a q-element since . For every -element , we have , the hypothesis of the theorem shows that , from which it follows that . Notice that , so , which implies that , consequently a p-complement of is a prime power order group, a contradiction.
Step 6 We may assume that a is a -element.
Let , where , are the q-component and -component of a, respectively. Notice that ; we have , where M is ever defined in Lemma 2.4. If , then, by Lemma 2.4, , in contradiction to Step 5.
Step 7 .
Suppose that there exists a non-central q-element , then , so we have by the hypothesis of the theorem. It follows that , which implies that because of . So , in contradiction to Step 5. Hence, , as required.
(8.1) Our immediate object is to show that . Otherwise, then there exists a non-central p-regular element . In view of Step 4, if , then we may assume that y is a q-element. So the q-element y lies in , in contradiction to Step 7. So another possibility must prevail, which implies that . Keeping in mind that , we have by the hypotheses, from which it follows that . Applying once again, we have , a contradiction, as wanted. Consequently, . Notice that and , so , as required.
(8.2) Next, the conclusion is to be dealt with. Obviously, by Lemma 2.1 in terms of , which implies that . This leads to , and hence . Notice that , and we have , as wanted.
Step 9 It followed that , where . And hence, if , is at most a -number.
Notice that and , and by the hypothesis that n divides , we have , where . On the other hand, . By what has already been proved, we find that, if , then is at most a -number, and so is .
Step 10 for any non-central p-regular element , and therefore is abelian.
If , then by Step 4. Obviously, is abelian. In addition, for any non-central p-regular element , we find that divides , application of Lemma 2.3 yields .
If , by Step 9, we know is at most a -number. Also, is a non-central p-regular element; we may assume that if necessary by a suitable conjugate.
If , then by Lemma 2.3 since divides .
If , notice that divides , then . However, , a contradiction.
This implies that .
So the argument on the above three cases forces .
Next, we show that is abelian when . Notice that , it is enough to show that is abelian. For any non-central element , we have . Otherwise, by the above, we have . Replacing b with x, we have , which contradicts Step 8. Therefore acts on the group fixed-point-freely, which implies that is cyclic or a generalized quaternion. So, if is not abelian, then and is a generalized quaternion group. Now but , so there exists an element and such that where . So , which indicates that , and of course we have , a contradiction. Therefore is abelian, as required.
Step 11 If d and t are two non-central p-regular elements of N such that , then:
(11.2) divides , and hence divides .
If , then by Step 4, which implies that . However, y is non-central, the hypotheses of the theorem shows that , from which it follows that . Notice that , we have by the hypotheses of the theorem, but this contradicts the fact that . Therefore .
Suppose that . Our immediate object is to show that t can be assumed to be a -element. In fact, let be the q-component of t. If , Step 10 implies that . Notice that , we have by Lemma 2.3, against the fact that . It follows that , so , where is the -component of t. Thus, without loss of generality, we may assume that t is a -element.
which implies that by the maximality of m. So , which shows that by Lemma 2.3(b), a contradiction. Thus, .
Next, in a similar manner as in Step (8.1), the equality in (11.1) follows.
Clearly, acts as a group on the set through the above action. Obviously, and , this shows that the group acts on the set fixed-point-freely. Therefore divides . Notice that , so we find that divides . By Step 9, divides , which is fairly straightforward. □
Corollary 1 Suppose that G is a group. Let be the two longest sizes of the non-central conjugacy classes of G. If , 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 , then G is p-solvable. Obviously, n divides . So, in Theorem A, by taking , we have if , from which it follows that . On the other hand, for any non-central element . In fact, if , we have , in contradiction to (iii) in Theorem A. It follows that is abelian. By Theorem A, 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.
the p-regular G-conjugacy class sizes of N are exactly 1, n and m.
Let x be a non-central p-regular element of N. If , then ; if , then .
A p-complement of N is a solvable quasi-Frobenius group with abelian kernel and complement.
The conjugacy class sizes of a p-complement of N are exactly 1, , and .
Now, we show the G-conjugacy class size of the p-regular element of N is 1, n or m. Obviously, is to be dealt with. Since n is a -number, by Theorem A, we find that n divides , forcing , as wanted.
Let be a non-central p-regular element. If , then , in contradiction to (iii) in Theorem A. So . By Theorem A once again, the conclusion (ii) is obtained.
Let , then is a p-complement of N. Now, () and are abelian, we find that H is solvable. Taking into account by Theorem A, we find that . For convenience, we employ ‘_’ to work in the factor group modulo . Notice that divides and divides , so we have .
so contains a non-central p-regular element.
- (iv)Let be a non-central element. We have or m by (i). If , then
So the conjugacy class sizes of H are 1, , and . □
Corollary 2 [, Theorem A]
the p-regular conjugacy class sizes of G are exactly 1, n, and m;
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 .
Proof Obviously, n divides . So, in Theorem B, by taking , the proof of this corollary is finished. □
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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