A note on the structure of a finite group G having a subgroup H maximal in 〈H, Hg〉

Yong Xu; Xianhua Li; Guiyun Chen

doi:10.1515/math-2019-0043

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A note on the structure of a finite group G having a subgroup H maximal in 〈H, H^g〉

Yong Xu , Xianhua Li and Guiyun Chen

Published/Copyright: June 4, 2019

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$Open Mathematics$

From the journal Open Mathematics Volume 17 Issue 1

Abstract

Let G be a finite group and H ≤ G. The authors study the structure of finite groups G having a subgroup H which is maximal in 〈H, H^g〉 for some g ∈ G. Some results on the structure of 〈H, H^g〉 and G are set up. Especially, a generalization of Baer′s theorem is established.

Keywords: finite group; chief factor; solvable; supersolvable

MSC 2010: 20D10; 20D15

1 Introduction

All groups considered are finite. Let π(G) be the set of all prime divisors of the order of a group G, S(G) the largest normal solvable subgroup of G. Let 𝓕 denote a formation, 𝓤 the formation of supersolvable groups. We use H ⋖ G to denote that H is a maximal subgroup of G. H Char G means that H is a characteristic subgroup of G. The other notations and terminologies are standard (Ref. to [1]).

Let H ≤ G and g ∈ G, then H ≤ 〈H, H^g 〉 ≤ 〈H, g〉. It is clear that H = 〈H, H^g〉 for all g ∈ G if and only if H ⊴ G. On the other hand, 〈H, H^g〉 = 〈H, g〉 for all g ∈ G if and only if H abn G ([2, p.247]). In [3], the famous Wielandt′s theorem shows that H ⊲ ⊲ 〈H, H^g〉 for all g ∈ G if and only if H ⊲ ⊲ G. In [4], H is called pronormal in G if H is conjugate to H^g in 〈H, H^g〉 for all g ∈ G. From these results we can see that the normalities of a subgroup H in G may be reflected from the normalities of H in 〈H, H^g〉; the closer 〈H, H^g 〉 is to 〈H, g 〉, the more H is non-normal; the closer 〈H, H^g 〉 is to H, the more H is normal. Hence 〈H, H^g〉 seems to be a “measurement” of normalities of H in G. These properties motivate us to investigate the properties of G from the size of H in 〈H, H^g〉. In [5], we investigated the structure of G by the index of H in 〈H, H^g〉. In this paper, we continue this research. We study the structure of a finite group G with a subgroup H which is maximal in 〈H, H^g〉 for some g ∈ G. We get a series of interesting results on 〈H, H^g〉 and G.

2 Preliminaries

In this section we include some lemmas from other sources that will be used in the following sections for the convenience of the reader.

Lemma 2.1

([6, Theorem 3.6]) Let G be a finite group. Then G is solvable if and only if for every cyclic subgroup H of G, 〈H, H^g〉 is solvable for all g ∈ G.

Lemma 2.2

([7, Theorem 4.3.4]) Let G be a finite group. Then G is solvable if and only if for every Sylow subgroup P of G, 〈P, P^g〉 is solvable for all g ∈ G.

Lemma 2.3

([8, Theorem 3.1]) Let 𝓕 be a saturated formation containing 𝓤 and G a group with a solvable normal subgroup N such that G/N ∈ 𝓕. Assume that every Sylow subgroup of F(N) is cyclic, then G ∈ 𝓕.

Lemma 2.4

(Wielandt) Let H be a nilpotent Hall subgroup of G and |π(H)| ≥ 2. Suppose that for every prime p dividing |H|, N_G(H_p) = H, where H_p ∈ Syl_p(H). Then there exists K ⊲ G such that G = HK and H ∩ K = 1.

Lemma 2.5

([9]) A finite group is solvable if it has a nilpotent maximal subgroup of odd order.

Lemma 2.6

([10]) Let G be a finite insolvable group with a nilpotent maximal subgroup. Let L = F(G). Then G/L has a unique minimal normal subgroup K/L, K/L is a direct product of copies of a simple group with dihedral Sylow 2-subgroups, and G/K is a 2-group.

Lemma 2.7

([11, Theorem 2]) If G is a simple group with dihedral Sylow 2-subgroups, then either G is isomorphic to PSL(2, q), q odd and q ≥ 5, or G is isomorphic to A₇.

Lemma 2.8

([12, Theorem 8.2.3]) Let A be a π^′-group that acts on the π-group G, p be a prime divisor of |G| and A or G solvable. Then

There exists an A-invariant Sylow p-subgroup of G.
The A-invariant Sylow p-subgroups of G are conjugate under C_G (A).

3 Main results

Theorem 3.1

If a minimal subgroup A of a finite group G is self-normalizing (or A < 〈A, A^g〉 for some g ∈ G), then A is a Sylow p-subgroup of G for a prime p and A^G = G.

Proof

Note that if the order of A is a prime p, then A is contained in a Sylow p-subgroup P and if A < P, then A is strictly contained in its normaliser, so A = P = N_G(P). It is well known that N_G(P) is abnormal, so g ∈ 〈A, A^g〉. In particular, G ≤ A^G.

Corollary 3.2

Let G be a non-trivial finite group. Suppose that every minimal subgroup A of G is a proper subgroup of 〈A, A^g〉 for all g ∈ G ∖ A, then G is a cyclic group of prime order.

Proof

Suppose that G is not a simple group, then there exists a normal subgroup N such that 1 < N < G. Let L be a minimal subgroup of N, of course, it is also a minimal subgroup of G. By the hypothesis and Theorem 3.1, we get G = L^G ≤ N < G, a contradiction. Thus G is a simple group. Since non-abelian simple groups cannot have all Sylow subgroups of prime order, we have this corollary.

Theorem 3.3

Let H be a solvable subgroup of G and T = 〈H, H^g〉. If H ⋖ T for some g ∈ G ∖ N_G(H), then T = LH, where L/Φ(L) is isomorphic to a chief factor of T and Φ(L) ≤ H.

Proof

If T = S(T)H, then T is solvable and S(T) = T. Let K/N be a complemented chief factor of H in T such that N ≤ H, K ≰ H, then T = HK and K ∩ H = N. Let L be a minimal supplement of N in K, then N ∩ L ≤ Φ(L) and T = HK = HLN = LH. Since L/N ∩ L ≅ K/N is an elementary abelian group, Φ(L) ≤ N ∩ L. Thus N ∩ L = Φ(L) and K/N ≅ L/Φ(L). The result is true.

If T ≠ S(T)H, then S(T) ≤ H. Let K/S(T) be a minimal normal subgroup of T/S(T), then T = KH. Let L be a minimal supplement of S(T) in K, then K = LS(T), where S(T) ∩ L ≤ Φ(L). Since K/S(T) ≅ L/S(T) ∩ L is a direct product of non-abelian simple groups, we have S(T) ∩ L = Φ(L). Thus K/S(T) ≅ L/Φ(L), T = HL and Φ(L) ≤ H.

Theorem 3.4

Let H be a non-trivial p-subgroup of a group G. If |H| = p^α(α ∈ N^*), H ⋖ 〈H, H^g〉 for some g ∈ G ∖ N_G(H), then T = 〈H, H^g〉 satisfies one of the following:

If T is solvable, then
1. if H ∉ Syl_p(T), then |T| = p^α+1;
2. if H ∈ Syl_p(T), then T = HQ, where Q is a Sylow q-subgroup of T with p ≠ q and Q is isomorphic to a chief factor of T.
If T is non-solvable, then H ∈ Syl_p(T), S(T) = O_p(T) ≤ H and T = HL, where L/Φ(L) is a chief factor of T, every simple direct factor of L/Φ(L) is either isomorphic to PSL(2, q), q odd and q ≥ 5, or A₇.

Proof

Assume that T is solvable. If |π(T)| ≥ 3, then there exists a Q ∈ Syl_q(T) such that HQ is a subgroup of T, where p ≠ q. So H < HQ < T contrary to the maximality of H in T. Therefore |π(T)| ≤ 2. If |π(T)| = 1, then 1 (a) is true. If |π(T)| = 2, then T = HQ, where Q is a Sylow q-subgroup of T with p ≠ q. By Theorem 3.3, Q/Φ(Q) is isomorphic to a chief factor of T and Φ(Q) ≤ H, thus |Φ(Q)| ∣ |H| = p^α. Hence Φ(Q) = 1 and so 1 (b) is true.

Now we suppose that T is non-solvable. Then S(T) ≤ H and S(T) = O_p(T). Let T = T/S(T), H = H/S(T). It is clear that H is a Sylow p-subgroup of the group T/O_p(T) and H ⋖ T. Let L be a minimal normal subgroup of T, then T = HL. By Lemma 2.5, H is a Sylow 2-subgroup of T. By Lemma 2.6, L = N₁ × N₂ × ⋯ × N_t, where N_i are non-abelian simple groups that are isomorphic to each other and H ∩ L is the direct product of dihedral 2-subgroups of L, which are Sylow 2-subgroups of each N_i. Hence by Lemma 2.7, we have that N_i ≅ PSL(2, q), q odd and q ≥ 5, or A₇. Thus our conclusion is true.

By Theorem 3.4, we have the following Corollary.

Corollary 3.5

Let H be a p-subgroup of a group G with |H| = p^α and G be dihedral free. If H ⋖ 〈H, H^g〉 for some g ∈ G ∖ N_G(H), then 〈H, H^g〉 is solvable and satisfies one of the following:

if H ∉ Syl_p(〈H, H^g〉), then |〈H, H^g〉| = p^α+1;
if H ∈ Syl_p(〈H, H^g〉), then 〈H, H^g〉 = HQ where Q is a Sylow q-subgroup of 〈H, H^g〉 with p ≠ q and Q is isomorphic to a chief factor of 〈H, H^g〉.

Theorem 3.6

Let G be dihedral free. If for every Sylow subgroup P of G, P ⋖〈P, P^g〉 for all g ∈ G ∖ N_G(P), then G is solvable.

Proof

By Corollary 3.5, we can get that 〈P, P^g〉 is solvable for all P ∈ Syl_p(G) and g ∈ G, then G is solvable by Lemma 2.2.

Theorem 3.7

Let H be a p-subgroup of G . If H ⋖ 〈H, H^g〉 and |π(〈H, H^g〉)| ≥ 2 for all g ∈ G ∖ N_G(H), then H is pronormal in G.

Proof

If g ∈ N_G(H), for any x ∈〈H, H^g〉, we have H^x = H^g. If g ∈ G ∖ N_G(H), then H ∈ Syl_p(〈H, H^g〉) by Theorem 3.4 and |π(〈H, H^g〉)| ≥ 2. Hence there exists some x ∈ 〈H, H^g〉 such that H^x = H^g. In a word, for every g ∈ G, there exists a x ∈ 〈H, H^g〉 such that H^x = H^g. Therefore, H is pronormal in G.

Theorem 3.8

Let H be an abelian p-subgroup of G with |H| = p^α. If H ⋖ 〈H, H^g〉 for some g ∈ G ∖ N_G(H), then 〈H, H^g〉 satisfies one of the following:

〈H, H^g〉 is a non-cyclic p-subgroup of order p^α+1;
〈H, H^g〉 = HL, where L is a minimal normal elementary abelian q-subgroup of 〈H, H^g〉 and q ≠ p.

Proof

Let T = 〈H, H^g〉. Since H is abelian, we have H ≤ C_T(H) ≤ N_T(H) ≤ T. Suppose that C_T(H) < N_T(H), then H = C_T(H) and N_T(H) = T. If H ∈ Syl_p(T), then H^g ∈ Syl_p(T), so H = H^g and H = 〈H, H^g〉 = T, a contradiction. Hence H ∉ Syl_p(T), thus T is a p-group and |T| = p^a+1. If T is a cyclic p-subgroup, then it has a unique maximal subgroup, so H = H^g, we also have H = T, a contradiction. Hence T is a non-cyclic p-subgroup. Therefore, we get claim (a).

Suppose that C_T(H) = N_T(H). It is clear that H ∈ Syl_p(T). By the Burnside′s theorem, we get that T is a p-nilpotent subgroup. Then there exists a normal subgroup K of T such that T = HK and H ∩ K = 1. Now consider the action of the p-subgroup H on the p^′-subgroup K. By Lemma 2.8 (a), there exists an H-invariant Sylow q-subgroup K_q of K and we conclude by maximality that T = HK_q and K_q ⊴ T. By the maximality of H in T, K_q is a minimal normal subgroup of T and Φ(K_q) = 1, then K_q is an elementary abelian subgroup. Thus we get (b).

Corollary 3.9

Let P be an abelian Sylow p-subgroup of G . If P ⋖ 〈P, P^g〉 for some g ∈ G ∖ N_G(P), then 〈P, P^g〉 = PL, where L is a minimal normal elementary abelian q-subgroup of 〈P, P^g〉 and q ≠ p.

Proof

It follows straight from Theorem 3.8.

Theorem 3.10

Let H be a cyclic subgroup of G. If H ⋖ 〈H, H^g〉 for some g ∈ G ∖ N_G(H), then there exists a supersolvable normal subgroup K such that 〈H, H^g〉 = KH.

Proof

Call T = 〈H, H^g〉. Then T has a cyclic maximal subgroup H. If H ⊴ T, then T/H is a cyclic group of prime order. Hence T itself is supersolvable, the theorem follows. If H ⋬ T, then there exists a g ∈ T ∖ N_T(H) such that T = 〈H, H^g〉. Thus we may assume that g ∈ T. Let T be a counterexample with minimal order. If there exists p ∈π(H) such that 1 ≠ H_p < T_p, where H_p ∈ Syl_p(H) and T_p ∈ Syl_p(T), then H < N_T(H_p). By the maximality of H in T, we have N_T(H_p) = T, so H_p ⊲ T. It is clear that H/H_p is a cyclic subgroup of T/H_p and H/H_p ⋖ T/H_p. Then the hypothesis is still true for (T/H_p, H/H_p), so T/H_p has a supersolvable normal subgroup L/H_p such that T/H_p = (L/H_p)(H/H_p). Obviously, L ⊲ T and T = LH. Since H_p is a cyclic subgroup, we have L is supersolvable by Lemma 2.3, a contradiction. Thus H_p = T_p for every p ∈π(H).

Hence H is a π-Hall subgroup of T, where π = π(H). Assume that |π(H)| = 1. By Theorem 3.8, the result is true. Assume that |π(H)| ≥ 2. Since H is a cyclic subgroup, we have H ≤ N_T(H_p) ≤ T for every p ∈ π(H). By the maximality of H in T, we get N_T (H_p) = T or N_T (H_p) = H. Assume that N_T (H_p) = T, by the proof of the first paragraph, we also get a contradiction. Thus N_T (H_p) = H for every p ∈π(H). By Lemma 2.4, there exists a normal subgroup K of T such that T = HK and H ∩ K = 1. Now the π-subgroup H acts on the π^′-subgroup K, by Lemma 2.8 (a), there exists an H-invariant Sylow q-subgroup K_q of K. By the maximality of H in T, we have T = HK_q, contrary to the fact that H has no supersolvable normal supplement in T. This contradiction completes the proof of this theorem.

Theorem 3.11

Let G be a group. If for every cyclic subgroup H of G, H ⋖ 〈H, H^g〉 for all g ∈ G ∖ N_G(H), then G is solvable.

Proof

By Theorem 3.10, for every cyclic subgroup H of G, we have that 〈H, H^g〉 is solvable for all g ∈ G. Then G is solvable by Lemma 2.1.

Now we give a very interesting result, which is a generalization of famous Baer′s theorem.

Theorem 3.12

Let H be a p-subgroup of G. Then H ≤ O_p(〈H, H^g〉) for all g ∈ G if and only if H ≤ O_p(G).

Proof

Suppose that H ≤ O_p(G), then 〈H, H^g〉 ≤ O_p(G) for all g ∈ G, so H ≤ O_p(〈H, H^g〉) for all g ∈ G. Conversely, for any g ∈ G, suppose that H ≤ O_p(〈H, H^g〉), then H ⊲ ⊲ 〈H, H^g〉 for all g ∈ G. By [3, Wielandt′s theorem], we have H ⊲ ⊲ G, then there exists a chain of subgroups U₀, U₁, …, U_r of G such that

H=U0⊴U1⊴⋯⊴Ur−1⊴Ur=G

Then U_i ⊴ U_i+1. Since O_p (U_i) is characteristic in U_i and U_i is normal in U_i+1, then O_p (U_i) is normal in U_i+1, hence O_p (U_i) ≤ O_p(U_i+1), furthermore, O_p (U_i) ⊴ O_p(U_i+1). Hence H = O_p(U₀) ⊴ O_p(U₁) ⊴… ⊴ O_p(U_r−1) ⊴ O_p(U_r) = O_p(G). Therefore, H ≤ O_p(G).

Corollary 3.13

([13, Baer′s theorem]) Let x be a p-element of G. Suppose that 〈x, x^g〉 is a p-subgroup for every g ∈ G, then x ∈ O_p(G).

Proof

It follows from Theorem 3.12 and 〈H, H^g〉 = 〈x, x^g〉 for H = 〈x〉.

Acknowledgement

The authors wish to express their gratitude to the referee for her/his valuable comments. This work was supported by the National Natural Science Foundation of China (Grant No. 11671324, 11601225, 11871360), the China Postdoctoral Science Foundation (N. 2015M582492) and the China Scholarship Council.

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Received: 2018-08-21

Accepted: 2019-03-18

Published Online: 2019-06-04

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Keywords for this article

finite group; chief factor; solvable; supersolvable

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