A Generalization of the Hughes Subgroup

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Theorem 1. Let G be a finite solvable group and set π π G . Then 1 Hπ G Hp G for = ( ) < ( ) < ( ) all p π if and only if G is a Frobenius group whose Frobenius kernel, F , is a nonabelian q- ∈ group such that 1 Hq F F , and whose Frobenius complement has prime order. In this case, < ( ) < Hπ G Hq F . ( )= ( ) At this point, it remains an open question whether such groups as in Theorem 1 actually exist. The existing body of work on the (original) Hughes subgroup provides a means to eliminate a few possible choices for the primes p and q. In particular, p 7 and q 5. ≥ ≥ We would like to thank Professor Khukhro for several helpful comments while writing this paper. 2. Preliminaries One can always find examples of proper and nontrivial generalized Hughes subgroups by con- sidering Frobenius actions. For example, the natural action of the multiplicative group of a field on the additive group of said field. For a given finite group G, we can restrict the sets of primes to consider in computing Hπ G . Clearly, if every element of G has prime order or π π G , ( ) ∩ ( )≠∅ then Hπ is either trivial or improper, respectively.

Fact 2. Let G be a ﬁnite group. If π1 and π2 are two sets of primes such that π1 π2, then ⊆ Hπ2 G Hπ1 G . ( )⩽ ( )

Fact 3. Let G be a ﬁnite group, π be a set of primes, and set π0 π π G . Then Hπ G Hπ0 G . = ∩ ( ) ( )= ( ) Consequently, we may assume π π G , and further, π 2; otherwise Hπ G coincides with ⊆ ( ) S S≥ ( ) the Hughes subgroup relative to the unique prime in π. From the deﬁnitions of Hπ and Hp, we obtain the following set of inclusions:

1 Hπ G Hp G G, (1) ⩽ ( )⩽ p∈π ( )⩽ which lead to questions regarding what conditions, if any, ensure that the inclusions in (1) are proper. To this end, we ﬁrst address intersections of Hughes subgroups. Proposition 4. Let G be a ﬁnite group with π G 2. Then S ( )S ≥

Hp G , for a unique prime p π G , Hp G ( ) ∈ ( ) (2) p∈π(G) ( ) = G, otherwise.

Proof. If Hp G G for all p π G , then there is nothing to show. Suppose there are distinct ( ) = ∈ ( ) primes p,q π G such that Hp G and Hq G, and note both Hughes subgroups are necessarily ∈ ( ) < < nontrivial. By a simple set theoretic argument, we have G Hp G Hq G Hp Hq . ( ∖ ) ∩ ( ∖ ) = ∖ ( ∪ ) Inasmuch as the only element x to simultaneously satisfy xq 1 xp is the identity, the left-hand- = = side is empty. Thus G Hp Hq which is impossible. = ∪ 2 We note that explicit examples exist where Hπ G is properly contained in a Hughes subgroup, ( ) albeit at the cost of being trivial. The simplest example is the Frobenius group G S3, where =

1 H 2,3 S3 H2 S3 S3. = { }( )< ( )< A second example is to take E GF 33 and consider the natural Frobenius action of the subgroup × = ( ) H of E of order 13 on the additive group N of E. The Galois group, G Gal E Z3 acts naturally = ( ~ ) on the resulting Frobenius group Γ0 NH, and so we can consider the semidirect product Γ Γ0G. = = In this example, we have

1 H 3,13 Γ H13 Γ0 N H3 Γ Γ0 Γ. = { }( )< ( )= < ( )= < 3. Results and Proof of Theorem

Next, we consider what influence Hπ G has on the structure of a finite group G. By definition, ( ) every element of G Hπ G has prime order for some prime in π. Thus, if Hπ G is trivial, then ∖ ( ) ( ) all nonidentity elements of G have prime order, and such groups were completely classified by Deaconescu [2], and Cheng, et. al. [1]. Otherwise, we fall under a situation investigated by Qian [13]:

Theorem 5 (“Theorem 1”, [13]). Let G be a ﬁnite group and N G such that every element of ◁ G N is of prime order. Then G has one of the following structures: ∖ i . G A5 and N 1. ) = = ii . G F A is a Frobenius group, where the complement, A, is of prime order, and the Frobenius ) = ⋊ kernel, F , is of prime power order, with N F . < iii . G is a p-group. ) iv . G L K A, where L K is nilpotent, A is of prime order p, K A Syl G , L G is ) = ( × )⋊ × ⋊ ∈ p( ) ◁ a Hall p′-subgroup of G, A acts ﬁxed-point-freely on L, and N L K. = ×

In the pursuit of cases where Hπ is nontrivial, we may assume G is solvable; otherwise, G A5 = by the above. Towards this end, we establish a minor relationship between Hπ G and Hughes ( ) subgroups.

Lemma 6. Let G be a ﬁnite solvable group, π be a nonempty subset of π G , and assume π G 2. ( ) S ( )S ≥ Then Hπ G G if and only if Hp G G for all p π. ( )= ( )= ∈

Proof. Observe that one direction is a triviality as Hπ G p∈π Hp G . ( )⩽ ⋂ ( ) Suppose Hπ G G and let M be a maximal normal subgroup of G containing Hπ G . Then ( )< ( ) G M p for some prime p π G and since exp G M q∈π q, we have p π. By hypothesis, S ∶ S= ∈ ( ) ( ~ ) S q∏ ∈ there exists a generator x of Hp G such that x M. Hence x 1 for some q π p . Consequently, ( ) ∉ = ∈ ∖{ } Mx is a q-element of the p-group G M implying x M, a contradiction. Therefore, Hπ G G. ~ ∈ ( )= We are now ready to prove the main theorem, which we restate below.

Theorem. Let G be a ﬁnite solvable group and set π π G . Then 1 Hπ G Hp G for all p π = ( ) < ( )< ( ) ∈ if and only if G is a Frobenius group whose Frobenius kernel, F , is a nonabelian q-group such that 1 Hq F F , and whose Frobenius complement has prime order. In this case, Hπ G Hq F . < ( )< ( )= ( ) 3 Proof. Suppose 1 Hπ G Hp G for all p π. By Lemma 6 and Proposition 4, there exists < ( ) < ( ) ∈ a unique prime p π such that Hp G G. It follows from ([13], Theorem 1) that G F A is a ∈ ( ) < = Frobenius group with complement, A, of prime order p, and kernel, F , a q-group. Now F Hp G = ( ) by ([8], Theorem 2) and so Hπ G F . In particular, we have Hπ G Hq F . If there exists ( q ) < q ( ) = ( ) z Z F Hπ G , then 1 zh h for all h Hπ G contrary to our assumption. Therefore, ∈ ( )∖ ( ) = ( ) = ∈ ( ) Z F Hπ G (in fact, Z F Hπ G by similar reasoning) and F satisﬁes the conclusion. ( )⩽ ( ) ( )< ( ) The converse follows immediately from the hypotheses.

The statement of Theorem 1 relied upon taking our set of primes to be precisely π G . The ( ) following Corollary addresses the natural question of what can be said of Hπ G when π ⊂ π G . ( ) ( ) Corollary 7. Let G be a ﬁnite solvable group, π G ≥ 2, π ⊂ π G , and suppose there exists S ( )S ( ) p ∈ π G such that Hp G < G. ( ) ( ) i . If p ∉ π, then Hπ G = G. ) ( ) ii . If p ∈ π, then Hπ G = Hp G . ) ( ) ( ) Proof. First, the results are trivial if π G = 2 and so we may assume π G > 2. Now (i) follows S ( )S S ( )S immediately from Proposition 4 and Lemma 6, whereas (ii) follows from Theorem 1 given our assumption.

4. Further Exploration

As indicated by Theorem 1, groups with proper and nontrivial Hπ-subgroups have a very restrictive structure, and explicit examples remain elusive. To perhaps stimulate some additional investigation into this speciﬁc case, we pose the following question:

Question. Does there exist a Frobenius group, F A, where the kernel, F , is a nonabelian q-group which satisﬁes 1 < Hq F < F , and the complement, A, is of prime order p? ( ) In the pursuit of an example, there are a few restrictions that can be made about the primes p and q. First, as F is nonabelian, the Frobenius complement A cannot be even ([10], Theorem 6.3); in particular, A cannot be a 2-group, ie. p ≠ 2. Next, we know the Hughes conjecture holds for the primes 2 and 3, so if q = 2, 3, then F Hq F = q. Since Hq F char F , F Hq F admits a S ∶ ( )S ( ) ~ ( ) Frobenius action by A implying p = A q 1. However, as A is nontrivial and not a 2-group, we S S U − must have q ≠ 2, 3. We would like to thank Evgeny Khukhro for inspiring the following by a helpful remark high- lighting the relationship between regular p-groups and the Hughes subgroup. Recall that a p-group p p p ′ G is called regular if, for all x,y ∈ G, x y = i z where zi ∈ x,y , and we shall rely on a few ∏ i ⟨ ⟩ results from Section III.10 of [9] to help prove the following:

Lemma 8. Let G be a regular p-group. Then Hp G = 1 or G. ( ) p Proof. By ([9], III.10.5 Hauptsatz), Ω1 G = g ∈ G g = 1 (note it is the set of such elements). ( ) { ∶ } Clearly G Ω1 G ⊆ Hp G and it follows that G = Hp G Ω1 G . Since G cannot be the union ∖ ( ) ( ) ( )∪ ( ) of two proper subgroups, either Hp G = G or Ω1 G = G forcing Hp G = 1. ( ) ( ) ( ) The following corollary is clear from ([9], III.10.2 Satz) and Lemma 8.

4 Corollary 9. If G is a p-group with 1 < Hp G < G, then G has nilpotence class, c G , at least p. ( ) ( ) Returning to the situation posed in our Question, we have from Corollary 9 that F must have nilpotence class at least q, where q > 3.

Lemma 10. Let G be a q-group with 1 < Hq G < G and admitting a ﬁxed-point-free (f.p.f) ( ) automorphism φ of prime order p. Then G cannot be metabelian.

Proof. Suppose G is metabelian. Since metabelian groups satisfy the Hughes conjecture ([5], The- orem 1), we have G Hq G = q. It follows that φ induces a f.p.f. automorphism of order p on S ∶ ( )S G Hq G . In particular, G Hq G φ is a Frobenius group and so p = φ G Hq G 1 = q 1. ~ ( ) ~ ( ) ⋊ ⟨ ⟩ S S U S ∶ ( )S− − Now a result of V.A. Kreknin and A.I. Kostrikin ([11], Theorem B) gives a bound on the nilpotence class of a group in terms of its derived length and the order of a ﬁxed-point-free automorphism of the group. In particular, as the derived length of G is 2, it follows from ([11], Theorem B):

p 1 2 1 p2 2p c G < ( − ) − = − = p ≤ q 1 < q. ( ) p 2 p 2 − − − However, this contradicts Corollary 9 and so G cannot be metabelian.

By the above result, we see that F cannot be metabelian. From this we can place additional constraints on the order of the Frobenius complement. If p = 3, then as F is ﬁnite and admits a ﬁxed-point-automorphism of order 3, we have by ([12], ‘Theorem’) that F has class at most 2, a contradiction. If p = 5, then by G. Higman’s work in ([4], Section 5), we have c F ≤ 6. Yet ( ) c F ≥ q, we have already shown q ≠ 2, 3, and we know q ≠ p, this yields another contradiction. ( ) Therefore, p ≠ 2, 3, or 5.

Corollary 11. In the situation of our Question, q cannot be any prime such that π F 1 ⊆ (S S− ) 2, 3, 5 . { } References

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MARK L. LEWIS MARIO SRACIC Department of Mathematical Sciences, Department of Mathematical Sciences, Kent State University, Kent State University, Kent, OH 44242, Kent, OH 44242, USA USA e-mail: [email protected] e-mail: [email protected]