p-GROUPS WITH CYCLIC OR GENERALISED QUATERNION HUGHES SUBGROUPS: CLASSIFYING TIDY p-GROUPS

Abstract Let G be a p-group for some prime p. Recall that the Hughes subgroup of G is the subgroup generated by all of the elements of G with order not equal to p. In this paper, we prove that if the Hughes subgroup of G is cyclic, then G has exponent p or is cyclic or is dihedral. We also prove that if the Hughes subgroup of G is generalised quaternion, then G must be generalised quaternion. With these results in hand, we classify the tidy p-groups.


Introduction
In this paper, all groups are finite.Given a group G and a prime p, Hughes considered the subgroup H p (G) generated by all elements of G whose order is not p.In [8], Hughes asked if it is always the case that when H p (G) is proper and nontrivial, then it has index p in G. Hughes proved that this is true for 2-groups in [7].Strauss and Szekeres proved it is true for 3-groups in [14], and Hughes and Thompson proved it is true when G is not a p-group in [9].However, the conjecture is not true in general.Wall published a counterexample for p = 5 [15].See the discussion in [6] for more background regarding the Hughes subgroup problem.
In this paper, our goal is quite modest.We wish to consider p-groups that have Hughes subgroups that are cyclic or generalised quaternion.We begin by considering p-groups with a cyclic Hughes subgroup.

THEOREM 1.1. Let G be a p-group. Then H p (G) is cyclic if and only if one of the following occurs:
(1) G has exponent p and Next, we consider a 2-group with a Hughes subgroup that is generalised quaternion.In this case, we prove that G must equal its Hughes subgroup.
Our interest in groups of prime power order with a Hughes subgroup that is cyclic or generalised quaternion arises in the context of tidy groups.For each element x in a group G, let Cyc G (x) = {g ∈ G | x, g is cyclic}.It is not difficult to find examples of a group G and an element x where Cyc G (x) is not a subgroup.In the literature, a group G is said to be tidy if Cyc G (x) is a subgroup of G for every element x ∈ G.As far as we can determine, tidy groups were introduced in [13] and in a second paper [12].We note that in [12], the authors define an object they call cycels, so the word 'cycels' in the title of that paper is not a typo.Tidy groups have been studied in [2][3][4][5].
In [12,Theorem 14], O'Bryant et al. prove that if G is a p-group, then G is tidy if and only if there is a normal subgroup H that is cyclic or generalised quaternion such that every element in G \ H has order p.It is not difficult to see that H must be the Hughes subgroup of G. Hence, the task of classifying the tidy p-groups becomes that of determining the p-groups whose Hughes subgroup is either cyclic or generalised quaternion.With that in mind, we obtain the following classification of tidy p-groups.THEOREM 1.3.Let G be a p-group for some prime p. Then the following are equivalent.
(1) G is a tidy group.
(2) The subgroup H p (G) is cyclic or generalised quaternion.
(3) One of the following occurs:

Results
To prove our results, we make use of the following classification of p-groups that have a cyclic maximal subgroup (see, for example, [10, Satz I.14.9]).

THEOREM 2.1. Let G be a nonabelian p-group for some prime p and assume that H = h is a cyclic maximal subgroup of G with | h | = p e . If H has a complement g in G, then one of the following situations occurs:
(1) p 2 and h g = h 1+p e−1 (for suitably chosen g); (2) p = 2 and h g = h −1 ; (3) p = 2, e ≥ 3 and h g = h −1+2 e−1 ; (4) p = 2, e ≥ 3 and h g = h 1+2 e−1 .
Theorem 2.1 depends on the structure of Aut(H), which we mention explicitly.If H is a cyclic p-group of order p e , where p is an odd prime, then Aut(H) is cyclic of order p e−1 (p − 1).If H is a cyclic 2-group of order 2 e , e ≥ 1, then Aut(H) is cyclic of order 2 e−1 for e ∈ {1, 2} and is isomorphic to Let G be a group and let p be a prime.We define the Hughes subgroup of G to be the subgroup generated by all of the elements of G whose order does not equal p.The Hughes subgroup of G with respect to the prime p is denoted by H p (G).Hence, When a p-group G is cyclic, then it will equal its Hughes subgroup.However, a p-group G has exponent p and order at least p 2 if and only if its Hughes subgroup is trivial.The following preliminary lemma about the Hughes subgroup is useful.

LEMMA 2.2. If G is a p-group for a prime p and H
The subgroup H p (G) has an element of order p 2 , say h.However, now, we deduce that the element hx has order p 2 and does not belong to H p (G), which is a contradiction.
We now prove the case when the Hughes subgroup is cyclic.PROOF OF THEOREM 1.1.Let H = H p (G) and assume that H is cyclic.If H = 1, then G has exponent p and G satisfies item (1).If H = G, then G satisfies item (2).We therefore proceed with the hypothesis that 1 < H < G.Note that |H| ≥ p 2 since H is nontrivial.

], G/H is isomorphic to a subgroup of Aut(H).
If p is odd, then Aut(H) is cyclic.Hence, the section G/H is also cyclic.Since every nonidentity element of G/H has order p, we conclude that |G : H| = p.In particular, H is a cyclic maximal subgroup of G.

Now, let H = h and write
and that g serves as a complement to H in G.By Theorem 2.1, h g = h 1+p e−1 (where g may have to be re-chosen).Observe that (h p ) g = (h g ) p = (h 1+p e−1 ) p = h p+p e = h p .Hence, h p ≤ Z(G) and it follows that |H : Z(G)| = p.Now, if |Z(G)| > p, then there would exist elements of order p 2 outside of H, which is a contradiction.We conclude that |Z(G)| = p, |G| = p 3 and the exponent of G is p 2 .Hence, G is extraspecial.Now G is extra-special of order p 3 and has exponent p 2 .This implies that G has nilpotence class 2. We claim that G is generated by elements of order p 2 .We know that G has an element a whose order is p 2 .It suffices to show that G \ a contains an element of order p 2 .Consider b ∈ G \ a , and assume b has order p.Then using induction, it is not difficult to compute that (ab) n = a n b n [b, a] (n−1)n/2 for every positive integer n.So, if p is odd, then (ab) p = a p b p [b, a] (p−1)p/2 = a p 1. Hence, ab has order p 2 .Thus, we conclude that G = H, which is a contradiction.In particular, if H is a nontrivial, proper cyclic subgroup of G, then p = 2.
So, assume that p = 2, while still operating under the assumption that H = H 2 (G) is cyclic.Again, fix g ∈ G \ H. Lemma 2.2 guarantees that h g h.Consider the subgroup X = h, g .We claim that g inverts h: that is, h g = h −1 .By Theorem 2.1 applied to X, the possibilities for h g are then there exist elements of order 4 in X \ h ⊆ G \ H (see [11,Problem 3A.1]), which is a contradiction.
The remaining possibility is, of course, that g inverts h.Indeed, g always inverts h, and so X is dihedral.As noted below Theorem 2.1, Then we can choose x, y ∈ G \ H such that Hx Hy.Both elements x and y are involutions and invert h.So h xy −1 = (h −1 ) y −1 = h.However, now xy −1 ∈ C G (H) = H and so Hx = Hy, which is a contradiction.This argument rules out the possibility that G C 2 × C 2 .Hence, |G : H| = 2 and G = X is dihedral, giving item (3).
Finally, if item (1), ( 2) or (3) occurs, then it is not difficult to see in each case that H is cyclic.
Finally, we consider the case when the Hughes subgroup is generalised quaternion.PROOF OF THEOREM 1.2.Assume that H = H 2 (G) is generalised quaternion.In this case, H is generated by elements x, y such that o then we are done.So, assume that H < G and fix s ∈ G \ H. Conjugation by s induces an automorphism of H.If s induces an inner automorphism of H, then, for all h ∈ H, h s = h t for some t ∈ H.However, then st −1 ∈ C G (H) ≤ H (using Lemma 2.2) and so s ∈ H, which is a contradiction.Hence, s induces an outer automorphism of H.
At this point, we recall a result mentioned previously.Reference [12, Theorem 14] says that if G is a p-group, then G is tidy if and only if there is a normal subgroup K that is cyclic or generalised quaternion such that every element in G \ K has order p.So, setting K = H in our present situation, we conclude that G is tidy.If a = 2, then the semi-direct product resulting from the action of s on H is necessarily semi-dihedral, which contradicts the fact that G is tidy.