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.

1 Introduction

In this paper, all groups are finite. Given a group G and a prime p, Hughes considered the subgroup $\mathrm {H}_{p}(G)$ generated by all elements of G whose order is not p. In [8], Hughes asked if it is always the case that when $\mathrm {H}_{p}(G)$ 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 $\mathrm {H}_{p}(G)$ is cyclic if and only if one of the following occurs:

1. (1) G has exponent p and $\mathrm {H}_{p}(G) = 1$ ;

2. (2) G is cyclic and $\mathrm {H}_{p}(G) = G$ ;

3. (3) $p = 2$ , G is a dihedral group and $\mathrm {H}_{2}(G)$ has index $2$ in G.

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.

Theorem 1.2. Let G be a $2$ -group. Then $\mathrm {H}_{2}(G)$ is generalised quaternion if and only if $G = \mathrm {H}_{2}(G)$ .

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 $\mathrm {Cyc}_G (x) = \{ g \in G \mid \langle x, g \rangle \mathrm {~is~cyclic} \}$ . It is not difficult to find examples of a group G and an element x where $\mathrm {Cyc}_G (x)$ is not a subgroup. In the literature, a group G is said to be tidy if $\mathrm {Cyc}_G (x)$ is a subgroup of G for every element $x \in G$ . 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 [2–5].

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 $G \setminus H$ 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. (1) G is a tidy group.

2. (2) The subgroup $\mathrm {H}_{p}(G)$ is cyclic or generalised quaternion.

3. (3) One of the following occurs:

   1. (a) G has exponent p;

   2. (b) G is cyclic;

   3. (c) $p = 2$ and G is dihedral or generalised quaternion.

2 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=\langle h\rangle $ is a cyclic maximal subgroup of G with $\left |\langle h\rangle \right |=p^{e}$ . If H has a complement $\langle g\rangle $ in G, then one of the following situations occurs:

1. (1) $p\neq 2$ and $h^{g}=h^{1+p^{e-1}}$ (for suitably chosen g);

2. (2) $p=2$ and $h^{g}=h^{-1}$ ;

3. (3) $p=2$ , $e\ge 3$ and $h^{g}=h^{-1+2^{e-1}}$ ;

4. (4) $p=2$ , $e\ge 3$ 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 $2^{e}$ , $e\ge 1$ , then $ {Aut}(H)$ is cyclic of order $2^{e-1}$ for $e\in \{1,2\}$ and is isomorphic to $C_{2}\times C_{2^{e-2}}$ for $e\ge 3$ .

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 $\mathrm {H}_{p}(G)$ . Hence,

$$ \begin{align*}\mathrm{H}_{p}(G)=\langle\, g\in G\mid o(g)\neq p\,\rangle.\end{align*} $$

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 $\mathrm {H}_{p}(G)\neq 1$ , then $\mathbf {C}_{G}({\mathrm {H}_{p}(G)}) \le \mathrm {H}_{p}(G)$ .

Proof. Suppose that $\mathbf {C}_{G}(\mathrm {H}_{p}(G))\not \le \mathrm {H}_{p}(G)$ and fix $x \in \mathbf {C}_{G}({\mathrm {H}_{p}(G)}) \setminus \mathrm {H}_{p}(G)$ . Note that $o(x) = p$ . The subgroup $\mathrm {H}_{p}(G)$ 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 $\mathrm {H}_{p}(G)$ , which is a contradiction.

We now prove the case when the Hughes subgroup is cyclic.

Proof of Theorem 1.1.

Let $H=\mathrm {H}_{p}(G)$ 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 $|H|\ge p^{2}$ since H is nontrivial.

Since H is cyclic, $H\le \mathbf {C}_{G}(H)$ . Using Lemma 2.2, we conclude that $H=\mathbf {C}_{G}(H)$ . By the normaliser/centraliser theorem [11, Corollary X.19], $G/H$ 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=\langle h\rangle $ and write $|\langle h\rangle |=p^{e}$ . Fix $g\in G\setminus H$ . Note that $|\langle g\rangle |=p$ and that $\langle g\rangle $ 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

$$ \begin{align*}(h^{p})^{g}=(h^{g})^{p}=(h^{1+p^{e-1}})^{p}=h^{p+p^{e}}=h^{p}.\end{align*} $$

Hence, $\langle h^{p}\rangle \le Z (G)$ 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 \setminus \langle a \rangle $ contains an element of order $p^2$ . Consider $b \in G \setminus \langle a \rangle $ , 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 $(ab)^p = a^p b^p [b,a]^{(p-1)p/2} = a^p \ne 1$ . 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 $H=\mathrm { H}_{2}(G)$ is cyclic. Again, fix $g\in G\setminus H$ . Lemma 2.2 guarantees that $h^{g}\neq h$ . Consider the subgroup $X=\langle h, g\rangle $ . 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 $h^{g}=h^{-1}$ , $h^{g}=h^{-1+2^{e-1}}$ or $h^{g}=h^{1+2^{e-1}}$ . If $h^{g}=h^{-1+2^{e-1}}$ , then there exist elements of order $4$ in $X\setminus \langle h\rangle \subseteq G\setminus H$ (see [11, Problem 3A.1]), which is a contradiction.

Assume that $h^{g}=h^{1+2^{e-1}}$ . Under this hypothesis, $Z(X)$ is cyclic of order $2^{e-2}$ (see [1, Exercise 8.2(1)]). If $|Z(X)|>2$ , then, as before, there exist elements of order $4$ in $X\setminus \langle h\rangle \subseteq G\setminus H$ , which is a contradiction. So $|Z(X)|=2$ and $e=3$ . Hence, $h^{g}=h^{5}$ . Note that $hg\in X\setminus H$ and so $(hg)^2=1$ . Now, $1 = (hg) (hg) = h (g^{-1} h g)= h h^5 = h^6$ , which is a contradiction to the fact that $o(h)=8$ .

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, $ {Aut}(H)\cong C_{2}\times C_{2^{e-2}}$ . As $G/H$ embeds in $ {Aut}(H)$ , we conclude that $G/H\cong C_{2}$ or $G/H\cong C_{2}\times C_{2}$ . Suppose that $G/H \cong C_2 \times C_2$ . Then we can choose $x,y\in G\setminus H$ such that $Hx\neq Hy$ . Both elements x and y are involutions and invert h. So $h^ {x y^{-1}} =(h^{-1})^{y^{-1}}=h$ . However, now $xy^{-1}\in \mathbf {C}_{G}(H)=H$ and so $Hx=Hy$ , which is a contradiction. This argument rules out the possibility that $G\cong C_2\times C_2$ . 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 $H=\mathrm {H}_{2}(G)$ is generalised quaternion. In this case, H is generated by elements x, y such that $o(x)=2^{a}$ , $o(y)=4$ , $x^{y}=x^{-1}$ , $x^{2^{a-1}}=y^{2}$ . Recall that $x^{2^{a-1}}=y^{2}$ is the unique involution of H. If $G=H$ , then we are done. So, assume that $H<G$ and fix $s\in G\setminus H$ . Conjugation by s induces an automorphism of H. If s induces an inner automorphism of H, then, for all $h\in H$ , $h^{s}=h^{t}$ for some $t\in H$ . However, then $st^{-1}\in \mathbf {C}_{G}(H)\le H$ (using Lemma 2.2) and so $s\in H$ , 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 $G \setminus K$ 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 $\langle s\rangle $ on H is necessarily semi-dihedral, which contradicts the fact that G is tidy.

Assume $a\ge 3$ . Note that $\langle x\rangle $ is characteristic in H and so s induces an automorphism of $\langle x\rangle $ . An analysis of the possibilities, similar to the argument in the proof of Theorem 1.1, shows that s acts as the inversion map on $\langle x\rangle $ .

Write $y^{s}=yx^{d}$ for $0\le d\le 2^{a}-1$ . Suppose that d is even and write $d=2b$ for $b\in \mathbb {Z}$ . However, now,

$$ \begin{align*}(yx^{b})^{s}=yx^{2b}x^{-b}=yx^{b}.\end{align*} $$

Thus, s and $yx^{b}$ commute. Since H is generalised quaternion and $yx^{b}$ does not lie in $\langle x \rangle $ , we see that $o (yx^b) = 4$ . Now, $o(syx^{b})=4$ and $syx^{b}\in G\setminus H$ , which is a contradiction.

We now suppose that d is odd. Observe that

$$ \begin{align*}(sy)^{2}=sysy=y^{s}y=yx^{d}y=y^{2}y^{-1}x^{d}y=y^{2}x^{-d}=x^{2^{a-1}-d}.\end{align*} $$

Since d is odd, $o((sy)^{2})=2^{a}$ . Hence, $o(sy)=2^{a+1}$ . Next, note that $\langle x\rangle \le \langle sy\rangle $ and that $|H\langle s \rangle :\langle sy\rangle |=2$ . Now, as $H\langle s \rangle $ contains subgroups of index $2$ that are generalised quaternion and cyclic, it can be deduced that $H\langle s\rangle $ is semi-dihedral (which is not tidy), in contrast to the fact that it is a subgroup of a tidy group. So, if H is generalised quaternion, then $G=H$ .

If G is generalised quaternion, it is not difficult to see that it is its own Hughes subgroup.

Combining all of the results, we see that Theorem 1.3 follows.