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.

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 [Reference Isaacs11, 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 [Reference Isaacs11, 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 [Reference Aschbacher1, 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.

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 [Reference O’Bryant, Patrick, Smithline and Wepsic12, 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.