Automorphisms of a Group fixing the Frattini Quotient

While reading a paper of HIgman, I encountered the following claim: "If $\alpha$ is an automorphism of $H$ which induces the identity on $H/\Phi(H)$ , so that $\alpha(h_i)=h_i x_i$ with $x_i \in \Phi(H)$ , and therefore central of order $p$, then $\alpha$ induces the identity also on $\Phi(H)$ "

Can someone please explain to me why does such an automorphism must also induces the identity on $\Phi(H)$ ? [$\Phi(H)$ is the Frattini subgroup , and $H$ is a $p$-group ] .

Thanks everyone!

• Post the paper please. – Alexander Gruber Feb 23 '13 at 9:53
• There are definitely some essential hypotheses missing, such as $\Phi(H)$ being central and of exponent $p$, and the $h_{i}$ being a minimal set of generators. Did I guess right? – Andreas Caranti Feb 23 '13 at 9:55

If, as I guessed in a comment, $\Phi(H)$ is central and of exponent $p$, and the $h_{i}$ are a minimal set of generators, here's a proof.
$\Phi(G) = G' [G, G]$. Now for $h, k \in H$ one has $\alpha(h^{p}) = \alpha(h)^{p} = (h x)^{p} = h^{p}$, and $\alpha([h, k]) = [\alpha(h), \alpha(k)] = [h x, k y] = [h, k]$, where $x, y$ are suitable elements of $\Phi(H)$, so $x, y \in Z(H)$ and $x^p=y^p = 1$.
• More generally, an automorphism of a finite $p$-group that induces the identity on $H/\Phi(H)$ will induce the identity on all layers of the lower $p$-central series of $H$. The order of such an automorphism must be a power of $p$. – Derek Holt Feb 23 '13 at 10:08
• @DerekHolt : can you give a reference for your claim regarding the identity on all layers of the lower $p$-central series? – theMissingIngredient Feb 23 '13 at 10:58
• Not really, but it's a straightforward induction, using the definition of the lower $p$-central series $P_1(G)=G$, $P_{k+1}(G) = [G,P_k(G)]P_k(G)^p$ (so, in particular, $P_2(G) = \Phi(G)$ for a finite $p$-group $G$). – Derek Holt Feb 23 '13 at 14:37
• Just to verify - we don't need the Frattini to be central and of exponent p in order for such a "fixation" will occur right? Andreas suggested a proof for the case the Frattini is central and of exponent p.My question is if we have a finite p-group and an automorphism inducing the identity on $H/\Phi(H)$ , does it also induce the identity on the Frattini itself? Thanks ! – theMissingIngredient Feb 23 '13 at 16:31
• @theMissingIngredient, any inner automorphism $\alpha$, conjugation by $x$, say, acts trivially on $H/\Phi(H)$. But for $\alpha$ to act trivially on $\Phi(H)$ you need $x$ to centralize $\Phi(H)$. So whenever $\Phi(H)$ is not central, there is an (inner) automorphism which acts trivially on $H/\Phi(H)$ but not trivially on $\Phi(H)$. – Andreas Caranti Feb 23 '13 at 16:36