# Another question about Higman's paper

In page 26 of his paper: http://plms.oxfordjournals.org/content/s3-10/1/24.full.pdf

Higman says the following: If $N_1 , N_2$ are subgroups of $\Phi(H)$ that are in the same equivalence classe under the automorphism group of $H$ , then the number in such an equivalence class is at most equal to the order of the automorphism group of $H/\Phi(H)$ . He is saying that such an automorphism must induce the identity on $H/\Phi(H)$ , but I can't understand this part...

Can someone please explain to me what exactly does Higman say in this paragraph? How does he count the number of groups in each equivalence class?

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The paper is behind a paywall. Is $H$ is an arbitrary finite group, and $\Phi(H)$ the Frattini subgroup? – Ted Feb 23 '13 at 21:29
$H$ is actually a $p$-group. I added a print screen of the relevant section of the paper... Thanks ! – theMissingIngredient Feb 23 '13 at 21:31
Another problem I have with his claims is as follows: If there exists an automorphism $\alpha$ of $H$ for which $\alpha(N_1)=N_2$ , then shouldn't $N_1 , N_2$ be isomorphic? – theMissingIngredient Feb 23 '13 at 21:37
There seems to be more context missing here. What is the theorem we are trying to prove? How are these $h_i$ and $k_i$ chosen? – Ted Feb 23 '13 at 21:39

This is the theorem in which Higman proves a lower bound on the number of isomorphism classes of $p$-groups $H$ in which $\Phi(H)$ is central and elementary abelian.

We have a $p$-group $H$ in which $H/\Phi(H)$ is elementary abelian of order $p^r$ and $\Phi(H)$ is central in $H$ and elementary abelian of order $p^R$.

We look at subgroups $N$ of $\Phi(H)$ of order $p^{R-s}$, and we are considering how many different isomorphism classes of quotients $H/N$ we get. To do this he splits the subgroups $N$ into equivalence classes, where each such class contains groups that give isomorphic quotients $H/N$. The displayed formula is $a/b$, where $a$ is the total number of subgroups of $\Phi(H)$ of order $p^{R-s}$ (which is the same as the number of order $p^s$), and $b$ is the order of the automorphism group of $H/\Phi(H)$, which he has proved is an upper bound on the order of each equivalence class.

To prove this upper bound, he says first that, if $H/N_1 \cong H/N_2$, then there is an automorphism of $H$ that maps $N_1$ to $N_2$. So that would give $|{\rm Aut}(H)|$ as an upper bound on the size of the equivalence classes. But, since any automorphism that induces the identity on $H/\Phi(H)$ must also induce the identity on $\Phi(H)$ and hence fix all of the subgroups $N$, we get the smaller upper bound $|{\rm Aut}(H/\Phi(H)|$ for the equivalence class sizes, and that is what he uses as $b$ in the formula.

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Dear @derekholt: Thanks a lot for your detailed answer! I still have a problem understanding your last statement: I understand that if we have an automorphism that induces the ideneity on $H/\Phi(H)$ , it must fix all subgroups $N$ . But why does this mean that we can actually look only at automorphisms of $H/\Phi(H)$ rather than automorphisms of $H$ ? [i.e.- we know that $H/N_1 = H/N_2$ iff we have an automorphism of $H$ that maps $N_1$ to $N_2$ . Why can we reduce the estimation to $Aut(H/\Phi(H) )$ ... Thanks a lot again! – theMissingIngredient Feb 24 '13 at 9:09
One more thing I can't understand: In page 25 of the paper, Higman says that if $g_1,...,g_r$ form a basis for $H/N$ , and $rank(H)=r$ , then we must have $N\subseteq \Phi(H)$ . Is this a general property of the Frattini subgroup? i.e.- if we have a quotient $G/N$ of the same rank as $G$ , we must have $N\subseteq \Phi(G)$ ? Thanks !!!!!!! – theMissingIngredient Feb 24 '13 at 9:39
For your first question, you are counting the number of images of $N$ under the action of ${\rm Aut}(H)$. Elements of ${\rm Aut}(H)$ that induce the identity on $H/\Phi(H)$ fix $N$, so the size of the orbit of $N$ under ${\rm Aut}(H)$ is at most $|{\rm Aut}(H/\Phi(H))|$ by the Orbit-Stabilizer Theorem. – Derek Holt Feb 24 '13 at 10:59
For your second question, yes, this property holds whenever $H$ is a $p$-group. This follows from the fact that all minimal generating sets of a $p$-group have the same size, whcih is the rank of $H/\Phi(H)$, and $\Phi(H)$ consists of the non-generators of $H$. – Derek Holt Feb 24 '13 at 11:02
Thanks a lot for all of your help ! It seems like I have some preliminary notions missing... So I'll have to learn a little in order to understand the paper. thanks anyway! – theMissingIngredient Feb 24 '13 at 11:35