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In page 26 of his paper: enter image description here

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?

Thanks in advance!

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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
up vote 3 down vote accepted

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

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