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This is from an article by Hall from 1961; it's probably one of the most trivial observations in that article, but I can't get the reasoning.

Let G be some group and let A be a normal abelian p-subgroup of G for some prime p. We want to show that $A^p$, the subgroup of p-th powers of elements of A, is contained in the Frattini subgroup of G.

So let M be a maximal subgroup in G which doesn't contain A and denote $N=M\cap A$. I have managed to show that N is normal in G and that A/N is a minimal normal subgroup of G/N. So A/N can't have any nontrivial proper characteristic subgroups. Therefore $A^pN=N$, in which case we are done, or $A^pN=A$. I don't know where to go from here. I suppose the fact that A is a p-group should come into play somewhere, since all of the above works for any exponent n instead of p.

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Can you point to the location in the paper where you find this claim (you probably should also give a citation for the paper, but I'll do that for you): Hall, P. Frattini Subgroups of Finitely Generated Groups, PLMS (3) 11, 1961, 327--352. –  user641 Apr 11 '11 at 17:11
Are you talking about Lemma 10? He gives a proof. Regarding your write-up above, $A/N$ is an elementary abelian p-group, so has exponent p; that is, $A^pN/N$ is trivial. –  user641 Apr 11 '11 at 17:20
Yes, it's Lemma 10 in that paper. The thing is, I don't see why $A/N$ should be elementary abelian. –  Miha Habič Apr 11 '11 at 18:07
$A/N$ is an abelian p-group with no non-trivial characteristic subgroups. So what does the subgroup of elements of order p look like? –  user641 Apr 11 '11 at 18:15
Right. The subgroup of elements of order p ($\Omega_1(A/N)$ I guess) is characteristic in $A/N$. If it were trivial, then $A/N$ would be trivial, otherwise it is the whole of $A/N$, which therefore is of exponent p. Great, thank you. Maybe you should post something as an answer. As a side note, am I right in thinking that we could have a trivial subgroup of elements of order n (n not a power of p) in $A/N$, and not have the whole quotient group trivial? –  Miha Habič Apr 11 '11 at 18:25
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2 Answers

up vote 2 down vote accepted

Let $G$ be a group with a normal abelian p-subgroup $A$. Let $M$ be any maximal subgroup of $G$ not containing $A$, then $N=M\cap A$ is normal in both $M$ and $A$ so is normal in $G=MA$. If there is a normal subgroup $B$ of $G$ with $N < B\le A$, then $B=B(A\cap M)=BM\cap A = A$. So $A/N$ is a minimal normal subgroup of $G/N$. Thus $A/N$ is an abelian p-group which has no non-trivial characteristic subgroups. Since $A/N$ certainly contains an element of order p, the subgroup of all elements of order p is $A/N$, and thus $(A/N)^p=A^pN/N$ is trivial, and so $A^p\le N\le M$.

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The Frattini subgroup of $G$ is also the intersection of the maximal (proper) normal subgroups of $G$. Let $M$ be such a subgroup. If $M \cap A = A$, no problem. So assume $M$ does not contain $A$. Let $\{g_1,\ldots,g_k\}$ be a maximal subset of $A$ such that $M' = \langle M , g_1, \ldots, g_k \rangle$ does not contain $A$. Then $M'$ is normalized by $A$ (the assumption that $A$ is abelian is used here), so for any $g \in A \setminus \left( M' \cap A\right)$, $\langle M',g \rangle = \left\{m'g^k\ |\ m' \in M',\ k \in \mathbb{Z} \right\}$. By maximality of our subset of $A$, for any $a \in A$, $a = m'g^k$ for some $m' \in M'$ and some $k \in \mathbb{Z}$. $m'=ag^{-k} \in A$, so we get that $M' \cap A$ is a maximal subgroup of $A$, so contains $A^p$ (the Frattini subgroup of a $p$-group $A$ is equal to $A^p[A,A]$, which is equal to $A^p$ in this case). Since $M'$ does not contain $A$, $M'$ is a proper subgroup of $G$, containing $A^p$ and $M$. Since $M = \cap_{g \in G} gM'g^{-1}$ by maximality, and $gA^pg^{-1}=A^p$ for any $g \in G$, $A^p \subset M$, and we win.

There must be an easier proof of this.

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Could you explain (or point to a reference) why $Frat(A)=A^p[A,A]$ for an arbitrary $p$-group $A$? I'm aware of this fact only in the case $A$ is finite, where the proof hinges on the fact that maximal subgroups have index $p$ in $A$. –  Miha Habič Apr 11 '11 at 17:03
Oops, shame on me, I thought the term "$p$-group" implied "finite"... and I didn't think to look up the article before posting. –  Plop Apr 11 '11 at 22:01
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