# Is this problem correct that $HG'=G$?

Here, I have the following homework:

Let $G$ is a finite $p-$group and let $H$ be a subgroup of it such that $HG'=G$. Prove that $H=G$ ($G'$ is the commutator subgroup).

I have tried to show that $G\subseteq H$ by taking an element in $G$ but this way seems to be weak here. Is it possible that this exercise is printed mistakenly? Thank you friends.

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This is in fact true for any nilpotent group, finite or not. – user641 Aug 24 '12 at 0:49

## 1 Answer

It suffices to prove that $G'$ is a subgroup of $\Phi(G)$, the Frattini subgroup of $G$. In the case of $p$-groups, you can do this by showing that $G/\Phi(G)$ is elementary abelian. (This is actually true for nilpotent groups in general, in which case $G/\Phi(G)$ just has to be abelian, though not necessarily elementary.)

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Thank you but as far as I know, we have always $G/\Phi(G)$ is an elementary abelian when $G$ is a $p-$ group. How is it linked to your first claim that is a famous lemma and is a way for proving this problem? – Nancy Rutkowskie Aug 23 '12 at 8:58
This is an excellent answer. However, I'd like to add an explanation just in case. Suppose $H \neq G$. There exists a maximal subgroup $M$ of $G$ containing $H$. Since $M \supset \Phi(G) \supset G'$, $M \supset HG'$. This is a contradiction. – Makoto Kato Aug 23 '12 at 9:00
@MakotoKato: Yes!. I got the point. Thanks both of you. – Nancy Rutkowskie Aug 23 '12 at 9:04