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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.2012-08-24

1 Answers 1

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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?2012-08-23
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    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.2012-08-23
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    @MakotoKato: Yes!. I got the point. Thanks both of you.2012-08-23