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Let $G$ be a $p$-group, where $p$ is prime, and let $H$ be a proper subgroup of $G$. I want to show that $H$ is contained in a normal subgroup of index $p$ in $G$.

I know that subgroups of index $p$ are normal, so all I have to do is show that $H$ is contained in such a subgroup. I also know that the normalizer $N(H)$ of $H$ in $G$ strictly contains $H$, so I was thinking of looking at the sequence of normalizers $N(N(H)), N(N(N(H))), \dots$, and arguing that this sequence eventually gives me a subgroup of index $p$. I'm not sure how to begin making this argument though.

Any help is appreciated!

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A $p$-group is a group with order $p^n$ for some positive integer $n$, so it does not follow that all subgroups must have order 1 or $p$: the cyclic group of order 8 has a subgroup of order 4. –  Jonas Nov 8 '12 at 0:06
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You want to show that any maximal subgroup has index $p$. Since $N(H)>H$, maximal subgroups are normal. If $M$ is maximal then $G/M$ is simple. What can you say about simple $p$-groups? –  Chris Godsil Nov 8 '12 at 0:14

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