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Let $p$ be a prime number, suppose $G$ is a finite p-group and let $H$ be a subgroup of $G$. Show that there is a composition series that contains $H$.

I have already shown that if $G$ if a finite $p$-group then G is solvable and has a composition series with $n$ distinct nontrivial subgroups. This means we have {$e$} $\triangleleft$ $G_1$ $\triangleleft$ $G_2$ ...$\triangleleft$ $G_n$ = $G$ where each $G_{i+1} $/$G_i$ is abelian and each $G_i$ has order $p^i$.

Since $H$ is a subgroup then by Lagrange's theorem the order of $H$ is $\lvert$$H$$\rvert$ =$p^1$ for $0$ $\le$ $l$ $\le$ $n$. Then there must exist an $i$ such that $\lvert$$H$$\rvert$ = $\lvert$$G_i$$\rvert$. If $H$ = $G_i$ then we are done. Otherwise by one of Sylow's theorems there exists $g$ $\in$ $G$ such that $H$ = $g$$G_i$$g^{-1}$.

Here is where I cannot make any more progress. Should I try to prove that {$e$} $\triangleleft$ $g$$G_1$$g^{-1}$ $\triangleleft$ $g$$G_2$$g^{-1}$ ...$\triangleleft$ $g$$G_n$$g^{-1}$ = $G$ is a composition series that contains $H$? I have a feeling that this is not true...

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  • $\begingroup$ Sylow's 2nd theorem states that any two Sylow $p$-subgroups must be conjugate. But here $H$ is not a Sylow $p$-subgroup if it's proper in $G$, just a $p$-subgroup. Two $p$-subgroups of same order needn't even be isomorphic (see the symmetric group $S_4$ and its subgroups of order $4$). $\endgroup$ – Leppala Nov 18 '14 at 8:43
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Lemma1: if $H$ is a proper subgroup of $G$, there exist subgroup $K$ s.t. $|K:H|=p$.

Lemma2: if $H$ is nontrivial subgroup of $G$ then there exist subgroup $K$ s.t. $|H:K|=p$.

You can show the lemma1 and lemma2 by induction and notice that when $|H:K|=p$ then $K$ is normal in $H$ then you can create a subnormal series including $H$ s.t. every quotient is cyclic.

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