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...

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.