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Show that if $G$ is a finite nonsolvable group then $G$ contains a nontrivial subgroup $H$ such that $\left[H,H\right]=H$.

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I assume that $H$ should be nontrivial, since $1$ holds this property in every group. – Alexander Gruber Oct 7 '12 at 2:37
Is $G$ assumed to be finite? – Berci Oct 7 '12 at 3:25
Yes, $G$ be finite – Muniain Oct 7 '12 at 3:27

Note that $[H,H]\leqslant H$ for any group $H$. Thus, by contrapositive, if $G$ contains no non-trivial subgroup $H$ so that $[H,H]=H$, then $[H,H]$ is a proper subgroup of $H$ for each nontrivial subgroup $H$ of $G$. In particular, every term of the derived series of $G$ is strictly smaller than the previous term, whence the series converges to $1$, so $G$ is solvable.

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why the derived series converges to $1$ – Muniain Oct 7 '12 at 2:52
The derived series is defined by $G^{(0)}=G$ and $G^{(k)}=[G^{(k-1)},G^{(k-1)}]$. Since we've assumed that $[G^{(k-1)},G^{(k-1)}]$ is strictly smaller than $G^{(k-1)}$, every term decreases in size. Since $G$ is finite, eventually this must reach the identity. – Alexander Gruber Oct 7 '12 at 2:57
Oh. I see. Thank you very much. – Muniain Oct 7 '12 at 3:04

Hint: If $\,G\,$ is simple (and non-abelian, of course) the claim is trivial (why?), so we can assume $\,G\,$ has a non-trivial normal subgroup. Now take a peek at a non-trivial normal minimal subgroup.

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