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Let $G$ be any finite group. I'm trying to show that $G$ has a nilpotent subgroup $H$ such that $H^G:=\langle H^g : g \in G \rangle = G$, i.e. the normal closure of $H$ in $G$ is the whole group.

I know that $H$ cannot have a trivial center since it is nilpotent. Also $H$ cannot be normal for otherwise $H^g=H$ for all $g \in G$.

Other than that, I'm stuck for ideas. How do I construct such a subgroup $H$?

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    Do it by induction on $|G|$. If $G$ has a maximal subgroup $H$ that is not normal in $G$, then the result follows from induction applied to $H$. Otherwise all maximal subgroups are normal, which implies that $G$ itself is nilpotent.2012-05-10
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    So, if $H$ is maximal but not normal in $G$, then there exists $K \leq H$ with $K$ nilpotent such that $K^H=H$. As $H, we have $K^H, and so by maximality $K^G=G$. Does that sound right?2012-05-10
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    Yes, that sounds good!2012-05-10

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This answer is based on the hint by Derek Holt.

Apply induction on $|G|$. The base case $|G|=1$ is trivial. If $G$ is nilpotent then we can simply take $H:=G$ and we are done. So suppose $G$ is not nilpotent. Then $\exists K such that $K$ is maximal but not normal in $G$. Then by the induction hypothesis, $\exists H \leq K$ with $H$ nilpotent such that $H^K=K$. Then $H^K ($H^K \neq H^G$ as $H^G$ is normal in $G$ but $H^K=K$ is not). But $K$ is maximal, so this forces $H^G=G$. Hence the result is true by induction.

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    Maybe you want to clarify why $H^K\neq H^G$2012-05-26