3
$\begingroup$

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 $H$?

  • 5
    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
  • 1
    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 $H$K^G=G$. Does that sound right? – 2012-05-10
  • 0
    Yes, that sounds good!2012-05-10

1 Answers 1

1

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

  • 0
    Maybe you want to clarify why $H^K\neq H^G$2012-05-26