I am trying to prove the following,
Let $G$ be a finite $p$-group and let $H$ be a proper subgroup. Then there exists a subgroup $H'$ such that
$$
H\lneq H'\leq G
$$
and $H\triangleleft H'$.
Obviously, the natural choice for $H'$ would be the normalizer $N_G(H)$ of $H$ in $G$. However, one needs to prove then that $H\lneq N_G(H)$ in finite $p$-groups. I am aware of a proof of this fact by induction on the order of $G$. However, I was wondering if there was another proof which only used group actions?
Best Answer
Yes, you can do it with group actions. Of course we only need to worry about the case that $H$ is a non-identity proper subgroup. Let $H$ act on the right cosets of $H$ in $G$ by right translation. Since H is proper, the number of such cosets is divisible by $p.$ At least one of these is fixed by $H,$ namely the coset $H.$ There must be another orbit of size prime to $p,$ but since orbit sizes in this situation are powers of $p,$ the orbit size must be $1$. Hence there is some $g \in G \backslash H $ such that $Hgh = Hg$ for all $h \in H.$ Then $gHg^{-1} \leq H,$ so that $gHg^{-1} = H$ as both these subgroups have the same order. Hence $g \in N_{G}(H) \backslash H$ and $N_{G}(H) > H.$ (Another standard proof not by induction is to use the upper central series).