# proper subgroups of finite p-groups are properly contained in the normalizer

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?

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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).

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Thanks! This is very helpful. –  Ler Jul 10 '12 at 18:53
Nice and simple. +1 –  DonAntonio Jul 10 '12 at 19:17
Note in fact that what is really proved above is that if $H$ is a non-trivial $p$-subgroup of a finite group $G$ (not necessaily a $p$-group itself), and $[G:H]$ is divisible by $p,$ then $N_{G}(H) >H.$ –  Geoff Robinson Jul 11 '12 at 9:10
In fact, for a finite group $\,G\,$, to be nilpotent is equivalent to the normalizer condition, namely: every proper subgroup $\,H\leq G\,$ is properly contained in its normalizer, and here enters what Geoff mentioned at the end of his answer.
Since a (non-trivial) finite $\,p-\,$group is trivially nilpotent (as its center is always non-trivial), we have $$1=:Z_0\leq Z_1\leq\ldots\leq Z_n=G\,\,,\,Z_i:=Z\left(G/Z_{i-1}\right)$$ the upper central series.
It follows, among other things, that $\,[G,Z_i]\leq Z_{i-1}\,\,,\,\forall i=1,2,...,n$
So, if $\,H\lneq G\,$ then there exists $\,0\leq i\leq n\,\,s.t.\,\,Z_{i-1}\leq H\lneq Z_i\,$ , so that $$\exists\,z\in Z_i-H\Longrightarrow [G:z]\in [G:Z_i]\leq [G:Z_{i-1}]\leq Z_{i-1}\leq H\Longrightarrow$$ $$\Longrightarrow\,\forall\,g\in G\,\,,\,[g,z]:=g^{-1}z^{-1}gz\in H\Longrightarrow h^{-1}z^{-1}hz\in H\,\,,\,\,\forall\,h\in H\Longrightarrow z\in N_G(H)$$ and from this it follows at once that $\,H\lneq N_G(H)\,$