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?


3 Answers 3


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).

  • $\begingroup$ Thanks! This is very helpful. $\endgroup$
    – CWcx
    Commented Jul 10, 2012 at 18:53
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    $\begingroup$ Nice and simple. +1 $\endgroup$
    – DonAntonio
    Commented Jul 10, 2012 at 19:17
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    $\begingroup$ 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 $N_{G}(H) >H.$ $\endgroup$ Commented Jul 11, 2012 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

$$\begin{align} \exists\,z\in Z_i-H&\Longrightarrow [G:z]\in [G:Z_i]\leq [G:Z_{i-1}]\leq Z_{i-1}\leq H\\ &\Longrightarrow\,\forall\,g\in G\, [g,z]:=g^{-1}z^{-1}gz\in H\\ &\Longrightarrow h^{-1}z^{-1}hz\in H\,\forall h\in H\\ &\Longrightarrow z\in N_G(H) \end{align}$$

and from this it follows at once that $H\lneq N_G(H)$


Consider the set of conjugates of $H$ in $G$, i.e. let $$X = \{H^{g} \ | \ g \in G \}$$ We have $|X| = |G : N_{G}(H)|$. If $|X| = 1$, there is nothing to show. So, assume $|X| \ne 1$. Then $p\ | \ |X|$.

Now $H$ acts on $X$ by conjugation. $H$ itself is a fixpoint for this action. Now as in Geoff Robinsons's answer we conclude that there is another orbit of size prime to $p$, and therefore another orbit of size $1$. Thus we have the existence of $g \in G \setminus N_{G}(H)$ with $H \le N_{G}(H^{g})$. Since obviously $H^{g} \le N_{G}(H^{g})$ we conclude that $H^{g} < N_{G}(H^{g})$. But $N_{G}(H^{g}) = g^{-1} N_{G}(H) \ g$, so $g^{-1} H \ g < g^{-1} N_{G}(H) \ g$, and therefore $H < N_{G}(H)$.


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