Let $H$ be a proper subgroup of $p$-group $G$. Show that the normalizer of $H$ in $G$, denoted $N_G(H)$, is strictly larger than $H$, and that $H$ is contained in a normal subgroup of index $p$.

Here's what I've got so far:

  • If $H$ is normal, $N_G(H)$ is all of $G$ and we are done.
  • If $H$ is not normal, then suppose for the sake of contradiction that $N_G(H)=H$. Then there is no element outside of $H$ that fixes $H$ by conjugation. But the center $Z(G)$ of $G$ does fix $H$, so $Z(G)$ must be a subgroup of $H$.

Don't know if I'm going on the right path or not, but either way can't really think my way out of this one... Any help would be appreciated.


You are on the right track. Now look at the subgroup $H/Z$ of $G/Z$. By induction, its normalizer is strictly larger than $H/Z$. Say it contains the residue class $\overline x$ of $x \in G$ where $\overline x \not\in H/Z$. Now show that $x$ also normalizes $H$ in $G$ to get a contradiction.

  • 1
    $\begingroup$ Could you clarify: 1) By induction from what? I thought we were assuming that N(H) is not larger than H. 2) By residue class do you just mean the elements outside of H/Z? $\endgroup$ – Benjamin Lu Nov 10 '12 at 21:19
  • $\begingroup$ 1) The induction is on the order of $G$. Since $G$ is a $p$-group its center $Z$ is non-trivial, so $G/Z$ is strictly smaller than $G$. 2) By the residue class of $x$ I mean its image in the quotient group $G/Z$, so the coset $xZ$ if you like. $\endgroup$ – marlu Nov 10 '12 at 21:28
  • $\begingroup$ Very clear. I will look into this, thank you sir. $\endgroup$ – Benjamin Lu Nov 10 '12 at 21:31
  • $\begingroup$ What about the second part, with H being contained in a normal subgroup index p? Any hints there? $\endgroup$ – Benjamin Lu Nov 11 '12 at 0:31
  • $\begingroup$ @BenjaminLu You know that taking normalizers strictly enlarges the subgroup. Suppose you repeat this a lot of times. What would happen? $\endgroup$ – Miha Habič Nov 11 '12 at 1:36

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