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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.

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1 Answer 1

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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.

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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? –  Benjamin Lu Nov 10 '12 at 21:19
    
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. –  marlu Nov 10 '12 at 21:28
    
Very clear. I will look into this, thank you sir. –  Benjamin Lu Nov 10 '12 at 21:31
    
What about the second part, with H being contained in a normal subgroup index p? Any hints there? –  Benjamin Lu Nov 11 '12 at 0:31
    
@BenjaminLu You know that taking normalizers strictly enlarges the subgroup. Suppose you repeat this a lot of times. What would happen? –  Miha Habič Nov 11 '12 at 1:36

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