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If $|G| = p^n$ , $p$ is a prime, and $H \neq G$ is subgroup of $G$ , then how to prove that $N_G(H) \supsetneq H $ ?

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You need to assume $H\neq G$. – Arturo Magidin Sep 23 '11 at 18:35
Yeah that's correct. – Mohan Sep 23 '11 at 18:37
A useful general method for proving statements about finite $p$-groups: (1) Prove the abelian case, then (2) Note that the center $Z(G)$ of a nonabelian $p$-group is non-trivial and proper. By induction, the statement is true for $Z(G)$ and $G/Z(G)$; combine these in some way to prove the statement for $G$. Arturo's proof below is an example of this. – Ted Sep 24 '11 at 2:10
up vote 3 down vote accepted

Induction on $n$.

If $n=1$ or $2$, then $G$ is abelian and so the normalizer is the whole group. Since $H\neq G=N_G(H)$, we are done.

Assume the result holds for groups of order less than $p^k$, and let $H$ be a proper subgroup of a group $G$ of order $p^{k+1}$ (the conclusion trivially holds if $H=\{e\}$.

The center of $G$ is nontrivial; if $Z(G)\subseteq H$, then consider $H/Z(G)$ as a subgroup of $G/Z(G)$. If $gZ(G)$ normalizes $HZ(G)$, then $g$ normalizers $H$: if $h\in H$, then $ghg^{-1}\in HZ(G)=H$. Thus, $N_{G/Z}(H/Z) = N_G(H)/Z$, so by the Isomorphism Theorems it follows that $N_G(H)\neq H$.

If $Z(G)$ is not contained in $H$, then let $z\in Z(G)$, $z\notin H$. Then $z$ normalizes $H$, so $H\neq N_G(H)$.

This finishes the induction. $\Box$

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