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Suppose $G$ is a group of order $p^n$ and $p$ is a prime number. Let $H$ be a subgroup of order $p^{n-2}$ and not normal. What can we say about $[G:N_G(H)]$? Is it equal to $p$?

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    Yes. Now you just need to show this. – 2017-02-21
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    $H$N_G(H)$ must be of order _$p^{n-1}$_? – 2017-02-21
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    In a non abelian group of order 8, it is possible for a subgroup $H$ of order 2 with $N_G(H) = H$. Hence the index can be $p^2$ as well. The symmetry group of a square is an example. (See for example, Baumslag and Chandler, Group Theory, page 152. – 2017-02-21
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    @Muralidharan, you are mistaken. A finite $p$-group $G$ is nilpotent, so if $H < G$, then $H < N_{G}(H)$. – 2017-02-21
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    @B.K-Theory, that's correct! – 2017-02-21
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    Yes....Thanks, I have misunderstood. – 2017-02-21

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We just have to prove that $N_G(H)\neq H$.

We can do this by induction, we prove that the normalizer of a proper subgroup $H$ of a group of order $p^n$ is not $H$.

The base case $n=1$ is trivial.

We now do the inductive step.

Notice that the center is not trivial, clearly if $H$ does not contain the center then $N_G(H)\neq H$ as the normalizer contains the center.

We conclude that $H$ contains the center. We conclude that $H$ is a union of right cosets of the center $Z$. We now work inside the group $G/Z$, by the induction hypothesis there is a $gZ$ such that $g\not\in H$ and such that $gZ(HZ)g^{-1}Z=HZ$

Since every element of $Z$ commutes with everything else it is easy to see that:

$gZ(HZ)g^{-1}Z=g(HZ)g^{-1}$ and since $Z\subseteq H$ we have that $HZ=H$.

So we have $gHg^{-1}=H$ for some $g$ not in $H$ as desired.