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Let G be a group of order $p^k$ where p is a prime and k is a positive integer. Let H be a proper subgroup of G and $$N(H) = \{a \in G|aHa^{-1} = H\}.$$ Show that $$N(H) \neq H.$$

I think I need to show that if $N(H) = H$, then $|H|$ will not divide $p^k$, but I have no idea how to show this.

Elementary answer that does not touch on group action or orbit is preferred.

Rudy the Reindeer
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user10024395
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2 Answers2

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Define an action of $\;H\;$ on $\;X:=H/G:=$ the set of left cosets of $\;H\;$ in $\;G\;$ by left shift:

$$h\cdot xH\mapsto (hx)H$$

Notice that the orbit of $\;1H=H\;$ is fixed by $\;H\;\implies\;|\mathcal O(H)|=1\;$, and since the sum of the sizes of the orbits is $\;|X|\;$ and this number is a power of $\;p\;$ , there must be another orbit which isn't divided by $\;p\;$, and the only option left is that its size is $\;1\;$ , again.

Thus, there exists $\;gH\in X\;,\;\;gH\neq H\iff g\notin H\;$ , such that

$$\forall\,h\in H\;,\;\;hgH=gH\implies H^g:=g^{-1}hgH\le H$$

and since $\;|H^g|=|H|<\infty\;$ , we get equality

$$H^g=H\implies g\in N_G(H)\setminus H\;$$

and we're done since clearly $\;H\le N_G(H)\;$ .

Timbuc
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Here's how you do the induction:

Assume that $G$ is the smallest finite $p$-group such that $\exists H < G$ s.t. $N_G(H) = H$ (that is, $G$ is a minimal counterexample).

Now $Z(G) > 1$ as $G$ is a $p$-group. Naturally $H^z = H \ \forall z \in Z(G)$, hence $Z(G) \leq N_G(H) = H$. Now consider the groups $H/Z(G) < G/Z(G)$. As $Z(G)$ is nontrivial, we have $|G/Z(G)| < |G|$, and thus by induction $N_{G/Z(G)}(H/Z(G)) > H/Z(G)$.

Then prove that if $xZ(G) \in N_{G/Z(G)}(H/Z(G)) \ \setminus \ H/Z(G)$, then $x \in N_G(H) \setminus H$ normalizes $H$. This is a contradiction as $N_G(H) = H$. Thus the assumption must be false, hence no such counterexample exists.

Leppala
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