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I'm sorry for my simple question, however i've started a sort of review on the theory of finite group, and i've to admit i'm really rusted, even though i've never really been brilliant in this subject.

However the question is the following:

Let $G$ be a finite group and $H\leq G$.

Show that $|N_G(H):H|$ is equal to the number of right cosets of $H$ in $G$ that are invariant under right multiplication by $H$. (i was able to solve this part)

Suppose now that $|H|$ is a power of the prime $p$ and that $|G:H|$ is divisible by $p$. Show that $|N_G(H):H|$ is divisible by $p$.

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up vote 3 down vote accepted

Denote by $E$ the set of right cosets of $H$ in $G$. As you said, $H$ acts by multiplication on the set $E$. Now write the class formula and reduce modulo $p$.

Since you need, more tips (i take the notations of the link of uforoboa) :

$H$ acts on the set $S$ of right cosets of $H$ in $G$. As you said, there are as much cosets invariant under this action as $\mid N_G(H):H\mid$. Hence the number of orbits of size one $S_0$ is exactly $\mid N_G(H):H\mid$. You thus have the class equality :

$$\mid S\mid=\mid N_G(H):H\mid + \sum_{i=1}^r\mid H\mid /\mid H_i\mid$$

since $\mid H \mid$ is a power of $p$ and the $H_i$'s are not equal to $H$ (recall the orbits are not reduced to one element on the very last term) any number $\mid H\mid/ \mid H_i\mid$ is divisible by $p$. Since also $\mid G:H \mid =\mid S \mid$ is by assumption you get that $p$ divides $\mid N_G(H):H \mid$.

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could you please add more details? i'm trying to figure out your hint, however i'm still not able to write down the solution, then i will accept your answer –  uforoboa Oct 7 '11 at 16:46
@ user15123: perhaps this link is somewhat illuminating groupprops.subwiki.org/wiki/Class_equation_of_a_group_action –  uforoboa Oct 7 '11 at 16:50
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