Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $p$.

share|cite|improve this question
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 $\mid N_G(H):H \mid$.

share|cite|improve this answer
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 – uforoboa Oct 7 '11 at 16:50

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.