I need to show that if $P$ is a Sylow p-subgroup of a group $G$, and if $M$ is a subgroup containing the normalizer of $P$, then The index of $M$ in $G$ is congruent to $1$ mod $p$. Here's what I've tried so far:
Let $P$ act on the left cosets of $M$ by left multiplication: $p*gM=pgM$. Then since $P$ is a $p$ group, the number of cosets, which is $\mid G : M \mid$, is congruent mod $p$ to the number of stable cosets. Clearly, $1M$ is stable, since $P \leq N_G(P) \leq M$. Now assume that the coset $gM$ is stable, with $g \not\in M$, meaning that $PgM=gM$, which implies that $g^{-1}Pg \in M$. Since $g$ is not in $M$, it also isn't in $P$'s normalizer, so $g^{-1}Pg \neq P$. This means that $M$ contains a 2nd Sylow p-subgroup distinct from $P$. Of course, I can't get a contradiction from this, since it may be true (for example, if $M=G$). I guess I could try to show that the nontrivial stable cosets come in multiples of $p$, but I'm not sure how to do that. Am I even using a good method to solve the problem?
Best Answer
Hint: Use Sylow's theorem in $M$ and $G$ and recall that index is multiplicative.