# P-Sylow subgroup congruence

1. Dec 8, 2008

### mathsss2

Let $$p$$ be a prime, $$G$$ a finite group, and $$P$$ a $$p$$-Sylow subgroup of $$G$$. Let $$M$$ be any subgroup of $$G$$ which contains $$N_G(P)$$. Prove that $$[G:M]\equiv 1$$ (mod $$p$$). (Hint: look carefully at Sylow's Theorems.)

2. Dec 8, 2008

### mathwonk

since the word N = normalizer occurs, one is led to look at the action on G on p by conjugation. G permutes subgroups of G and we look at the orbit of P. this orbit contains kp+1 subgroups, so the theorem holds if N = M. then what?