Sylow $p$-subgroup where $[G:H] \equiv 1$ (mod $p$)

Question

If $K$ is a Sylow $p$-subgroup of $G$, and H is a subgroup that contains $N(K)$. Prove that $[G:H]=1$(mod$p$).

I believe this has to do with the second Sylow theorem, but I am unsure of how to proof this is true. I do know that if H and K are subgroups. The number of distinct $H$-conjugates of $K$ is $[H:H \cap N(K)]$ and thus divides $|H|$.

Any help would be appreciated.

Answer 1

Hint: $H \cap N(K)=N(K)$ and $[G:N(K)]=[G:H].[H:N(K)]$. Now take this equation mod $p$.