How do i prove $|G:H|\equiv |N_G(H):H| \pmod p$? Reference: How do i prove this statement about a p-subgroup?
Let $G$ be a finite group and $p$ be a prime.
Let $H$ be a p-subgroup of $G$.
How do i prove $|G:H|\equiv |N_G(H):H| \pmod p$?
 A: You use the orbit decomposition theorem. Let a group $G$ act on a finite set $X$; then 
$$|X|=|Fix(X)|+\sum [G:C(x)]$$ where $C(x)=\{g \in G : gx=x\}$ for every $x \in X$, and the sum is over a system of representatives of all distinct nontrivial $G$-orbits on $X$.
In this case let $H$ act on the set of left cosets of $H$, by left multiplication. Let us find the fixed set. $gH$ is fixed if for all $h \in H$ we have 
$$hgH=gH$$ which means that
$$g^{-1}hgH=H$$ so
$$g^{-1}hg\in H$$ and clearly this holds for all $h \in H$ if and only if $g \in N_G(H)$.
Now of course not all $gH$ are different, $gH=g_1H$ iff $g_1^{-1}g \in H$ so the number of fixed cosets is $[N_G(H):H]$.
So we have from orbit decomposition, 
$$[G:H]=[N_G(H):H]+\sum [H:C(x)]$$ now every element in the sum is non trivial divisor of $|H|$ and thus divisible by $p$. The congruence follows.
A: This is clearly true when $H$ is  Sylow $p$-subgroup of $G,$ for then $[G:N_{G}(H)] \equiv 1$ (mod $p$). Otherwise, let $P$ be a Sylow $p$-subgroup of $G$ containing $H$ with $P \neq H.$ Then $[G:H] \equiv 0$ (mod $p$) and $N_{P}(H) >H$ ( there are many ways to prove the last fact). This gives sufficient information to answer the question, though I don't give all details.
