$p$-subgroup of $G$ with $p$ a prime.

$p$-subgroup of $G$ with $p$ a prime.

If $H$ is a $p$-subgroup of a finite group $G$ then $[N_G(H) : H] ¡Õ
[G:H]\operatorname{mod} p$.
Proof:
Let S be the set of left cosets of H in G. H acts on S by left
translation, $h ¡¤ aH = haH$, then $|S| = [G : H]$. So $xH¡ôS_0 ⇐¢¡
hxH=xH ¢£h¡ôH ⇐¢¡ x^{−1}hxH=H ¢£h¡ôH ⇐¢¡
x^{−1}Hx=H ⇐¢¡ x¡ô N_G(H)$ $|S_0| = [N_G(H) : H]$ By the fact
that $ |H|=p^n$ and H acts on S we have $|S|¡Õ|S_0|$ mod p. So
$[G:H]¡Õ[N_G(H):H]$ modp.
Is it correct my proof?.
And If p divides $[G : H]$ then $N_G(H) ¡Á H $
Proof:
$0¡Õ[G:H]¡Õ[N_G(H):H]¡Ã1$. So $[N_G(H) : H] ¡Ã 1$, thus $ N_G(H) ¡Á H$ .
Is it correct?
Please help!!! T_T