Congruence condition on index of subgroup containing Sylow-normalizer

Statement

Suppose $G$ is a finite group, $p$ is a prime number, $P$ is a $p$-Sylow subgroup, and $H$ is a subgroup of $G$ such that $N_{G}(P)\leq H$. In other words, $H$ contains a $p$-Sylow normalizer (?). Then, the index of $H$ in $G$ is congruent to $1$ modulo $p$.

Facts used

1. Congruence condition on Sylow numbers
2. Sylow satisfies intermediate subgroup condition: Any $p$-Sylow subgroup of a group is also a $p$-Sylow subgroup in any intermediate subgroup.
3. Index is multiplicative

Proof

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Given: A finite group $G$, a prime $p$, a $p$-Sylow subgroup $P$, a subgroup $H$ of $G$ containing $N_{G}(P)$.

To prove: $[G:H]\equiv 1\mod p$.

Proof:

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	$[G:N_{G}(P)]\equiv 1{\pmod {p}}$	Fact (1)	$P$ is $p$-Sylow in $G$		direct
2	$[H:N_{G}(P)]\equiv 1{\pmod {p}}$	Facts (1), (2)	$P$ is $p$-Sylow in $G$, $N_{G}(P)\leq H\leq G$		$P\leq N_{G}(P)\leq H\leq G$ as given. Thus, by Fact (2), $P$ is a $p$-Sylow subgroup of $H$. Also, since $N_{G}(P)\leq H$, we have $N_{G}(P)=N_{H}(P)$. Thus, applying fact (1) to the group $H$, we get $[H:N_{G}(P)]=[H:N_{H}(P)]\equiv 1\mod p$.
3	$[G:N_{G}(P)]=[G:H][H:N_{G}(P)]$	Fact (3)	$N_{G}(P)\leq H\leq G$		Fact-direct
4	$[G:H]\equiv 1{\pmod {p}}$			Steps (1), (2), (3)	Go mod $p$ in Step (3) and plug in Steps (1) and (2) into the right side to obtain the conclusion.