# Congruence condition on index of subgroup containing Sylow-normalizer

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## 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 $N_G(P) \le H$. 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 $G$, $N_G(P) \le H \le G$ $P \le N_G(P) \le H \le G$ as given. Thus, by Fact (2), $P$ is a $p$-Sylow subgroup of $H$. Also, since $N_G(P) \le H$, 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) $N_G(P) \le H \le G$ 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.