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.