Every p-subgroup is conjugate to a p-subgroup whose normalizer in the Sylow is Sylow in its normalizer

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A related survey article is: finding conjugate subgroups

Statement

Suppose G is a finite group and P is a p-Sylow subgroup of G. Suppose H is a subgroup contained inside P. Then, there exists a subgroup K of P, such that H and K are conjugate subgroups inside G, and such that N_P(H) is a Sylow subgroup of N_G(H).

Related facts

Facts used

  1. Sylow subgroups exist
  2. Sylow implies order-dominating

Proof

Given: A finite group G, a p-Sylow subgroup P, a subgroup H contained inside P.

To prove: There exists a subgroup K of P such that H and K are conjugate subgroups inside G, and such that ,math>N_P(H)</math> is p-Sylow inside N_G(H).

Proof:

  1. Let L be a p-Sylow subgroup of N_G(H). Such a L exists by fact (1) in the subgroup N_G(H).
  2. Let g \in G be such that L \le g^{-1}Pg. Note that such a g exists by fact (2) in the group G.
  3. Let K = gHg^{-1}. Then, gLg^{-1} is a ,math>p</math>-Sylow subgroup of N_G(K). By construction, gLg^{-1} \le P, so a p-Sylow subgroup of N_G(K) is contained in P. Thus, N_P(K) is p-Sylow in N_G(K).