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[parent] subgroups containing the normalizers of Sylow subgroups normalize themselves (Corollary)

Let $ G$ be a finite group, and $ S$ a Sylow subgroup. Let $ M$ be a subgroup such that $ N_G(S)\subset M$. Then $ M=N_G(M)$.

Proof. By order considerations, $ S$ is a Sylow subgroup of $ M$. Since $ M$ is normal in $ N_G(M)$, by the Frattini argument, $ N_G(M)=N_G(S)M=M$. $ \qedsymbol$



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Cross-references: Frattini argument, normal, order, subgroup, Sylow subgroup, finite group
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This is version 3 of subgroups containing the normalizers of Sylow subgroups normalize themselves, born on 2002-12-14, modified 2003-12-16.
Object id is 3763, canonical name is SubgroupsContainingTheNormalizersOfSylowSubgroupsNormalizeThemselves.
Accessed 1794 times total.

Classification:
AMS MSC: 20D20 (Group theory and generalizations :: Abstract finite groups :: Sylow subgroups, Sylow properties, $\pi$-groups, $\pi$-structure)
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