Sylow implies automorph-conjugate

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This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property must also satisfy the second subgroup property
View all subgroup property implications | View all subgroup property non-implications
This fact is an application of the following pivotal fact/result/idea: Sylow's theorem
View other applications of Sylow's theorem OR Read a survey article on applying Sylow's theorem


In a finite group, any Sylow subgroup is an automorph-conjugate subgroup.

Definitions used

Sylow subgroup

Further information: Sylow subgroup

Automorph-conjugate subgroup

Further information: Automorph-conjugate subgroup


Hands-on proof

Given: A finite group G, a p-Sylow subgroup S

To prove: For any automorphism \sigma of G, S and \sigma(S) are conjugate

Proof: The key thing to observe is that \sigma(S) is also a p-Sylow subgroup. Hence, S and \sigma(S) are p-Sylow subgroups, so by the conjugacy part of Sylow's theorem, they are conjugate.