Sylow implies order-conjugate

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 (i.e., Sylow subgroup) must also satisfy the second subgroup property (i.e., order-conjugate subgroup)
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Statement

Statement with symbols

There are two formulations:

• Any two $p$-Sylow subgroups of a finite group are conjugate subgroups.
• Suppose $P$ is a $p$-Sylow subgroup of a finite group $G$, and $Q$ is a subgroup of $G$ of the same order as $P$. Then $P$ and $Q$ are conjugate in $G$.

The equivalence of the formulations follows from the fact that the condition of being a $p$-Sylow subgroup is completely determined by the order.

Related facts

Other parts of Sylow's theorem

• Sylow subgroups exist
• Sylow implies order-dominating: A stronger formulation of the result, that can be used to prove it.
• Congruence condition on Sylow numbers: Another corollary of one of the proofs of this result.
• Divisibility condition on Sylow numbers

All these facts together are known as Sylow's theorem.

Applications

• Sylow implies isomorph-conjugate
• Sylow implies intermediately isomorph-conjugate
• Sylow implies automorph-conjugate
• Sylow implies intermediately automorph-conjugate
• Sylow implies procharacteristic
• Sylow implies pronormal
• Sylow of normal implies pronormal

Facts used

1. Sylow implies order-dominating: This states that if $P$ is $p$-Sylow and $Q$ is a $p$-subgroup, there exists a conjugate $gPg^{-1}$ of $P$ such that $Q \le gPg^{-1}$.
2. Order-dominating implies order-conjugate in finite

Proof

Proof using order-domination

This proof follows directly from facts (1) and (2).

Proof using a group action

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