# Conjugacy-closed and Hall not implies retract

This article gives the statement and possibly, proof, of a non-implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., conjugacy-closed Hall subgroup) need not satisfy the second subgroup property (i.e., Hall retract)
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## Statement

A conjugacy-closed Hall subgroup (i.e., a Hall subgroup that is conjugacy-closed: any two elements in the subgroup that are conjugate in the whole group are conjugate in the subgroup) need not be a Retract (?). Specifically, there need not be a Normal Hall subgroup (?) that is a complement to it.

## Proof

### Example of Sylow complements in symmetric groups

Further information: Symmetric group on subset is conjugacy-closed

If $S \subseteq T$ are sets, then the symmetric group on $S$ embeds as a conjugacy-closed subgroup of the symmetric group on $T$. In other words, if two elements of $\operatorname{Sym}(S)$ are conjugate in $\operatorname{Sym}(T)$, they are also conjugate in $\operatorname{Sym}(S)$.

Let $p$ be a prime equal to $5$ or more. Then, let $S = \{ 1,2,3,\dots,p-1 \}$ and $T = \{ 1,2,3,\dots,p \}$. Define $S_{p-1} = \operatorname{Sym}(S)$ and $S_p = \operatorname{Sym}(T)$. Then:

• conjugacy-closed: The group $S_{p-1} = \operatorname{Sym}(S)$ is a conjugacy-closed subgroup of the group $S_p = \operatorname{Sym}(T)$ by the above fact.
• Hall: $S_{p-1}$ is also a Hall subgroup (in fact, it is a Hall $p'$-subgroup) of $S_p$.
• Not a retract: For this, observe that for $p \ge 5$, the only normal subgroups of $S_p$ are the whole group, the trivial subgroup, and the alternating group. None of these have order $p$, and thus, $S_{p-1}$ cannot be realized as a retract.