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.

Related facts

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 $p\geq 5$ , 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.