Conjugacy-closed and Hall not implies retract

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 fact about::retract. Specifically, there need not be a fact about::normal Hall subgroup that is a complement to it.

Related facts

 * Conjugacy-closed and Sylow implies retract
 * Conjugacy-closed Abelian Sylow implies retract
 * Conjugacy-closed Abelian Hall implies retract

Example of Sylow complements in symmetric groups
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.