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)
View a complete list of subgroup property non-implications | View a complete list of subgroup property implications
Get more facts about conjugacy-closed Hall subgroup|Get more facts about Hall retract

EXPLORE EXAMPLES YOURSELF: View examples of subgroups satisfying property conjugacy-closed Hall subgroup but not Hall retract|View examples of subgroups satisfying property conjugacy-closed Hall subgroup and Hall 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 Retract (?). Specifically, there need not be a 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

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 $\mathrm{S}\mathrm{y}\mathrm{m}(S)$ are conjugate in $\mathrm{S}\mathrm{y}\mathrm{m}(T)$, they are also conjugate in $\mathrm{S}\mathrm{y}\mathrm{m}(S)$.

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 $S_{p-1}=\mathrm{S}\mathrm{y}\mathrm{m}(S)$ and $S_p=\mathrm{S}\mathrm{y}\mathrm{m}(T)$. Then:

• conjugacy-closed: The group $S_{p-1}=\mathrm{S}\mathrm{y}\mathrm{m}(S)$ is a conjugacy-closed subgroup of the group $S_p=\mathrm{S}\mathrm{y}\mathrm{m}(T)$ by the above fact.
• Hall: $S_{p-1}$ is also a Hall subgroup (in fact, it is a Hall $p^{\prime }$-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.