Pronormality is not commutator-closed

This article gives the statement, and possibly proof, of a subgroup property (i.e., pronormal subgroup) not satisfying a subgroup metaproperty (i.e., commutator-closed subgroup property).
View all subgroup metaproperty dissatisfactions | View all subgroup metaproperty satisfactions|Get help on looking up metaproperty (dis)satisfactions for subgroup properties
Get more facts about pronormal subgroup|Get more facts about commutator-closed subgroup property|

Statement

It is possible to have a group $G$ and pronormal subgroups $H,K$ of $G$ such that the commutator $[H,K]$ is also a pronormal subgroup.

Related facts

• Normality is commutator-closed
• Characteristicity is commutator-closed
• Endo-invariance implies commutator-closed
• Join of subnormal subgroups is subnormal iff their commutator is subnormal

Proof

Example of the symmetric group of degree four

Further information: symmetric group:S4

Let $G$ be the symmetric group on the set $\{1,2,3,4\}$. Let $H=K$ be a $2$-Sylow subgroup of $G$, say:

$H=\{(),(1,2,3,4),(1,3)(2,4),(1,4,3,2),(1,3),(2,4),(1,2)(3,4),(1,4)(2,3)\}$.

Then, $H$ is isomorphic to a dihedral group of order eight. $[H,K]=[H,H]=\{(),(1,3)(2,4)\}$.

• $H$ is pronormal in $G$: In fact, $H$ is a Sylow subgroup of $G$, and Sylow implies pronormal.
• $[H,H]$ is not pronormal in $G$: This subgroup is conjugate to $\{(),(1,2)(3,4)\}$ by the permutation $(2,3)$, but these two subgroups are not conjugate in the subgroup they generate.