# 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).
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## 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.

## 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.