Pronormality is not commutator-closed

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

Example of the symmetric group of degree four
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.