Pronormality is not centralizer-closed

This article gives the statement, and possibly proof, of a subgroup property (i.e., pronormal subgroup) not satisfying a subgroup metaproperty (i.e., centralizer-closed subgroup property).
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Statement

It is possible to have a group $G$ and a pronormal subgroup $H$ of $G$ such that the centralizer $C_{G}(H)$ is not pronormal.

Related facts

• Pronormality is normalizer-closed: In fact, normalizer of pronormal implies abnormal
• Pronormality is not commutator-closed
• Automorph-conjugacy is centralizer-closed
• Isomorph-conjugacy is not centralizer-closed

Proof

Example of the symmetric group

Further information: symmetric group:S4, dihedral group:D8

Suppose $G$ is the symmetric group on the set $\{1,2,3,4\}$ and $H$ is a $2$-Sylow subgroup of $G$, given by:

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

Then, $H$ is a pronormal subgroup of $G$, because it is a Sylow subgroup and Sylow implies pronormal. On the other hand, we have:

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

This is not pronormal, because it is conjugate to the subgroup $\{(),(1,2)(3,4)\}$ via the permutation $(2,3)$, but these are not conjugate in the subgroup they generate.

Let $G$ be the symmetric group on the set $\{1,2,3,4\}$.