Pronormality does not satisfy transfer condition

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

It is possible to have a group $G$ , a pronormal subgroup $H$ of $G$ , and a subgroup $K$ of $G$ , such that $H\cap K$ is not pronormal in $K$ .

Related facts

Related metaproperties satisfied by pronormality

Related metaproperties not satisfied by pronormality

Other facts relating pronormality and the transfer condition

Normality satisfies transfer condition

Related properties

Transfer-closed pronormal subgroup: A subgroup whose intersection with any other subgroup is pronormal in that other subgroup.

Facts used

Join with any distinct conjugate is the whole group implies pronormal

Proof

Example of the symmetric group

Further information: symmetric group:S4

Let $G$ be the symmetric group on the set $\{1,2,3,4\}$ . Consider subgroups $H,K$ of $G$ as follows:

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

Then, we have:

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

Note that:

$H\cap K$ is not pronormal in $K$ : Indeed, the subgroup $\{(),(1,4)(2,3)\}$ is conjugate to it in $K$ , but not in the subgroup they generate. Another way of seeing this is that $H\cap K$ is 2-subnormal but not normal in $K$ , hence it cannot be pronormal in $K$ .
$H$ is pronormal in $G$ : $H$ is a subgroup whose join with any distinct conjugate is the whole group. Thus, by fact (1), $H$ is pronormal in $G$ .