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 $K$.

Facts used

1. 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 $G$.