Pronormality does not satisfy transfer condition

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 metaproperties satisfied by pronormality

 * Pronormality satisfies image condition
 * Pronormality satisfies intermediate subgroup condition
 * Pronormality is quotient-transitive

Related metaproperties not satisfied by pronormality

 * Pronormality is not finite-intersection-closed
 * Pronormality is not finite-join-closed

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

 * 1) uses::Join with any distinct conjugate is the whole group implies pronormal

Example of the symmetric group
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 $$G$$.