Pronormality is not finite-intersection-closed

Statement
An intersection of two pronormal subgroups of a group need not be pronormal.

Pronormal subgroup
A subgroup $$H$$ of a group $$G$$ is termed pronormal in $$G$$ if, given any $$g \in G$$, $$H$$ and $$g^{-1}Hg$$ are conjugate in the subgroup they generate.

Facts used

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

The example of the symmetric group
Let $$G = S_4$$ be the symmetric group on the set $$\{1,2,3,4\}$$. Let $$K = \{, (1,2)(3,4), (1,3)(2,4), (1,4)(2,3) \}$$ be the subgroup comprising the identity and double transpositions. Let $$H = \{, (1,2), (3,4), (1,2)(3,4) \}$$ be the subgroup generated by two disjoint single transpositions. Then, $$H \cap K = \{, (1,2)(3,4) \}$$ is a two-element subgroup.


 * $$K$$ is pronormal: In fact, $$K$$ is a normal subgroup of $$G$$.
 * $$H$$ is pronormal: Any conjugate of $$H$$ is either equal to $$H$$ or intersects $$H$$ trivially, in which case they generate the whole group (in other words, $$H$$ is a subgroup whose join with any distinct conjugate is the whole group). Thus, $$H$$ is pronormal in $$G$$ (See fact (1)).
 * $$H \cap K$$ is not pronormal: Indeed, this subgroup and a conjugate of it generate the subgroup $$K$$, within which they are not conjugate.