# Pronormality is not finite-intersection-closed

This article gives the statement, and possibly proof, of a subgroup property (i.e., pronormal subgroup) not satisfying a subgroup metaproperty (i.e., finite-intersection-closed subgroup property).This also implies that it does not satisfy the subgroup metaproperty/metaproperties: Intersection-closed subgroup property (?), .
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## Statement

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

## Definitions used

### Pronormal subgroup

Further information: 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. Join with any distinct conjugate is the whole group implies pronormal

## Proof

### The example of the symmetric group

Further information: symmetric group:S4

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.