Pronormal not implies NE

This article gives the statement and possibly, proof, of a non-implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., pronormal subgroup) need not satisfy the second subgroup property (i.e., NE-subgroup)
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Statement

A pronormal subgroup of a group need not be a NE-subgroup.

Facts used

1. Sylow implies pronormal
2. Sylow not implies NE

Proof

Examples of Sylow subgroups

The proof that pronormal subgroups need not be NE follows from facts (1) and (2). Further, any example of a Sylow subgroup that is not NE gives an example of a pronormal subgroup that is not NE. Two examples of situations where Sylow subgroups are not NE are the $2$-Sylow subgroup and the $5$-Sylow subgroup in the alternating group of degree five.

Further information: Sylow not implies NE

Example of the symmetric group

Further information: symmetric group:S4

Let $G$ be the symmetric group on the set $\{1,2,3,4\}$, and $H$ be the four-element subgroup $\{(),(1,2,3,4),(1,3)(2,4),(1,4,3,2)\}$. Then, $H$ is a pronormal subgroup, because the subgroup generated by $H$ and any other conjugate of it is the whole group. On the other hand, we have $H^{G}=G$ and $N_{G}(H)$ is a dihedral group of order eight, so $H^{G}\cap N_{G}(H)$ is a dihedral group of order eight, which is strictly bigger than $H$.