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.

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 $H$.