Sylow 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., Sylow subgroup) need not satisfy the second subgroup property (i.e., NE-subgroup)
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Statement

A Sylow subgroup of a finite group need not be a NE-subgroup: it need not equal the intersection of its normalizer and normal closure in the whole group.

Facts used

A5 is simple

Related facts

Proof

Example of the alternating group of degree five

Further information: alternating group:A5

Let $G$ be the alternating group on ${1, 2, 3, 4, 5}$ . Then:

Let $H$ be a $2$ -Sylow subgroup of $G$ , for instance, $H = {(), (1, 2) (3, 4), (1, 3) (2, 4), (1, 4) (2, 3)}$ . Since $G$ is simple, the normal closure of $H$ in $G$ is $G$ . The normalizer of $H$ in $G$ is the alternating group on ${1, 2, 3, 4}$ , which is strictly bigger than $H$ . Thus, the intersection of the normalizer and normal closure is strictly bigger than $H$ .
Let $K$ be a $5$ -Sylow subgroup of $G$ , for instance, $K = {(), (1, 2, 3, 4, 5), (1, 3, 5, 2, 4), (1, 4, 2, 5, 3), (1, 5, 4, 3, 2)}$ . Then, the normalizer of $K$ in $G$ is a dihedral group of order ten, which in particular includes double transpositions such as $(2, 5) (3, 4)$ . The normal closure of $K$ in $G$ is $G$ . Thus, the intersection of the normalizer and normal closure is strictly bigger than $K$ .