# 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

1. A5 is simple

## 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 $K$.