Sylow not implies NE

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) uses::A5 is simple

Related facts

 * Pronormal not implies NE
 * Weakly normal not implies NE

Example of the alternating group of degree five
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 $$K$$.