Classification of alternating groups that are N-groups

Statement
Suppose $$n$$ is a positive integer greater than or equal to 3. The alternating group $$A_n$$ is a  N-group if and only if $$n \in \{ 3,4,5,6,7 \}$$.

Similarly, the finitary alternating group on an infinite set is never a N-group.

Related facts

 * Classification of symmetric groups that are N-groups

Proof of failure for larger $$n$$
For $$n \ge 8$$, denote by $$A_n$$ the alternating group on the set $$\{ 1,2,3,\dots,n\}$$. Let $$x = (1,2,3)$$. The normalizer $$N_G(\langle x \rangle)$$ contains the centralizer of $$x$$, which in turn contains the alternating group on the set $$\{4,5,\dots,n\}$$, which is isomorphic to $$A_{n-3}$$, which is simple non-abelian since $$n \ge 8 \implies n - 3 \ge 5$$ (see alternating groups are simple). Thus, $$N_G(\langle x \rangle)$$ is not solvable, so $$A_n$$ is not a N-group.