I got stuck on the following exercise:
Find a group $G$, a subgroup $N$, a subgroup of this subgroup $H$ such that $$ H \lhd N \lhd G \quad \text{but} \quad H \ntriangleleft G $$
This is what I tried to solve it. I knew that
- $\{e\} \ \neq \ H \neq N \neq G $
- $G$ is not abelian.
I didn't know if a had to look for a finite group or rather an infinite one. I was thinking about taking $G= S_n$, $H = A_n$, and $|H| = \frac{n!}{4}$ for some convenient $n$. I hoped it would be easy to look for an even element $h$ of order $\frac{n!}{4}$, so that we can put $H = \langle h \rangle$. Do you think that it would be a good approach?