$H \lhd N \lhd G$ but $H \ntriangleleft G$ [duplicate]

Question

This question already has an answer here:

• Are normal subgroups transitive? 3 answers

I got stuck on the following exercise:

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Find a group $G$, a subgroup $N$, a subgroup of this subgroup $H$ such that $$ H \lhd N \lhd G \quad \text{but} \quad H \ntriangleleft G $$

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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?

Answer 1

Consider $S_3×S_3$ $$H⊲A_3×A_3⊲S_3×S_3$$ Where $$H=\{(1,1),((123),(123)),((132),(132))\}$$ is a simple counter example.

Answer 2

On top of my head:

1. $A_4=(\Bbb Z/2\times \Bbb Z_2)\rtimes \Bbb Z/3$ with $G=A_4$, $N=\Bbb Z/2\times \Bbb Z/2$, $N=\Bbb Z/2$.

2. The Heisenberg group: $G=\langle a,b: a^p=b^p=[a,b]^p=[a,[a,b]]=[b,[a,b]]=1\rangle$ with $N=\langle a,[a,b]\rangle$ (abelian) and $H=\langle a\rangle$.

Answer 3

Try in $S_4$, $H=\{(1),(12)(34)\} \lhd V_4 = N = \{(1),(12)(34),(13)(24),(14)(23)\} \lhd S_4$. $H$ is certainly not normal in $S_4$. Can you show that?