If $N$ is a normal subgroup of semidirect product group of $N$ and $H$ where $H$ is generated by $a$ and $b$, is $ab=ba$?

Question

I need help to get the idea of the presentation of semidirect products groups. I got this question from an example the $A_4$ is a semidirect product of $N=\langle a,b\rangle$ and $\langle (1 2 3)\rangle$.

If $N$ is a normal subgroup of semidirect product group of $N$ and $H$ where $N$ is generated by $a$ and $b$, is $ab=ba$? If so why?

I know if $H$ and $N$ are both normal and their intersection is trivial, then $hn=nh$ for all $h$ in $H$ and $n$ in $N$, but not sure why in semidirect case.

regards

Answer 1

In the interest of adding a proper answer so that future users know how to spend their time:

The symmetric group $S_5 = C_2 \ltimes A_5$ is a counterexample, since $A_5 = N = \langle(1,2,3), (1,2,3,4,5)\rangle$ yet is nonabelian.

---

I hope this helps ^_^

Answer 2

Let $n\ge3$.

$S_n=\langle(12),(123\dots n)\rangle$.

And any index two subgroup is normal.

So $S_n\rtimes\Bbb Z_2$ will do as a counterexample.