Why $( Z_3\rtimes Z_2)\times Z_2 \cong (Z_3\times Z_2)\rtimes Z_2$?

Question

I got an explanation, it says as $Z_2$ is in the kernel of the homomorphism. But I can't understand from that. Also can you tell me why $Z_3\rtimes Z_2\cong S_3$ ? Thank you.

Answer 1

$G=(Z_3\rtimes Z_2)\times Z_2\cong S_3\times Z_2$, Now Let $H=A_3 \times Z_2$ which is clearly a subgroup of $G$ and it is also normal.

Set $K=<(1,2)>\times1$ then you can observe that $H\cap K=e$ and $G=HK$ so $G\cong H \rtimes K$. We are done.

Note: If $G=H\rtimes K$ then semidirect product depends on the function $\phi: K\to Aut(H)$ (So it is normally mentioned)

Why $S_3=Z_3\rtimes Z_2$ ?

It normally depends on $\phi$ as I said but assume $\phi$ is nontrivial then such $\phi$ is uniqe as $\phi : Z_2\to Aut(Z_3)\cong Z_2$.

Now, $A_3\cap <(1,2)>=e$ and $A_3$ is normal in $S_3$ and $S_3\cong A_3 \rtimes<(1,2)>$ and since $\phi$ is uniqe, this group is $S_3$.