Why is $V_{4}$ the semi direct product of $Z_{2}$× $Z_{2}$

Question

I'm trying to understand what is a semi direct product , so by the definition semi-direct product of G , I'd need two groups , $N$ and $H$ , where :

1. $H∩N$ = {e}
2. $H \cdot N$ = $G$

If $H=N=Z_{2}$ , then : $Z_{2}∩Z_{2}≠{e}$ .

Which contradicts $H∩N$ = {e} . So what am I missing here ?

Why does $V_{4}$ is indeed the semi direct product of $Z_{2}$× $Z_{2}$ ?

Thanks

Answer 1

Actually, here, the idea could be stated more concisely this way:

Show that there exist subgroups $H,K\leq V_4$ such that $H,K\cong \mathbb{Z}_2$ and $V_4$ is the semi-direct product of $H$ and $K$.

Answer 2

I think if you consider the presentation of Klein four-group $V$ as follows, then the problematic points get solved easily. This is one presentation of it: $$V=\langle a,b\mid a^2=b^2=1,(ab)^2=1\rangle$$ The relation $(ab)^2=1$ can be regarded as $abab=1$ or $aba^{-1}=b^{-1}=b$. It shows that if we set $H=\langle b\rangle$ then $H$ is normal in $V$ (However, we already know that this group is abelian and maybe we don't need this result anymore). Now $V/H$ is defined as $$V/H=\langle a,b\mid a^2=b^2=1,(ab)^2=1, b=1\rangle=\langle a\mid a^2\rangle=K$$ so $K$ is a complement of $H$ in $G$ and since $H\cap K=1$ and $HK=V$ so we have $$V=H\times K\cong\mathbb Z_2\times \mathbb Z_2$$