Exact sequence splits? I am stuck in the following problem:
Show that every group of order 4 is an extension of $\mathbb{Z}_{2}$ by $\mathbb{Z}_{2}$. Which of the exact sequences splits?
Ok, i have to consider two cases for a group of order 4: $\mathbb{Z}_{4}$ and $\mathbb{Z}_{2}\times \mathbb{Z}_{2}$, so i get two exact sequences:
$0\rightarrow \mathbb{Z}_{2}\overset{\psi }{\rightarrow}\mathbb{Z}_{4}\overset{\varphi }{\rightarrow}\mathbb{Z}_{2}\rightarrow 0$  and $0\rightarrow \mathbb{Z}_{2}\overset{\tilde{\psi} }{\rightarrow}\mathbb{Z}_{2}\times \mathbb{Z}_{2}\overset{\tilde{\varphi} }{\rightarrow}\mathbb{Z}_{2}\rightarrow 0$
For both cases i checked the conditions of exact sequence and proved that these are exact sequences. Now i have a problem how to show if they split or not. 
We say that an exact sequence splits if for the first sequence there exists a group homomorphism $\pi :\mathbb{Z}_{2}\rightarrow \mathbb{Z}_{4}$ with $\varphi \circ \pi =id_{\mathbb{Z}_{2}}$ and for the second $\tilde{\varphi} \circ \tilde{\pi} =id_{\mathbb{Z}_{2}}$, right? But how to show if these group homomorphisms exist or not?
Can anybody help me, please? I would appreciate any hints and comments.
Thank you in advance!

Suppose that the exact sequence: $$0\rightarrow \mathbb{Z}_{2} \overset{\psi }{\rightarrow} \mathbb{Z}_{4} \overset{\varphi }{\rightarrow} \mathbb{Z}_{2}\rightarrow 0$$
is split, then the group $\mathbb {Z}_{4}$ must be the semi-direct product of $\mathbb{Z}_{2}$ by $\mathbb{Z}_{2}$. Since the only automorphism of $\mathbb{Z}_{2}$ is the identity so the action would be trivial. Hence $\mathbb {Z}_{4}$ becomes  the direct product $\mathbb{Z}_{2}\times \mathbb{Z}_{2}$.But this is impossible.
