Exact sequence splits?

Let's take $$0\rightarrow \mathbb{Z}_{2} \overset{\psi }{\rightarrow} \mathbb{Z}_{4} \overset{\varphi }{\rightarrow} \mathbb{Z}_{2}\rightarrow 0$$ and assume there is a splitting homomorphism $\pi:\Bbb Z_2 \to \Bbb Z_4$. What are the possible $\pi$? Well, $\pi(0) = 0$ is a given, and $\pi(1) \neq 0$ if we want $\varphi\circ\pi = id_{\Bbb Z_2}$. But we need to have $\pi(1) + \pi(1) = \pi(1+1) = \pi(0) = 0$ in $\Bbb Z_4$, which forces $\pi(1) = 2$.

Now, what does this say about the possibility of a splitting? Well, we want $$1 = id_{\Bbb Z_2}(1) = \varphi(\pi(1)) = \varphi(2) = \varphi(1+1) = \varphi(1) + \varphi(1) $$ (as elements of $\Bbb Z_2$). But this is impossible. This means that our assumption that there is such a $\pi$ must be false, and the sequence does not split.

On the other hand, for the sequence with the direct sum, it does split. Exactly how it splits depends on the details of $\bar\psi$ and $\bar\varphi$, but let's say that $\bar\psi(1) = (1, 0)$ and $\bar\phi(a, b) = b$. In that case, $\bar\pi(1) = (0, 1)$ works as a splitting homomorphism.

In general, when dealing with abelian groups, an exact sequence splits iff the middle term is the direct product of the right and left terms, $\psi$ is the corresponding inclusion and $\varphi$ is the corresponding projection. $\pi$ will be the corresponding inclusion ("opposite" of $\psi$).

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.