Exact sequence splits?

Question

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!

Answer 1

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 $\psi$).

Answer 2

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.