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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!

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    $\begingroup$ Write down a set-theoretic splitting, with the constraint of preserving the identity element, and see if this defines a homomorphism of groups. $\endgroup$ – Espen Nielsen Oct 13 '15 at 22:15
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    $\begingroup$ For $\pi$ to be a homomorphism you need the element of order $2$ to be sent to an element of order $1$ or $2$. It needs to be an inclusion if the composition is to be the identity, so the element of order $2$ needs to go to an element of order $2$. Think about what happens to elements of order $4$. $\endgroup$ – Mark Bennet Oct 13 '15 at 22:16
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    $\begingroup$ The group $\mathbb{Z}_4$ cannot be the direct sum of proper subgroups, because it has a single minimal subgroup. $\endgroup$ – egreg Oct 13 '15 at 22:17
  • $\begingroup$ Thank you all for for your comments! $\endgroup$ – Lullaby Oct 14 '15 at 19:06
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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 $\psi$).

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  • $\begingroup$ Dear $Arthur$, than you very much for the great help! $\endgroup$ – Lullaby Oct 14 '15 at 19:06
  • $\begingroup$ Where is the proof for the assertion about abelian groups made in the last paragraph? Also, do you mean $\psi$ instead of $\phi$? $\endgroup$ – Junglemath Apr 24 '19 at 20:22
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    $\begingroup$ @Junglemath I believe I do mean $\psi$ rather than $\phi$. As for your other concern, clearly the direct product splits, so the crux is the other direction. One proof would be to use $\psi$ and $\varphi$, along with the two corresponding splitting homomorphisms to show that the middle element satisfies the universal property of direct products. (In fact, that sounds like a rather fun exercise.) So it isn't necessarily the direct product, and $\psi$ and $\varphi$ are not necessarily the canonical inclusions and projections. But up to unique isomorphism, they are. $\endgroup$ – Arthur Apr 24 '19 at 20:37
  • $\begingroup$ @Junglemath If you are very concerned about this, you could always just post it as a new question. $\endgroup$ – Arthur Apr 24 '19 at 20:47
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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.

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