Does the short exact sequence $0\rightarrow \mathbb{Z}_3 \rightarrow G \rightarrow \mathbb{Z}_2\oplus \mathbb{Z} \rightarrow 0$ always split?

Question

Does the short exact sequence $0\rightarrow \mathbb{Z}_3 \rightarrow G \rightarrow \mathbb{Z}_2\oplus \mathbb{Z} \rightarrow 0$ always split?

Answer 1

Let $G=S_6\oplus\mathbb{Z}$ then $\mathbb{Z}_3 < S_6$ is a subgroup index $2$ and so normal with quotent $\mathbb{Z}_2$.

Thus $$0\rightarrow \mathbb{Z}_3\rightarrow S_6\oplus\mathbb{Z} \rightarrow \mathbb{Z}_2\oplus\mathbb{Z} \rightarrow 0$$ is exact but not split since $S_6$ is not Abelian.

Answer 2

In the category of abelian groups this short exact sequence always splits. This is because all SESs of the forms $0\to\Bbb{Z}_3\to A\to\Bbb{Z}_2\to0$ and $0\to\Bbb{Z}_3\to A\to\Bbb{Z}\to0$ always split. The former because $\gcd(3,2)=1$. The latter because $\Bbb{Z}$ is free, hence projective. Also the bifunctor $E(A,B)$ classifying SESs of abelian groups of the form $0\to A\to G\to B\to0$ is additive in both $A$ and $B$.

See for example Hilton-Stammbach or any other textbook on homological algebra for the details.

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In this simple case we can also dispense with the general theory.

It is easy to see that $G$ is necessarily finitely generated, so $G\cong \Bbb{Z}^r\oplus \Bbb{Z}_{n_1}\oplus \cdots\oplus \Bbb{Z}_{n_k}$ with $n_{i+1}\mid n_i$ for all $i$ by the structure theorem of f.g. abelian groups. Here $G$ must have a subgroup isomorphic to $\Bbb{Z}_3$, so $3\mid n_1$. Moding out a finite subgroup won't affect the free part, so $r=1$. A homomorphic image of a torsion element is necessarily torsion, so the torsion part of $G$ will necessarily be mapped into the torsion part of $\Bbb{Z}_2\oplus\Bbb{Z}$, i.e. $\Bbb{Z}_2$. This implies that $G_{torsion}$ is of order three or six. In either case it is cyclic, so $k=1$. Let $x\in G$ be an element mapping onto $(\overline1,0)$. Then $2x$ is in the kernel and, by exactness, has to come from $\Bbb{Z}_3$. This implies that $x$ is a torsion element itself and, furthermore, $2\mid |G_{torsion}|$. Thus $G\cong \Bbb{Z}\oplus\Bbb{Z}_6.$

Shortcuts probably available at selected steps.