# An exact sequence of abelian groups

Consider an exact sequence of abelian groups $$0 \to A \to \mathbb{Z} \oplus \mathbb{Z} \to \mathbb{Z} \to B \to 0,$$ where we make no assumption on the map $$\mathbb{Z} \oplus \mathbb{Z} \to \mathbb{Z}$$. Is it possible to determine $$A$$ and $$B$$ up to isomorphism of abelian groups from this data?

No, $$A$$ and $$B$$ depend on the map $$\phi:\mathbb{Z}\oplus\mathbb{Z}\to\mathbb{Z}$$.
If $$\phi$$ is trivial then $$(A,B)=(\mathbb{Z}\oplus\mathbb{Z},\mathbb{Z})$$ and if for example $$\phi$$ is the projection onto one copy of $$\mathbb{Z}$$ then $$(A,B)=(\mathbb{Z},1)$$ (where $$1$$ is the trivial group).
These are not the only possibilities, we can take any homomorphism $$\phi$$, then $$A\cong\ker(\phi)$$ and $$B\cong\mathbb{Z}/{\rm im}(\phi)$$
• ker$(\phi)$ is a subgroup of $\mathbb{Z}\oplus \mathbb{Z}$, does this imply that it is necessarily free abelian of rank at most two? In that case, the kernel is either $0, \mathbb{Z}$, or $\mathbb{Z} \oplus \mathbb{Z}$. It can't be $0$ since the kernel of $\mathbb{Z} \to B$ would have to be $\mathbb{Z}\oplus \mathbb{Z}$, which is not a subgroup of $\mathbb{Z}$. – Max Schattman Feb 25 at 15:35
• This is correct and $B$ can be any cyclic group – Robert Chamberlain Feb 25 at 16:02