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 $B\cong\mathbb{Z}/{\rm im}(\phi)$

  • $\begingroup$ 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 $\mathbb{Z}$. $\endgroup$ – Max Schattman Feb 25 at 15:35
  • 1
    $\begingroup$ This is correct and $B$ can be any cyclic group $\endgroup$ – Robert Chamberlain Feb 25 at 16:02

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