Similar short exact sequences where the first abelian group is different

Question

I am trying to get a better understanding of short exact sequences of abelian groups. I know that if

$0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$

and

$0 \rightarrow A' \rightarrow B' \rightarrow C' \rightarrow 0$

are short exact sequences, having two 'column isomorphisms' (i.e. $A$ isomorphic to $A'$, $B$ isomorphic to $B'$, or $C$ isomorphic to $C'$) does not imply that we also have the isomorphism that is missing. However, I am struggling to find a counterexample for the case where $B$ is isomorphic to $B'$ and $C$ is isomorphic to $C'$. What are two groups $A$ and $A'$ such that they complete the short exact sequences while not being isomorphic?

Answer 1

Any abelian group with two non-isomorphic subgroups having isomorphic quotients would work.

This could happen even for finite abelian groups.

Let $B=\mathbb{Z}_4 \times \mathbb{Z}_2$ and $A$ and $A'$ be the subgroups $\mathbb{Z}_4 \times \{0\}$ and $2\mathbb{Z}_4 \times \mathbb{Z}_2$ respectively. Then, $B/A$ and $B/A'$ are both isomorphic to $\mathbb{Z}_2$, but $A$ and $A'$ are not themselves isomorphic.

So, the short exact sequences $0 \to \mathbb{Z}_4 \times \{0\} \to \mathbb{Z}_4 \times \mathbb{Z}_2 \to \mathbb{Z}_2 \to 0$ (with the second map being the projection map) and $0 \to 2\mathbb{Z}_4 \times \mathbb{Z}_2 \to \mathbb{Z}_4 \times \mathbb{Z}_2 \to \mathbb{Z}_2 \to 0$ (with the second map reducing the first coordinate mod $2$) are what you are looking for.