I am trying to solve the following problem:
Determine which abelian groups $A$ can appear as central terms in a short exact sequence $\mathbb{Z} \to A \to \mathbb{Z} \oplus \mathbb{Z}_5$
What I've done so far is the following:
Calling the first map $i$ and the second $\pi$ I know that im($i$)=ker($\pi$) so if I call $i(\mathbb{Z})=H \subset A$ I know that
$$ \mathbb{Z} \oplus \mathbb{Z}_5 \cong A/H $$
This condition is clearly satisfied considering $A = \mathbb{Z} \oplus \mathbb{Z} \oplus \mathbb{Z}_5$ but I don't know how to determine if this is the only group which makes the sequence exact or if there are more.
I thought of trying to get conditions to know if the sequence splits or not but I haven't been able to get far.
How should I proceed?
Thanks to @DanielFischer comment I was able to get the answer by myself.
Considering $i(1)=(5,0)$ then the $A= \mathbb{Z} \oplus \mathbb{Z}$ is another valid option. These are the two only valid options. Let's see the intuition behind this.
$A = LT(A) \oplus T(A)$ where $T(A)$ is the torsion part of $A$. We can racionalize the sequence and taking into account that racionalizing preserves exact sequences we get:
$$ \mathbb{Q} \to A_\mathbb{Q} \to \mathbb{Q}$$
The exactness of this sequence gives us that the rank of A must be equal to the rank of $\mathbb{Z} \oplus \mathbb{Z}$.
If the torsion part of $A$ is 0 we have to generate it when taking the quotient with respect to $i(\mathbb{Z})$ this allows us to prove that if $i(1) \neq 5$ we don't get $\mathbb{Z}_5$.
On the other hand, if $T(A) \neq 0$ then when taking the quotient with respect to $\mathbb{Z}$ we get the first group proposed in the answer as the only chance.