This is an exercise from Algebraic Topology book of Hatcher:
Exercise 2.1.14, pg. 132: Determine whether there is a short exact sequence $ 0 \rightarrow \mathbb{Z}_4 \xrightarrow{f} \mathbb{Z}_8 \oplus \mathbb{Z}_2 \xrightarrow{g} \mathbb{Z}_4 \rightarrow 0$. More generally, which abelian groups $A$ fit into a short exact sequence $0 \rightarrow \mathbb{Z}_{p^m} \xrightarrow{f} A \xrightarrow{g} \mathbb{Z}_{p^n} \rightarrow 0 $ where $p$ prime. What about $0 \rightarrow \mathbb{Z} \xrightarrow{f} A \xrightarrow{g} \mathbb{Z}_n \rightarrow 0$.
The first part of this question was profoundly discussed on Determining whether there is a short exact sequence and there is a proof here.
Now, I have first tried to clarify the algebraic explanation of this question: This exercise can be totally solved when the following two lemmas are proved.
Lemma 1. For an abelian group $A$, t.f.a.e.
(i) $0 \rightarrow \mathbb{Z}_{p^m} \to A \to \mathbb{Z}_{p^n} \to 0 $.
(ii) $A \cong \mathbb{Z}_{p^k} \times \mathbb{Z}_{p^{m+n-k}}$ where $0 \leq k \leq min(m,n)$.
Lemma 2. For an abelian group $A$, t.f.a.e.
(i) $0 \to \mathbb{Z} \to A \to \mathbb{Z}_n \to 0 $.
(ii) $A \cong \mathbb{Z}_d \times \mathbb{Z}$ where $d \vert n$.
The sketch of the proof of Lemma 2: (i) $\Rightarrow$ (ii) By the pullback along the $\mod d$ reduction homomorphism and Snake lemma, we obtain a short exact sequence $$0 \rightarrow \mathbb{Z} \xrightarrow{\left[\begin{matrix} r \\ n \end{matrix}\right]} \mathbb{Z} \times \mathbb{Z} \to A \to 0 $$ for some integer $r$. Since the matrix $\left[\begin{matrix} d \\ 0 \end{matrix}\right]$ where $d=gcd(r,n)$, $A \cong \mathbb{Z}_d \times \mathbb{Z}$.
(i) $\Leftarrow$ (ii) If $d$ divides $n$, $$0 \rightarrow \mathbb{Z} \xrightarrow{\left[\begin{matrix} 1 \\ n/d \end{matrix}\right]} \mathbb{Z}_d \times \mathbb{Z} \to \mathbb{Z}_n \to 0 $$ is a short sequence. Therefore;
$$\frac{\mathbb{Z}_d \times \mathbb{Z}}{(1,n/d)\mathbb{Z}} \cong \mathbb{Z}_n.$$
As far as I understood, this question seems only algebraic meaning. Can anyone see and say something more?