Determine which abelian groups $A$ fit into a short exact sequence ${0\rightarrow \mathbb{Z}_{p^m} \xrightarrow{\phi} A \xrightarrow{\psi} \mathbb{Z}_{p^n} \rightarrow 0}$ with $p$ prime.
At the moment, I'm not sure how to rule out the possibility that $A$ is infinite. There are other proofs online, but I want to see if this one works. I'm most concerned about my computation of the Smith normal form because I wasn't aware of the technique until a few hours ago.
In order for the sequence to be exact, we need $\phi$ to be injective and $Im(\phi)=ker(\psi)$, which means that $A$ must contain $\mathbb{Z}_{p^m}$ as a subgroup of index $p^n$ and $A/\mathbb{Z}_{p^m}$ must be isomorphic to $\mathbb{Z}_{p^n}$.
If $A$ is finite abelian, then we immediately know that $|A|=p^{m+n}$. We can immediately throw out any groups comprised of $n$ summands where $n\geq 3$ because taking the quotient by the image of our injective map from a cyclic group will eliminate at most one maximal orbit (By "maximal orbit," I mean an orbit which is not properly contained in another), leaving the remaining $n-1$ disjoint orbits be, meaning the quotient group is not cyclic by definition. So we need only concern ourselves with groups of two summands or less.
Let $A=min(a,b)$ and suppose $min(m,n)<A<m+n-min(m,n)$. Suppose there is an element $(a,b)\in \mathbb{Z}_{p^{A}}\oplus \mathbb{Z}_{p^{m+n-A}}$ such that $\langle (a,b)\rangle\cong \mathbb{Z}^m$, then it follows that $$ap^m \equiv p^{A} \equiv 0 \;\; \Rightarrow \;\; a \equiv p^{A-m \; (\text{mod}\; A)}$$ $$bp^m\equiv p^{n+m-A} \equiv 0 \;\; \Rightarrow \;\; b=p^{n-A\; (\text{mod}\; m+n-A)}$$ Henceforth, we drop the "mod" notation in the exponents. It's clear that neither $(p^{A-m \; (\text{mod}\; A)},0)$ nor $(0, p^{n-A\; (\text{mod}\; m+n-A)})$ will give the desired quotient, so we turn our eyes to the final possibility: $(a,b)\equiv(p^{A-m},p^{n-A})$. Let's denote the group it generates by $G$. We now use the Smith normal form to compute $$\mathbb{Z}_{p^{A}}\oplus \mathbb{Z}_{p^{m+n-A}}/\langle (p^{A-m},p^{n-A}) \rangle $$ so we can see if this quotient is $\mathbb{Z}_{p^n}$.
$$ \begin{pmatrix} p^A & 0 \\ 0 & p^{m+n-A} \\ p^{A-m} & p^{n-A} \end{pmatrix} \xrightarrow{\text{R1-}p^m\text{R3}} \begin{pmatrix} 0 & -p^{m+n-A} \\ 0 & p^{m+n-A} \\ p^{A-m} & p^{n-A} \end{pmatrix} \xrightarrow{\text{swap R1 and R3}} \begin{pmatrix} p^{A-m} & p^{n-A}\\ 0 & p^{m+n-A} \\ 0 & -p^{m+n-A} \end{pmatrix} $$
$$ \xrightarrow{\text{R2+R3}} \begin{pmatrix} p^{A-m} & p^{n-A}\\ 0 & p^{m+n-A} \\ 0 & 0 \end{pmatrix} \xrightarrow{R1-p^mR2} \begin{pmatrix} p^{A-m} & 0\\ 0 & p^{m+n-A} \\ 0 & 0 \end{pmatrix} $$
This final matrix corresponds to the group $\mathbb{Z}_{p^{A-m}}\oplus \mathbb{Z}_{p^{m+n-A}}$, which is decidedly not isomorphic to $\mathbb{Z}_n$. Hence, the sequence cannot be made exact for $min(n,m)<A<m+n-min(m,n)$.
However, if we require $0 \leq A \leq min(m+n)$, we can define the injective homomorphism $$\phi: \mathbb{Z}_{p^m} \rightarrow \mathbb{Z}_{p^A}\oplus \mathbb{Z}_{p^{m+n-A}}$$ $$\phi(1)=(1,p^{n+m-min(m,n)-A})$$ For brevity we denote $G=\langle (1,p^{n+m-min(m,n)-A})\rangle$ We can show that the quotient group is cyclic by letting $(a,b)$ be some element of ${\mathbb{Z}_{p^{A}}\oplus \mathbb{Z}_{p^{m+n-A}}}$ whose associated coset is not trivial and considering the difference $$(a,b)-b(1,1)=(a-b,0)$$ We then determine whether there is some $K$ such that $$(a-b,0)\equiv (Kp^A+a-b, (Kp^A+a-b)p^{m+n-min(m,n)-A})$$ solving for $K$ gives us $K=p^{m+n-min(m,n)-2A}$, so $(a-b,0)\in G$. This means that every coset is generated by $(1,1)+G$, which means that the quotient of $\mathbb{Z}_{p^A}\oplus \mathbb{Z}_{p^{m+n-min(m,n)-A}}$ is cyclic by definition.