(a) Consider $0 \rightarrow \mathbb{Z}_4 \xrightarrow{\alpha} \mathbb{Z}_8 \oplus \mathbb{Z}_2 \xrightarrow{\beta} \mathbb{Z}_4 \rightarrow 0$. We wish to determine if there is a choice of $\alpha$ and $\beta$ that makes this into a short exact sequence.

Indeed, pick $\alpha$ to be the map sending $1 \mapsto (2,1)$. Then, $H = \im \alpha = \set{(0, 0), (2, 1), (4, 0), (6, 1)}$. Notice then that $(\mathbb{Z}_8 \oplus \mathbb{Z}_2) / H$ is group of order 4. One can check that the element $(1, 0) + H$ has order 4, so it must generate the group and the quotient is isomorphic to $\mathbb{Z}_4$. This lets us define $\beta$ as the quotient map, giving a short exact sequence.

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(b) Now consider $0 \rightarrow \mathbb{Z}_{p^m} \xrightarrow{\alpha} A \xrightarrow{\beta} \mathbb{Z}_{p^n} \rightarrow 0$, for $p$ prime. We wish to find which abelian groups $A$ admit maps that make the sequence exact. First note that, for this sequence to be exact, we need $\mathbb{Z}_{p^n} \cong A / \mathbb{Z}_{p^m}$. Thus, $A$ must have order $p^n p^m = p^{m+n}$.

A certain class of groups of order $p^{m+n}$ can be constructed as follows: $A_k = \mathbb{Z}_{p^{m+n-k}} \oplus \mathbb{Z}_{p^k}$ for $0 \leq k \leq \min\set{m,n}$. Notice when $k$ is minimal we get $\mathbb{Z}_{p^{m+n}}$ and when $k$ is maximal we get $\mathbb{Z}_{p^m} \oplus \mathbb{Z}_{p^n}$ (or the other way around).

The maps can be explicitly constructed. For $A_k$, let $\alpha$ be the map sending $1 \mapsto (p^{n-k}, 1)$. Then with $H = \im \alpha$ (which is of order $p^m$), we must have $\abs{A_k / H} = p^n$. We claim that $(1,0) + H$ is an element of order $p^n$, and thus the quotient is isomorphic to $\mathbb{Z}_{p^{n}}$. To see this, consider an arbitrary coset in $A_k / H$ represented by $(a,b)$. Notice that $(a,b) - b(p^{n-k}, 1)$ represents the same coset, since we are subtracting a multiple of an element of $H$. But, we have:

$$(a, b) - b(p^{n-k}, 1) = (a-bp^{n-k}, 0) = (a-bp^{n-k})(1, 0)$$

What we've shown is that every coset, and therefore element of $A_k / H$, is represented by a multiple of $(1, 0)$. So, $(1,0) + H$ generates the (cyclic) quotient group, so it must be isomorphic to $\mathbb{Z}_{p^n}$. Then, let $\beta$ be the quotient map and we have a short exact sequence.

So all of the groups $A_k$ are acceptable. To show that these are all the groups (up to isomorphism, of course), we note that $A$ is generated by two (non-unique) elements: $a$ and $b$ such that $\alpha(1) = a$ and $\beta(b) = 1$ (this implies that $A$ must be isomorphic to direct product of two cyclic groups, since $a$ and $b$ necessarily generate cyclic subgroups):

To see this, let $x \in A$ be arbitrary. There are two possibilities: either $x \in \im \alpha$ or not.

Suppose $x \in \im \alpha$. Then $x$ is a multiple of $a$ (since $\im \alpha$ is a cyclic subgroup).

Suppose $x \not\in \im \alpha$. Then we can also conclude that $x \not\in \ker \beta$. This means that $\beta(x)$ is a non-trivial element, and thus a multiple of $\beta(b)$, i.e. $\beta(x) = k \beta(b) = \beta(kb)$. This means that $\beta(x - kb) = 0$, so that $x - kb \in \ker \beta$. Then, $x - kb \in \im \alpha$, so we have $x - kb = ta$. We now conclude that $x = ta + kb$ for some $t, k$. So in both cases, $x$ is expressed as a combination of $a$ and $b$.

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(c) Finally, consider $0 \rightarrow \mathbb{Z} \xrightarrow{\alpha} A \xrightarrow{\beta} \mathbb{Z}_n \rightarrow 0$.

Here, we show explicitly that $A_d \cong \mathbb{Z} \times \mathbb{Z}_d$ for all $d$ such that $d \mid n$ are acceptable. Set $q = n/d$. Let $\alpha$ be the map

$$t \mapsto (t, [tq])$$

Let $\beta$ be the map

$$(x,[y]) \mapsto [y - xq]$$

Since $\beta(t, [tq]) = tq - tq = 0$, we have $\im \alpha \subseteq \ker \beta$.

Let $(t, [y]) \in \ker \beta$. Then, $y-tq = kn$ and $y=kn + tq = k(qd) + tq$. Then, $[y] = [kqd + tq] = [tq]$ (modulo $d$). So, $(t,[y]) = (t, [tq]) \in \im \alpha$. So we have $\ker \beta \subseteq \im \alpha$

Then $\ker \beta = \im \alpha$, and $A_d$ with this choice of $\alpha$ and $\beta$ makes the sequence exact.