Explicit extensions of $\mathbb{Z}/m$ by $\mathbb{Z}/n$ I am trying to determine the extensions of $\mathbb{Z}/m$ by $\mathbb{Z}/n$ for coprime $m,n$. I am unsure as to how to go about finding the extensions explicilty. What's the general procedure? Or what is the procedure demonstrated on some example like $\mathbb{Z}/7$ by $\mathbb{Z}/4$? What about when $m=n$, say m=3?
 A: Any group $E$ that can be place at
$$0\to\mathbb{Z}_m\stackrel{\alpha}\to E\stackrel{\beta}\to\mathbb{Z}_n\to0$$
where the $\alpha$ is injective and $\beta$ is surjective, can be consider an extension of $\mathbb{Z}_m$ by $\mathbb{Z}_n$ and there are, at least, two kinds of solutions, their direct sum
$E=\mathbb{Z}_m\times\mathbb{Z}_n$,
and semi-direct products $E=\mathbb{Z}_m\times_{\varphi}\mathbb{Z}_n$.
A: Welcome to MSE!
Recall an extension $0 \to K \to G \to Q \to 0$ depends not only on the class in $\text{Ext}^1(Q,K)$, but also on the action $Q \curvearrowright K$.
So we first want to understand how $Q$ can act on $K$. For us, that means $\mathbb{Z}/n = \langle x \mid x^n = 1 \rangle$ acting on $\mathbb{Z}/m = \langle y \mid y^m = 1 \rangle$. Of course, what is an action? It's not hard to see that an automorphism of $\mathbb{Z}/m$ is determined by the image of $y$. That is, it must look like $y \mapsto y^k$ for $k$ coprime to $m$.1
Next we look at $\text{Ext}^1(\mathbb{Z}/m,\mathbb{Z}/n)$, which (as we've said) is actually equal to $0$. You can find a proof of this here, but you can find many more examples of this computation by googling around.
Now, the $0$ element of $\text{Ext}^1(Q,K)$ corresponds to the split extension $Q \ltimes K$. Notice again that we had to fix an action of $Q$ on $K$ earlier, so this semidirect product makes sense!
You might leave things here. After all, semidirect products of cyclic groups are well understood. But we can also finish up the computation and get an extremely concrete answer:
If our automorphism is the map $y \mapsto y^k$, then we get the group
$$\langle x, y \mid x^n = 1, y^m = 1, x^{-1}yx = y^k \rangle.$$
Notice our action of $\mathbb{Z}/n$ on $\mathbb{Z}/m$ has become the conjugation action! This is standard. Whenever we have an extension $0 \to K \to G \to Q \to 0$, the action of $Q$ on $K$ becomes the conjugation action in $G$. This is worth remembering when you're doing concrete calculations with group extensions.
Lastly, as the other answerer notes in the comments, we can simplify slightly when we work with abelian groups. If we want the end group $G$ to be abelian, we need to know that $yx = xy$. That is, $x^{-1}yx = y$. So the only option available to us in that setting is the trivial action! Thus the only abelian extension is the direct product $\mathbb{Z}/m \times \mathbb{Z}/n$.

1: Of course, not every $k$ works. For this to be an action of $\mathbb{Z}/n$, we need to know that $\mathbb{Z}/n \to \text{Aut}(\mathbb{Z}/m)$ sends $x \mapsto (y \mapsto y^k)$, which can only happen if $(y \mapsto y^k)$ has order dividing $n$.

I hope this helps ^_^
