Calculating $\operatorname{Ext}$ in special cases. Is there a set of "methods" for calculating $\operatorname{Ext}$ in some special cases? For instance, I would be interested in calculating $\operatorname{Ext}_{\mathbb{Z}}^n (\mathbb{Z}/4\mathbb{Z}, \mathbb{Z}/3\mathbb{Z})$. Is there a general way for $\operatorname{Ext}_{\mathbb{Z}}^n (\mathbb{Z}/k\mathbb{Z}, \mathbb{Z}/m\mathbb{Z})$?
 A: Consider the exact sequence
$$\def\Z{\mathbb{Z}}\def\H{\operatorname{Hom}_{\Z}}
0\to \Z\xrightarrow{\mu_k}\Z\to\Z/k\Z\to0
$$
where $\mu_k$ is “multiplication by $k$”. Applying the functor $\H(-,\Z/m\Z)$ gives the exact sequence
\begin{multline}
0\to\H(\Z/k\Z,\Z/m\Z)\to
\H(\Z,\Z/m\Z)\xrightarrow{\mu_k}\H(\Z,\Z/m\Z)\to\\
\to\operatorname{Ext}_{\Z}^1(\Z/k\Z,\Z/m\Z)\to
0=\operatorname{Ext}_{\Z}^1(\Z,\Z/m\Z)
\end{multline}
that can be rewritten as
$$
0\to\H(\Z/k\Z,\Z/m\Z)\to
\Z/m\Z\xrightarrow{\mu_k}\Z/m\Z\to
\operatorname{Ext}_{\Z}^1(\Z/k\Z,\Z/m\Z)\to0
$$
Now it's just a matter of computing the cokernel of multiplication by $k$ on $\Z/m\Z$. Its image is $(m\Z+k\Z)/m\Z=d\Z/m\Z$, where $d=\gcd(k,m)$, so the cokernel is $\Z/d\Z$. Thus
$$
\operatorname{Ext}_{\mathbb{Z}}^1(\Z/k\Z,\Z/m\Z)
\cong
\Z/\gcd(k,m)\Z
$$
In the particular case of $k=4$ and $m=3$, the Ext group is zero: any extension by $\Z/4\Z$ of $\Z/3\Z$ is split, which can also be checked directly.
The higher Ext groups are zero, since $\Z$ is hereditary.
