Let $G$ be an abelian torsion group. I have to compute the group $$ \text{Ext}^{1}_{\mathbb{Z}}(G, \mathbb{Z}) $$ and to prove the isomorphism $$ \text{Ext}^{1}_{\mathbb{Z}}(G, \mathbb{Z}) \simeq \text{Hom}_{\mathbb{Z}} (G ,\mathbb{Q} /\mathbb{Z} ).$$ I am stucked trying to find a projective resolution of the $\mathbb{Z}$- module $G$ or, equivalently, a short exact sequence involving $G$. Possibly, the key is to use the torsion of the group to find such sequence.
Any hints? Thanks in advance.
Part 1
I don't think there's a clean description of $\operatorname{Ext}_\mathbb{Z}^1(G, \mathbb{Z})$ for arbitrary torsion abelian group $G$. I believe that the isomorphism in the second part is in general the best one can do, but this question has a nice discussion on a structure theorem that might be able to piece it together. But maybe there is a nice answer to what this Ext group is that someone a bit more knowledgeable in algebra than I am can provide.
In the case that $G$ is finitely generated, by the fact that $\operatorname{Ext}_\mathbb{Z}(-, \mathbb{Z})$ splits up over direct sums, it's sufficient to compute $\operatorname{Ext}_\mathbb{Z}(\mathbb{Z}/k\mathbb{Z}, \mathbb{Z})$. I'll let you do this one yourself if you haven't already; start with the projective resolution $0 \to \mathbb{Z} \xrightarrow{k} \mathbb{Z} \to \mathbb{Z}/k\mathbb{Z} \to 0$.
Part 2
Let $\mathbb{Q}/\mathbb{Z} = T$. This isomorphism follows from properties of $\operatorname{Ext}$ plus the assumption that $G$ is torsion:
Consider the short exact sequence $0 \to \mathbb{Z} \to \mathbb{Q} \to T \to 0$. Then it induces a long exact sequence for $\operatorname{Ext}_\mathbb{Z}^*(G, -)$, the first 6 terms of which are
$0 \to \operatorname{Hom}_\mathbb{Z}(G, \mathbb{Z}) \to \operatorname{Hom}_\mathbb{Z}(G, \mathbb{Q}) \to \operatorname{Hom}_\mathbb{Z}(G, T) \to \operatorname{Ext}_\mathbb{Z}^1(G, \mathbb{Z}) \to \operatorname{Ext}_\mathbb{Z}^1(G, \mathbb{Q}) \to \cdots$
Now, $\mathbb{Q}$ is a divisible abelian group, which is equivalent to being an injective $\mathbb{Z}$-module, so the last displayed term is zero. Also, since $G$ is torsion, both $\operatorname{Hom}_\mathbb{Z}(G, \mathbb{Z})$ and $\operatorname{Hom}_\mathbb{Z}(G, \mathbb{Q})$ are zero (although it's really only relevant that the latter is zero). So exactness implies $\operatorname{Hom}_\mathbb{Z}(G, T) \cong \operatorname{Ext}_\mathbb{Z}^1(G, \mathbb{Z})$.
