11.50 (1990)

Let $A, C$ be abelian groups. If $A[n] = 0$, i. e. for $a \in A, na = 0$ implies $a = 0$, then the sequence
$$\frac{\operatorname{Hom}(C, A)}{n\operatorname{Hom}(C, A)} \rightarrowtail \operatorname{Hom}\left(\frac{C}{nC}, \frac{A}{nA}\right) \twoheadrightarrow \operatorname{Ext}(C, A)[n]$$ is exact. Given $\displaystyle f \in \operatorname{Hom}\left(C, \frac{A}{nA}\right)=\operatorname{Hom}\left(\frac{C}{nC}, \frac{A}{nA}\right)$ the corresponding extension $X_f$ is obtained as a pull-back:
$$\begin{array}{ccccc} A & \rightarrowtail & X_f & \twoheadrightarrow & C \\ \downarrow & & \downarrow & & \downarrow \\ A & \rightarrowtail & A & \twoheadrightarrow & \frac{A}{nA} \end{array}.$$ Use this scheme to classify certain extensions of $A$ by $C$. The case $nC = 0, A$ being torsion-free is interesting. Here $\operatorname{Ext}(C, A)[n] = \operatorname{Ext}(C, A)$. (See E. L. Lady, A. Mader, J. Algebra, 140 (1991), 36–64.)