Kourovka Notebook Problem 11.50

Definitions and mathematical statement

Let $A$ and $C$ be abelian groups, written additively. The source uses a scalar $n$ without specifying its range; the formulas below interpret $n$ as an integer, with its intended range left unresolved. For an abelian group $G$, write \[ G[n]=\{g\in G:ng=0\},\qquad nG=\{ng:g\in G\}. \] Assume $A[n]=0$, meaning that $na=0$ implies $a=0$ for every $a\in A$. Write $\operatorname{Hom}(C,A)$ for the group of homomorphisms and $\operatorname{Ext}(C,A)=\operatorname{Ext}^{1}_{\mathbb Z}(C,A)$ for the group of equivalence classes of abelian-group extensions \[ 0\longrightarrow A\longrightarrow X\longrightarrow C\longrightarrow 0. \] Here equivalence means an isomorphism of extensions inducing the identity on both $A$ and $C$.

The source supplies the exact sequence \[ 0\longrightarrow \frac{\operatorname{Hom}(C,A)}{n\operatorname{Hom}(C,A)} \xrightarrow{\rho} \operatorname{Hom}(C/nC,A/nA) \xrightarrow{\delta} \operatorname{Ext}(C,A)[n] \longrightarrow 0. \] The reduction map is \[ \rho([\varphi])(c+nC)=\varphi(c)+nA. \] Since $n(A/nA)=0$, precomposition with $C\to C/nC$ gives the canonical identification \[ \operatorname{Hom}(C/nC,A/nA)\cong\operatorname{Hom}(C,A/nA). \] For $f\in\operatorname{Hom}(C,A/nA)$, let $q:A\to A/nA$ be the quotient map and form the pull-back \[ X_f=\{(a,c)\in A\times C:q(a)=f(c)\}. \] The corresponding extension is \[ 0\longrightarrow A\xrightarrow{i_f}X_f\xrightarrow{p_f}C\longrightarrow 0, \qquad i_f(a)=(na,0),\quad p_f(a,c)=c. \] Thus $\delta(f)$ is the class of this extension, using the displayed identification of Hom groups. The pull-back diagram is \[ \begin{array}{ccccccccc} 0&\longrightarrow&A&\xrightarrow{i_f}&X_f&\xrightarrow{p_f}&C&\longrightarrow&0\\ &&\downarrow\scriptstyle{\mathrm{id}_A}&&\downarrow\scriptstyle{(a,c)\mapsto a}&&\downarrow\scriptstyle{f}&&\\ 0&\longrightarrow&A&\xrightarrow{n}&A&\xrightarrow{q}&A/nA&\longrightarrow&0. \end{array} \]

Requested classification. Use this scheme to classify the extensions intended by the source. The source says only "certain extensions of $A$ by $C$" and does not specify their subclass or the desired classification criterion. The displayed diagram fixes the orientation as kernel $A$ and quotient $C$; the classification request remains incomplete beyond that orientation.

The source singles out the case $nC=0$ with $A$ torsion-free, where torsion-free means that $ma=0$, for $m\in\mathbb Z\setminus\{0\}$, implies $a=0$. It states that in this case \[ \operatorname{Ext}(C,A)[n]=\operatorname{Ext}(C,A). \] This special case is described as interesting, without a more specific classification target.

Short English statement

For abelian groups $A,C$ with $A[n]=0$, the supplied exact sequence relates homomorphisms modulo $n$ to extension classes annihilated by $n$, represented by explicit pull-backs. A. Mader asks that this construction be used to classify certain extensions, particularly when $nC=0$ and $A$ is torsion-free. The intended subclass and classification criterion are not specified.

Sources

• Supplied local UnsolvedMath registry record, permanent ID $700742$, exact source problem code catalogue entry, title Kourovka Notebook Problem 11.50. The registry labels the problem open; this is supplied metadata, not an independently verified mathematical or literature assessment.
• The Kourovka Notebook (Unsolved Problems in Group Theory), Problem 11.50, proposed by A. Mader, 11th issue (1990). The supplied background identifies the source edition as the 21st issue edition (2026), arXiv:1401.0300v48. The supplied collection description names E. I. Khukhro and V. D. Mazurov as editors.
• E. L. Lady and A. Mader, J. Algebra 140 (1991), 36--64, cited in the supplied statement. No article title, DOI, or URL is supplied.