Kourovka Notebook Problem 10.60

Definitions and mathematical statement

Let $B$ be an abelian group, written additively, and let $A\leq\operatorname{Aut}(B)$. Under the usual fixed-point-free interpretation of a group of regular automorphisms, the hypothesis is \[ \forall a\in A\setminus\{1\},\qquad \{b\in B:a(b)=b\}=\{0\}. \] The source does not define "regular," so this interpretation requires confirmation. A group $A$ is periodic if every element has finite order. For a prime $p$, a periodic $p$-group is a group in which every element has order $p^n$ for some integer $n\geq0$. Its centre is \[ Z(A)=\{a\in A:ax=xa\text{ for every }x\in A\}. \] The question is whether, for every abelian group $B$ and every periodic subgroup $A\leq\operatorname{Aut}(B)$ satisfying the displayed fixed-point condition, one necessarily has \[ Z(A)\neq\{1\}. \] The parenthetical wording "$p$-group" also raises the corresponding question with $A$ restricted to a periodic $p$-group, for an arbitrary prime $p$. The record does not specify whether this is a separate restricted question or the intended domain of the whole question. It also does not explicitly require $A$ or $B$ to be nontrivial; the intended nontriviality hypotheses require clarification and have not been silently added.

The supplied editors' comment, dated 2001, reports an affirmative answer under the additional hypothesis that $A$ contains an element of order $2$ or $3$, attributing this result to A. Kh. Zhurtov. This is a supplied historical qualification, not an independently verified theorem or a resolution of the unrestricted question.

Short English statement

The problem asks whether every periodic group of regular automorphisms of an abelian group has a nontrivial centre, and mentions a $p$-group version parenthetically. Interpreting regularity as the absence of nonzero fixed points for every nonidentity automorphism requires confirmation, as do the intended nontriviality hypotheses. A supplied 2001 comment reports an affirmative answer when the automorphism group contains an element of order $2$ or $3$.

Sources

• Local UnsolvedMath registry, permanent ID $700675$, source problem code catalogue entry, title "Kourovka Notebook Problem 10.60." The supplied registry labels the problem open; this label is not an independent verification of its present status.
• A. I. Sozutov, problem 10.60, The Kourovka Notebook (Unsolved Problems in Group Theory), 10th issue (1986). The supplied background identifies the 21st issue edition (2026), arXiv:1401.0300v48. The collection description names E. I. Khukhro and V. D. Mazurov as editors. These bibliographic details are supplied attribution and have not been independently checked.
• Editors' comment (2001), citing A. Kh. Zhurtov, Siberian Math. J., volume 41, no. 2 (2000), 268--275, for the reported affirmative answer when $A$ contains an element of order $2$ or $3$. No article title is supplied.