Kourovka Notebook Problem 21.97

Definitions and mathematical statement

Shared notation.

$\mathbb N=\{1,2,\ldots\}$, and $\mathbb Z,\mathbb Q,\mathbb R,\mathbb C$ denote the integers, rationals, reals and complex numbers. Any other conventions are specified locally.

For a finite group $G$, $\operatorname{Aut}(G)$ is the group of all bijective homomorphisms from $G$ to itself. Its order and $|G|$ are positive integers. No commutativity, simplicity, or restriction on the order of $G$ is imposed.

Statement recorded in the supplied source.

Is \[ \left\{\frac{|\operatorname{Aut}(G)|}{|G|}:G\text{ is a finite group}\right\} =\mathbb Q_{>0}? \] Equivalently, for every positive rational $r$, does some finite group satisfy $|\operatorname{Aut}(G)|/|G|=r$ exactly? Density of this set in the nonnegative real numbers would be a weaker assertion. [S2]

Short English statement

Can every positive rational occur exactly as a finite group’s automorphism-group order divided by its own order?

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, UnsolvedMath: A Curated Collection of Open Mathematics Problems, record 2606 (KOU-21.97). CC BY 4.0. https://huggingface.co/datasets/ulamai/UnsolvedMath.
• S2 M. Tărnăuceanu and R. Molinier, Problem 21.97, in E. I. Khukhro and V. D. Mazurov (eds.), The Kourovka Notebook: Unsolved Problems in Group Theory, 21st edition, arXiv:1401.0300v48 (6 October 2026). This version records an affirmative answer; the displayed question is the historically registered target. https://arxiv.org/abs/1401.0300v48.