Kourovka Notebook Problem 21.2

Definitions and mathematical statement

For an element $x$ of a finite group, let $o(x)$ be its order, the least positive integer $m$ with $x^m=1$. A simple group is a nontrivial group whose only normal subgroups are $1$ and the whole group; normality means invariance under conjugation by every group element. Cyclic groups of prime order are included in this definition.

Let $S$ be finite simple and $G$ any finite group. Suppose there exists a bijection of their underlying sets $f:G\to S$ such that \[ o(x)\mid o(f(x))\qquad\text{for every }x\in G. \] Here divisibility is divisibility of positive integers. The question is whether $G$ must itself be simple.

The map $f$ is not required to preserve multiplication, inverses, conjugacy classes, or the identity. In fact the hypotheses imply $f(1_G)=1_S$: the preimage of $1_S$ must have order dividing one. The bijection also implies $|G|=|S|$, but the requested conclusion is simplicity, not an asserted isomorphism $G\cong S$.

The condition matches individual elements without repetitions: each element of $S$ can serve as the image of only one element of $G$. It is stronger than merely demanding that every element order in $G$ divide some element order in $S$, because multiplicities must be matched. A negative answer would require a nonsimple group of the same order as some simple group together with such a complete matching of elements.

Short English statement

If the elements of a finite group can be bijectively matched to those of a simple group with each order dividing its matched order, must the first group be simple?

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, supplied problems.json, record 2511 (KOU-21.2), Kourovka Notebook - New Problems, Issue 21. Catalogue identity and source transcription; CC BY 4.0. https://huggingface.co/datasets/ulamai/UnsolvedMath.
• S2 The Kourovka Notebook, 21st edition, new problems, Problem 21.2. Expanded formulation of the supplied catalogue statement. https://arxiv.org/pdf/1401.0300v47.