Kourovka Notebook Problem 21.95

Definitions and mathematical statement

A finite group $S$ is nonabelian simple if it is nonabelian, nontrivial, and has no proper nontrivial normal subgroup. Its center is trivial, so identify $S$ with its inner automorphism subgroup inside $\operatorname{Aut}(S)$. A finite group $G$ is almost simple if for some such $S$ it can be identified with a subgroup satisfying $S\le G\le\operatorname{Aut}(S)$. Require in this problem that $S<G$, so that $G$ itself is not simple.

For any finite group $X$, form a graph with vertex set the primes dividing $|X|$, joining distinct $p,q$ when $X$ has an element of order $pq$. Denote its unlabelled isomorphism type by $\overline{\operatorname{GK}}(X)$: graph isomorphisms may permute the prime labels freely.

Does there exist an almost simple but nonsimple finite group $G$ such that \[ \forall\text{ finite groups }H,\quad \overline{\operatorname{GK}}(H)\cong \overline{\operatorname{GK}}(G)\ \Longrightarrow\ H\cong G? \] Thus the requested group must be uniquely determined by the abstract graph alone. The comparison is against every finite group, not only other extensions of the same simple subgroup $S$. The existence quantifier permits any nonabelian simple $S$ and any proper intermediate extension up to its full automorphism group.

Short English statement

Can an almost simple but nonsimple finite group be uniquely identified by the unlabelled shape of its prime graph?

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, supplied problems.json, record 2604 (KOU-21.95), 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, version 47 (30 September 2026), new problems, Problem 21.95. Expanded formulation of the supplied catalogue statement. https://arxiv.org/pdf/1401.0300v47.