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Claim. The answer to Problem 1098 is yes. Let $G$ be a group and $\Gamma(G)$ its non-commuting graph. Neumann proved (Theorem 6, p. 470) that $\Gamma(G)$ contains no infinite complete subgraph if and only if the center $Z(G)$ has finite index in $G$. When $|G:Z(G)|=n$ is finite, any $n+1$ elements of $G$ include two in the same coset of the center, and those two commute, so no complete subgraph of $\Gamma(G)$ has more than $n$ vertices; p. 471 sharpens the bound to $n-1$ for non-abelian $G$, since a complete subgraph with at least two vertices contains no central element and meets each of the other $n-1$ cosets at most once (for abelian $G$, where $n=1$, a single vertex is a complete subgraph); the bound $n-1$ is attained by the quaternion group and the dihedral group of order $8$. So a group whose non-commuting graph has no infinite complete subgraph has a finite bound on the size of its complete subgraphs, which is the question.

Proof shape. The forward direction has four steps, summarized on the source card: by Ramsey's theorem, a group without an infinite pairwise non-commuting set has every conjugacy class finite (Lemma 1); such a group with an abelian subgroup of finite index has a center of finite index (Lemma 2 and Corollary 3); and in a group with finite conjugacy classes whose center has infinite index, pairwise non-commuting sequences of every finite length extend by one more element (Lemma 4), which produces an infinite complete subgraph by a limit argument (Corollary 5). The converse is the coset count above. Neumann also notes that the proof gives $\log n=O(m^2)$ for the index $n$ of the center in terms of the largest complete subgraph order $m$, and that every finite $m\neq2$ occurs as the exact maximum.

Source. B. H. Neumann, A problem of Paul Erdős on groups, J. Austral. Math. Soc. Ser. A 21 (1976), no. 4, 467--472, doi:10.1017/S1446788700019303; received 24 January 1975, with a note added 18 July 1975. The issue is dated June 1976 and carries no day, so this page's date is the first of that month. The paper reports that Erdős asked the question at the 15th Summer Research Institute of the Australian Mathematical Society in 1975. The added note records that Ralph N. McKenzie had obtained the same results by much the same methods two or three months earlier, and that McKenzie and Vance Faber hoped to publish extensions to higher cardinals and to cancellation semigroups; no posting of McKenzie's proof is on record. The source card records the reading depth, the statements of Lemma 1 through Theorem 6 clause by clause; nothing on this page is independently reviewed by this project.

Acceptance. Refereed: the result is a journal paper in the Journal of the Australian Mathematical Society, Series A. Reviewed: the curator of erdosproblems.com, T. F. Bloom, marks Problem 1098 proved and credits Neumann's paper as the solution (problem page last edited 18 October 2025, accessed 2026-10-07).

Formalizations. A Lean 4 development by John Jennings with the AI system Aristotle (Harmonic), posted in the problem's discussion thread on 2026-04-25, declares itself a formalization of Neumann's solution. Its main theorem states that for any group $G$, if every pairwise non-commuting subset of $G$ is finite, then some natural number bounds the cardinality of every finite pairwise non-commuting subset. Neither copy contains sorry. A modified copy of the file, which wraps it in a namespace and adds a header, is archived in Boris Alexeev's lean-proofs repository and closes with a #print axioms line whose recorded output is propext, Classical.choice and Quot.sound; the formal-conjectures entry for the problem (category research solved) names that copy as its formal proof, and the Lean suffix of the site's label rests on that development. This project has not built, replayed or audited either copy, so the development is recorded as a formalization link and gives no formalized evidence here; the acceptance rests on the refereed paper and the curator's credit.