Erdős Problem #1098

Erdős Problem 1098: noncommuting sets in a group

Statement and attribution

For an arbitrary group $G$, assume that no infinite set of its elements is pairwise noncommuting. We prove that $G/Z(G)$ is finite, and hence obtain a finite bound on every noncommuting set. This reconstructs B. H. Neumann's proof, A problem of Paul Erdős on groups, J. Austral. Math. Soc. 21 $1976$, 467--472, https://doi.org/10.1017/S1446788700019303. No novelty is claimed.

First step: every conjugacy class is finite

Suppose, to the contrary, that $g$ has infinitely many conjugates. Choose a countably infinite set $T$ such that $t^{-1}gt$ are distinct for $t\in T$. Apply the infinite two-colour Ramsey theorem to pairs from $T$, colouring according to whether they commute. The noncommuting homogeneous outcome contradicts the hypothesis. Thus there is an infinite commuting subset $U$ of $T$.

For distinct $u,v\in U$, the elements $gu,gv$ cannot commute. Indeed, their commuting would imply $ugv=vgu$ after cancelling an initial $g$. Since $u,v$ commute, multiplication on the left by $(uv)^{-1}$ gives \[ v^{-1}gv=u^{-1}gu, \] contrary to the choice of $T$. Hence $gU$ is an infinite noncommuting set, again a contradiction.

Every centralizer $C_G(g)$ therefore has finite index, since its cosets correspond to the conjugates of $g$. In particular $C_G(F)$ has finite index for every finite $F\subset G$: the intersection of finitely many finite-index subgroups has index at most the product of their indices. The only combinatorial input used here is infinite Ramsey for pairs. It follows, for completeness, by choosing successively a vertex and an infinite subset on which all its outgoing edges have one colour, then retaining an infinite subsequence of chosen vertices with the same chosen colour.

Second step: finite-index abelian subgroups

In a group with finite conjugacy classes, an abelian subgroup $A$ of finite index forces $Z(G)$ to have finite index. Choose a finite right-coset transversal $R$ for $A$. The group is generated by $A\cup R$. Consequently \[ A\cap C_G(R)\subset Z(G). \] Both factors on the left have finite index, proving the assertion. Thus, if $Z(G)$ has infinite index, every finite-index subgroup of $G$ is nonabelian.

Building an infinite noncommuting set

Assume now that $Z(G)$ has infinite index. We construct pairs $a_i,b_i$ such that \[ [a_i,a_j]\ne1\ (i\ne j),\qquad [a_i,b_j]=1\ (i\ne j),\qquad [a_i,b_i]\ne1,\qquad [b_i,b_j]=1. \tag{1} \] A noncommuting pair starts the construction. Suppose the first $n$ pairs have been chosen. Their common centralizer \[ H=C_G(\{a_1,\ldots,a_n,b_1,\ldots,b_n\}) \] has finite index and is nonabelian by the preceding step. Choose $a,b\in H$ with $ab\ne ba$, and put \[ a_{n+1}=a b_1\cdots b_n,\qquad b_{n+1}=b. \] For a fixed $i\le n$, all factors in $a_{n+1}$ except $b_i$ commute with $a_i$; those other factors also commute with $b_i$. Therefore $a_{n+1}$ commutes with $a_i$ if and only if $b_i$ does, which is false. Each old $b_i$ commutes with $a$ and all $b_j$, so it commutes with $a_{n+1}$. The element $b$ commutes with all old elements, but fails to commute with $a b_1\cdots b_n$ because it fails to commute with $a$. These observations establish all four conditions $1$ at the next stage. The resulting infinite sequence $(a_i)$ is pairwise noncommuting, the required contradiction.

Finite bound and conclusion

It follows that $[G:Z(G)]=m<\infty$. Two elements in the same coset of $Z(G)$ commute: writing them as $xz,xw$, their products both equal $x^2zw$. A pairwise noncommuting set has at most one element in each coset, so has size at most $m$. This also proves the converse: finite central index rules out an infinite clique.

The equivalence and the requested bound are fully proved. The argument applies to infinite and uncountable groups; the countable choices occur only inside an assumed infinite set. Source statement: https://www.erdosproblems.com/1098.

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