Abstract harmonic analysis Amenability and invariant means Counterexample

Ol'shanskii's counterexample to the von Neumann conjecture

Builds on Tarski monster group Amenable group Free group von Neumann conjecture

Counterexample

Let $p$ be a large prime and let $G$ be a Tarski monster group for $p$ of the kind constructed by Ol'shanskii. So $G$ is infinite, and every subgroup of $G$ other than $\set{1}$ and $G$ is cyclic of order $p$ . Ol'shanskii proved in 1980 that such groups can be chosen not amenable. On the other hand, $G$ has no subgroup isomorphic to the free group $F_2$ . So $G$ is a non-amenable group without free subgroups of rank $2$ , and the von Neumann conjecture fails.

Remarks

Why $G$ has no free subgroup: every element $g \ne 1$ of $G$ lies in a proper nontrivial subgroup, which is cyclic of order $p$ , so $g^p = 1$ . A free group of rank $2$ has elements of infinite order, so it cannot embed in $G$ . The non-amenability is the hard part; the proof uses Grigorchuk's cogrowth criterion, a form of Kesten's criterion.

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Amenable group concept
Free group concept
Subgroup concept
Tarski monster group object
von Neumann conjecture conjecture
Ol'shanskii's counterexample to the von Neumann conjecture counterexample

Tag	`XA7YE70A6P`
URI	`m2://XA7YE70A6P`
Kind	counterexample
Writing	W0 stub
MSC	43A07, 20F50, 20F65
Uses	Tarski monster group, Amenable group, Free group, von Neumann conjecture
Changed	3 November 2026

Ol'shanskii's counterexample to the von Neumann conjecture

Counterexample

Remarks

Reading path

Further reading

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