Non-amenability of free groups

Remarks

For part 1, let $W(x)$ be the set of reduced words that begin with the letter $x \in \set{a, a^{-1}, b, b^{-1}}$. Then $F_2 = W(a) \sqcup a W(a^{-1})$ and $F_2 = W(b) \sqcup b W(b^{-1})$. An invariant finitely additive probability $\mu$ would give $\mu(W(a)) + \mu(W(a^{-1})) = 1$ and $\mu(W(b)) + \mu(W(b^{-1})) = 1$. The four sets are disjoint, so their total measure is at most $1$, a contradiction. For part 2, every subgroup $H$ of an amenable group $G$ is amenable: choose a set $R$ of representatives of the right cosets $Hg$, and define $\nu(A) = \mu(AR)$ for $A \subseteq H$.

Example: $\mathrm{SL}(2, \Z)$ contains the free subgroup generated by $\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$ and $\begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix}$, so it is not amenable. The free groups $F_k$ with $k \ge 2$, and $F_\infty$, all contain $F_2$ and are not amenable, while $F_1 = \Z$ is amenable.

Going deeper

For a student

This is the basic example of non-amenability, and the source of the Banach–Tarski paradox, since the rotation group $\mathrm{SO}(3)$ contains $F_2$. The decomposition above is a paradoxical decomposition of $F_2$, and Tarski's theorem shows that such a decomposition is the only obstruction to amenability. The von Neumann conjecture asked whether containing $F_2$ is also necessary for non-amenability; it is not.