Tarski's theorem on paradoxical groups

Remarks

That 1 and 2 cannot both hold is direct: the pieces of a paradoxical decomposition, chosen with $g_i A_i$ disjoint and $h_j B_j$ disjoint, would give $\mu(E) \ge 2\mu(E)$ with $\mu(E) = 1$. The hard direction builds $\mu$ when $E$ is not paradoxical. Tarski's proof studies the commutative semigroup of equidecomposability types and uses a cancellation law in it, with the Hahn–Banach theorem or a similar extension argument. These arguments use the axiom of choice.

Example: the free group $F_2$ is paradoxical, so it carries no invariant finitely additive probability on its subsets. The group $\Z$ is amenable, so it is not paradoxical.

Going deeper

For a student

The theorem explains why paradoxes like the Banach–Tarski paradox appear exactly for non-amenable groups. In dimensions $1$ and $2$ the isometry groups are solvable, hence amenable, so no paradoxical decomposition of the interval or the disc by isometries exists. In dimension $3$, the rotation group contains a free group of rank $2$, and the ball becomes paradoxical. The minimal number of pieces in a paradoxical decomposition, the Tarski number, is a finer invariant of non-amenable groups.