Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 143 / 4

An amenable group is a group $G$ admitting a left-invariant finitely additive probability measure $m:\mathcal{P}(G)\to [0,1]$. For a $G$-set $X$, subsets $A,B\subseteq X$ are equidecomposable subsets under a group action if there are finite partitions $A={⨆}_i{A}_i$, $B={⨆}_i{B}_i$ and elements $g_i\in G$ such that $g_i{A}_i=B_i$. The action is a paradoxical group action if $X$ contains two disjoint subsets, each $G$-equidecomposable with $X$.

To prove the nonamenability of a nonabelian free group, write $F_2=⟨a,b⟩$ and let $W(s)$ be the set of nonempty reduced words beginning with $s\in \{a,a^{-1},b,b^{-1}\}$. Cancellation of the first letter gives the disjoint decompositionsIf an invariant measure $m$ existed, these would implyEvery singleton has measure zero: invariance gives all singletons the same measure, and finite additivity over arbitrarily many distinct points forces that measure to vanish. The four sets $W(s)$ partition $F_2\setminus \{1\}$, so their measures sum to one. The two displayed equations say that the same sum is two, a contradiction. Hence

For a finite generating set $S$, the Følner condition requires that for every $\epsilon >0$ there be a nonempty finite $F\subseteq G$ such thatwhere $△$ denotes symmetric difference. Choose a Følner sequence $(F_n)$ and define normalized counting functions on all subsets $A\subseteq G$ byBy compactness of the product $[0,1{]}^{\mathcal{P}(G)}$, some subnet converges pointwise to a function $m$. The identities ${\mu }_n(G)=1$ and finite additivity on disjoint subsets pass to the limit, so $m$ is a finitely additive probability measure.

For a fixed $g=s_1\cdots s_{\ell }$, the triangle inequality for symmetric differences givesConsequentlyso $m(gA)=m(A)$. The limit is left invariant and the Følner condition implies amenability. Thus every finitely generated group satisfying the Følner condition is amenable.

For the standard generator $1$ of the additive group $\mathbb{Z}$, take the intervalsOnly the two endpoints change under translation by $1$, and henceThe same follows for every fixed integer translation by iteration. Therefore these sets are a Følner sequence for the integers, and $\mathbb{Z}$ satisfies the Følner condition.

Yes, but there is only one nontrivial isomorphism type that works. If $G=C_2$, thenThe infinite dihedral group contains its infinite cyclic rotation subgroup with index two, so it is amenable by virtually abelian groups are amenable.

Suppose now that $G\ncong C_2$. The action of $G\ast G$ on its Bass-Serre tree is non-elementary: one vertex degree is greater than two, so there are hyperbolic elements with disjoint pairs of endpoints. Suitable powers satisfy the ping-pong lemma on the boundary and generate a copy of $F_2$. Since a subgroup of an amenable group is amenable, an amenable group cannot contain this nonamenable subgroup. The amenability of a free product therefore yieldsfor nontrivial finitely generated $G$.