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Every subgroup of an amenable group is amenable. One proof restricts an invariant mean after extending bounded functions from the subgroup to the whole group along a set of coset representatives. Consequently, a group containing a nonamenable subgroup is nonamenable.

Normalized counting measures on a Følner sequence have a convergent subnet in the compact product $[0,1{]}^{\mathcal{P}(G)}$. The limit is a left-invariant finitely additive probability measure, so every group satisfying the Følner condition is an amenable group.

The free group $F_2=⟨a,b⟩$ is not an amenable group. Partitioning reduced words by their first letters gives two translated decompositions of $F_2$ whose invariance equations add to $2=1$, contradicting the existence of an invariant mean.

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

Every abelian group is amenable, and amenability is preserved by finite extensions. Hence every group with an abelian finite-index subgroup is an amenable group.