Amenable Groups and Folner Criteria

Suppose $m$ is left invariant. For bounded $f$, define $If(x)=f(x^{-1})$ and $m_r(f)=m(If)$. Then $m_r$ is a mean, and for every $g\in G$ one has $I(f\cdot g)=g^{-1}\cdot If$, so left invariance of $m$ gives $m_r(f\cdot g)=m_r(f)$. Thus a left-invariant mean produces a right-invariant mean.

Replacing $g$ by $g^{-1}$ in the same computation shows that inversion also carries right-invariant means back to left-invariant means. Hence the two notions are equivalent.

Define $m(f)=\mathrm{\mid }G{\mathrm{\mid }}^{-1}{\sum }_{x\in G}f(x)$ on ${\mathrm{\ell }}^{\mathrm{\infty }}(G)$. This is linear, positive, and satisfies $m({\mathbf{1}}_G)=1$, so it is a mean.

For every $g\in G$, the map $x\mapsto g^{-1}x$ is a permutation of $G$, so ${\sum }_{x\in G}f(g^{-1}x)={\sum }_{x\in G}f(x)$. Therefore $m(g\cdot f)=m(f)$, and [L1] makes $G$ amenable.

Suppose first that $A$ is finitely generated. By [L1], write $A\cong {\mathbb{Z}}^r\oplus T$ with $T$ finite. Let $S\subseteq A$ be finite and let $\epsilon >0$. If $r=0$, then $A=T$ is finite and $F=A$ satisfies $sF=F$ for every $s\in S$, so the Folner condition is immediate. Assume now that $r\ge 1$. Transport $S$ across the isomorphism, and let $M$ be the maximum of the ${\mathrm{\ell }}^{\mathrm{\infty }}$-norms of the ${\mathbb{Z}}^r$-components of the transported elements. For $n\ge 1$, put $B_n=[-n,n{]}^r\times T$. Then every translate by an element of $S$ changes only the $M$-thick boundary layers of the box, so $\mathrm{\mid }(s+B_n)\mathrm{△}{B}_n\mathrm{\mid }=O(n^{r-1})$ uniformly in $s\in S$, while $\mathrm{\mid }{B}_n\mathrm{\mid }=(2n+1{)}^r\mathrm{\mid }T\mathrm{\mid }$. For large $n$ this gives $\mathrm{\mid }(s+B_n)\mathrm{△}{B}_n\mathrm{\mid }<\epsilon \mathrm{\mid }{B}_n\mathrm{\mid }$ for every $s\in S$. Thus finitely generated abelian groups satisfy the Folner condition.

By [L2], every finitely generated abelian group is therefore amenable.

Now let $A$ be arbitrary. Given a finite subset $S\subseteq A$ and $\epsilon >0$, the subgroup $⟨S⟩$ is finitely generated and abelian, so step 2.1 makes it amenable. Applying [L2] inside $⟨S⟩$ yields a finite nonempty set $F\subseteq ⟨S⟩$ with $\mathrm{\mid }sF\mathrm{△}F\mathrm{\mid }<\epsilon \mathrm{\mid }F\mathrm{\mid }$ for every $s\in S$. The same set $F$ witnesses the Folner condition in $A$. Since $S$ and $\epsilon$ were arbitrary, [L2] shows that every abelian group is amenable.

Let $H\le G$. To show that $H$ is amenable, by [L3] it is enough to verify the Folner condition in $H$. Fix a finite subset $S\subseteq H$ and $\epsilon >0$. If $S=\mathrm{\varnothing }$, then $\{e\}\subseteq H$ is already an $(S,\epsilon )$-Folner set. Assume now that $S\mathrm{\ne }\mathrm{\varnothing }$, and put $\delta =\epsilon \mathrm{/}\mathrm{\mid }S\mathrm{\mid }$. Since $G$ is amenable, [L3] gives a finite nonempty set $F\subseteq G$ with $\mathrm{\mid }sF\mathrm{△}F\mathrm{\mid }<\delta \mathrm{\mid }F\mathrm{\mid }$ for every $s\in S$. Write $F={⨆}_{j=1}^m{E}_j{t}_j$, where the $E_j$ are the nonempty intersections of $F$ with the finitely many right $H$-cosets that meet $F$, transported back into $H$. For each $s\in S$, left translation by $s$ preserves every right $H$-coset $H{t}_j$, so $sF\mathrm{△}F={⨆}_{j=1}^m((s{E}_j)t_j\mathrm{△}{E}_j{t}_j)$ and hence ${\sum }_{j=1}^m\mathrm{\mid }s{E}_j\mathrm{△}{E}_j\mathrm{\mid }=\mathrm{\mid }sF\mathrm{△}F\mathrm{\mid }$. Summing over $s\in S$ gives ${\sum }_{j=1}^m{\sum }_{s\in S}\mathrm{\mid }s{E}_j\mathrm{△}{E}_j\mathrm{\mid }<\mathrm{\mid }S\mathrm{\mid }\delta \mathrm{\mid }F\mathrm{\mid }=\epsilon \mathrm{\mid }F\mathrm{\mid }=\epsilon {\sum }_{j=1}^m\mathrm{\mid }{E}_j\mathrm{\mid }\mathrm{.}$ Therefore some $j$ satisfies ${\sum }_{s\in S}\mathrm{\mid }s{E}_j\mathrm{△}{E}_j\mathrm{\mid }<\epsilon \mathrm{\mid }{E}_j\mathrm{\mid }$, and then each summand is itself $<\epsilon \mathrm{\mid }{E}_j\mathrm{\mid }$. So $E_j$ is an $(S,\epsilon )$-Folner set in $H$. Since $S$ and $\epsilon$ were arbitrary, $H$ satisfies the Folner condition, and [L3] makes $H$ amenable.

Let $N⊴G$, and let $q:G\to G\mathrm{/}N$ be the quotient map from [L2]. For bounded $u:G\mathrm{/}N\to \mathbb{R}$, define $m_{G\mathrm{/}N}(u)=m(u\circ q)$ using a left-invariant mean $m$ on $G$. Then $m_{G\mathrm{/}N}$ is a mean, and for $\bar{g}=gN$ one has $(\bar{g}\cdot u)\circ q=g\cdot (u\circ q)$, so left invariance of $m$ implies $m_{G\mathrm{/}N}(\bar{g}\cdot u)=m_{G\mathrm{/}N}(u)$. Therefore the quotient is amenable.

Steps 1.1 and 1.2 prove the two permanence statements.

Let $m_N$ be a left-invariant mean on $N$. For bounded $f:G\to \mathbb{R}$ and $g\in G$, define ${\mathrm{\Phi }}_f(gN):=m_N(n\mapsto f(gn))$. If $g^{\mathrm{\prime }}=gh$ with $h\in N$, then the integrand for $g^{\mathrm{\prime }}$ is $n\mapsto f(ghn)$, which is the left translate of $n\mapsto f(gn)$ by $h^{-1}$ in the $N$-variable. Thus the value of ${\mathrm{\Phi }}_f(gN)$ does not depend on the chosen representative of the right coset $gN$. The same formula also shows $\mathrm{\mid }{\mathrm{\Phi }}_f(gN)\mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$, so ${\mathrm{\Phi }}_f$ is bounded on $G\mathrm{/}N$.

Let $m_Q$ be a left-invariant mean on $G\mathrm{/}N$, and set $m_G(f)=m_Q({\mathrm{\Phi }}_f)$. Positivity and $m_G({\mathbf{1}}_G)=1$ are immediate. For $x\in G$, one has ${\mathrm{\Phi }}_{x\cdot f}(gN)=m_N(n\mapsto f(x^{-1}gn))={\mathrm{\Phi }}_f(x^{-1}gN)$, so ${\mathrm{\Phi }}_{x\cdot f}=xN\cdot {\mathrm{\Phi }}_f$. Therefore $m_G(x\cdot f)=m_Q(xN\cdot {\mathrm{\Phi }}_f)=m_Q({\mathrm{\Phi }}_f)=m_G(f)$.

Thus $m_G$ is a left-invariant mean on $G$, so [L1] makes $G$ amenable.

Let $S\subseteq G$ be finite and let $\epsilon >0$. Because the family is directed and ${\bigcup }_i{H}_i=G$, some index $i$ satisfies $S\subseteq H_i$. Since $H_i$ is amenable, [L1] gives a finite nonempty set $F\subseteq H_i$ with $\mathrm{\mid }sF\mathrm{△}F\mathrm{\mid }<\epsilon \mathrm{\mid }F\mathrm{\mid }$ for every $s\in S$.

The set $F$ from step 1.1 is also an $(S,\epsilon )$-Folner set when viewed inside $G$. Since $S$ and $\epsilon$ were arbitrary, $G$ satisfies the Folner condition. Therefore [L1] makes $G$ amenable.

Let $G$ be solvable. If its derived length is $0$, then $G$ is trivial and hence finite, so [L4] applies. If the derived length is positive, then $G^{\mathrm{\prime }}=[G,G]$ has smaller derived length by [L1], and the quotient $G\mathrm{/}{G}^{\mathrm{\prime }}$ is abelian. Inducting on derived length and applying [L5] shows that every solvable group is amenable.

Let $G$ be locally finite. The family of finitely generated subgroups of $G$ is directed by inclusion, its union is all of $G$, and every member is finite by [L2]. Hence each member is amenable by [L4], and [L3] gives amenability of $G$.

Steps 1.1 and 1.2 prove the two claims.

For each $s\in S$, left translation by $s$ is a bijection of $G$, so $\mathrm{\mid }sF\mathrm{\mid }=\mathrm{\mid }F\mathrm{\mid }$ and therefore $\mathrm{\mid }sF\mathrm{△}F\mathrm{\mid }=\mathrm{\mid }sF\setminus F\mathrm{\mid }+\mathrm{\mid }F\setminus sF\mathrm{\mid }=2\mathrm{\mid }sF\setminus F\mathrm{\mid }$. Hence the symmetric-difference and one-sided boundary formulations differ only by the factor $2$.

Inversion is a bijection $F\mapsto F^{-1}$ with $\mathrm{\mid }(sF)\mathrm{△}F\mathrm{\mid }=\mathrm{\mid }(F^{-1}{s}^{-1})\mathrm{△}{F}^{-1}\mathrm{\mid }$. Thus the left-translate and right-translate versions are equivalent after replacing $S$ by $S^{-1}$.

Assume $G$ satisfies the Folner condition. Let $D$ be the directed set of triples $(S,n,F)$ with $S\subseteq G$ finite, $n\ge 1$, and $F$ an $(S,1\mathrm{/}n)$-Folner set, ordered by enlarging $S$ and $n$. For $d=(S,n,F)$, define $m_d(f)=\mathrm{\mid }F{\mathrm{\mid }}^{-1}{\sum }_{x\in F}f(x)$. These $m_d$ are means, and [L3] implies that if $g\in S$ then $\mathrm{\mid }{m}_d(g\cdot f)-m_d(f)\mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{/}n$. The cofinal tails $D_{g,N}=\{(S,n,F):g\in S,n\ge N\}$ have the finite intersection property, so by [A1] some ultrafilter on $D$ contains all of them. The ultrafilter limit of the bounded family $(m_d(f){)}_{d\in D}$ is therefore a left-invariant mean on $G$.

Assume instead that $G$ is amenable but not Folner. Then some finite $S\subseteq G$ and $\epsilon >0$ satisfy: for every finite nonempty $F\subseteq G$, some $s\in S$ has $\mathrm{\mid }sF\mathrm{△}F\mathrm{\mid }\ge \epsilon \mathrm{\mid }F\mathrm{\mid }$. Put $S_0=S\cup \{e\}$ and $\delta =\epsilon \mathrm{/}2$. Since $\mathrm{\mid }sF\mathrm{\mid }=\mathrm{\mid }F\mathrm{\mid }$, [L3] gives $\mathrm{\mid }sF\setminus F\mathrm{\mid }\ge \delta \mathrm{\mid }F\mathrm{\mid }$ for that $s$, and hence $\mathrm{\mid }{S}_{0}F\mathrm{\mid }\ge (1+\delta )\mathrm{\mid }F\mathrm{\mid }$.

Choose $m\ge 1$ with $(1+\delta {)}^m\ge 2$ and put $K=S_0^m$. Applying step 1.2 successively to $F,S_{0}F,\dots ,S_0^{m-1}F$ gives $\mathrm{\mid }KF\mathrm{\mid }\ge 2\mathrm{\mid }F\mathrm{\mid }$ for every finite nonempty $F\subseteq G$.

Form the bipartite graph with left vertices $G\times \{1,2\}$, right vertices $G$, and edges $(g,r)\sim kg$ for $k\in K$. If $P$ is a finite set of left vertices and $Q$ is its projection to $G$, then $N(P)=KQ$ and $\mathrm{\mid }P\mathrm{\mid }\le 2\mathrm{\mid }Q\mathrm{\mid }\le \mathrm{\mid }KQ\mathrm{\mid }$ by step 2.1. Thus [L4] gives a matching saturating every prescribed finite left set.

Let $\mathcal{M}$ be the set of finite partial matchings in this graph, and for finite $P\subseteq G\times \{1,2\}$ let $X_P$ be the set of members of $\mathcal{M}$ whose domains contain $P$. Step 3.1 shows that the family $(X_P)$ has the finite-intersection property. By [A1], an ultrafilter $\mathcal{U}$ on $\mathcal{M}$ contains every $X_P$. For a left vertex $\mathrm{\ell }$, the set $X_{\{\mathrm{\ell }\}}$ is the disjoint union of the finitely many sets on which the partial matching assigns a fixed neighbour $y\in N(\mathrm{\ell })$. Exactly one such cell belongs to $\mathcal{U}$; call its neighbour $\mathrm{\Phi }(\mathrm{\ell })$. If distinct left vertices had the same $\mathrm{\Phi }$-value, the two corresponding cells would have empty intersection, contradicting closure of $\mathcal{U}$ under intersections. Hence $\mathrm{\Phi }:G\times \{1,2\}\to G$ is injective and satisfies $\mathrm{\Phi }(g,r)\in Kg$.

Write ${\mathrm{\Phi }}_r(g)=\mathrm{\Phi }(g,r)$ and, for $r\in \{1,2\}$ and $k\in K$, put $D_{r,k}=\{g:{\mathrm{\Phi }}_r(g)=kg\}$. For each fixed $r$, the sets $D_{r,k}$ partition $G$, while the sets $k{D}_{r,k}$ partition the range $R_r$ of ${\mathrm{\Phi }}_r$; injectivity of $\mathrm{\Phi }$ makes $R_1$ and $R_2$ disjoint. If $m_G$ is a left-invariant mean, write $m_G(E)=m_G({\mathbf{1}}_E)$. Finite additivity and invariance give $m_G(R_r)={\sum }_{k\in K}{m}_G(k{D}_{r,k})={\sum }_{k\in K}{m}_G(D_{r,k})=m_G(G)=1$ for $r=1,2$. But $R_1\cap R_2=\mathrm{\varnothing }$, so positivity gives $2=m_G(R_1)+m_G(R_2)=m_G(R_1\bigsqcup R_2)\le m_G(G)=1$, a contradiction.

Therefore an amenable group cannot fail the Folner condition. Together with step 1.1, this proves the equivalence.

For each $n\ge 1$, apply [L2] to the finite test set $S_n=\{g_1,\dots ,g_n\}$ and the tolerance $1\mathrm{/}n$. This gives a finite nonempty set $F_n\subseteq G$ with $\mathrm{\mid }{g}_i{F}_n\mathrm{△}{F}_n\mathrm{\mid }<\mathrm{\mid }{F}_n\mathrm{\mid }\mathrm{/}n$ for every $i\le n$.

Fix $i$. Once $n\ge i$, step 1.1 gives $\mathrm{\mid }{g}_i{F}_n\mathrm{△}{F}_n\mathrm{\mid }\mathrm{/}\mathrm{\mid }{F}_n\mathrm{\mid }<1\mathrm{/}n$, which tends to $0$. Therefore $(F_n)$ is a Folner sequence in the sense of [L1].

Let $B_n$ be the word-metric ball of radius $n$ about the identity. If some $\delta >0$ satisfied $\mathrm{\mid }S{B}_n\mathrm{\mid }\ge (1+\delta )\mathrm{\mid }{B}_n\mathrm{\mid }$ for every $n$, then iterating would give $\mathrm{\mid }{B}_n\mathrm{\mid }\ge (1+\delta {)}^n$ up to multiplicative constants, contradicting the subexponential alternative in [L2]. Therefore for every $\epsilon >0$ there exists $n$ with $\mathrm{\mid }S{B}_n\setminus B_n\mathrm{\mid }<(\epsilon \mathrm{/}2)\mathrm{\mid }{B}_n\mathrm{\mid }$.

For such an $n$, every $s\in S$ satisfies $\mathrm{\mid }s{B}_n\setminus B_n\mathrm{\mid }\le \mathrm{\mid }S{B}_n\setminus B_n\mathrm{\mid }<(\epsilon \mathrm{/}2)\mathrm{\mid }{B}_n\mathrm{\mid }$, so the boundary formulation of [L3] makes $B_n$ an $(S,\epsilon )$-Folner set. Hence $G$ satisfies the Folner condition, and [L3] shows that $G$ is amenable.

Assume toward contradiction that $m$ is a left-invariant mean on $G$. By [L2] and positivity, finite additivity on indicator functions gives $1=m({\mathbf{1}}_G)={\sum }_{i}m({\mathbf{1}}_{A_i})+{\sum }_{j}m({\mathbf{1}}_{B_j})$.

The translated families from [L2] also partition $G$, so left invariance gives $1=m({\mathbf{1}}_G)={\sum }_{i}m({\mathbf{1}}_{a_i{A}_i})={\sum }_{i}m({\mathbf{1}}_{A_i})$ and likewise $1={\sum }_{j}m({\mathbf{1}}_{B_j})$. Combining these equalities with step 1.1 yields $1=2$, a contradiction.

If $G$ is paradoxical, then [L1] says that no left-invariant mean exists, so $G$ is not amenable. This proves the forward implication of the statement.

Assume now that $G$ is not amenable. By contrapositive use of [A1], there is a finite set $K\subseteq G$ with $e\in K$ such that $\mathrm{\mid }KF\mathrm{\mid }\ge 2\mathrm{\mid }F\mathrm{\mid }$ for every finite nonempty $F\subseteq G$. Indeed, if no such $K$ existed, then for a finite test set $S$ and $\epsilon >0$ one could put $S_0=\{e\}\cup S\cup S^{-1}$, choose $m$ with $(1+\epsilon \mathrm{/}2{)}^m>2$, and find finite nonempty $F$ with $\mathrm{\mid }{S}_0^{m}F\mathrm{\mid }<2\mathrm{\mid }F\mathrm{\mid }$. Among the layers $F_j=S_0^{j}F$, some ratio $\mathrm{\mid }{F}_{j+1}\mathrm{\mid }\mathrm{/}\mathrm{\mid }{F}_j\mathrm{\mid }$ would then be below $1+\epsilon \mathrm{/}2$, giving $\mathrm{\mid }s{F}_j\mathrm{△}{F}_j\mathrm{\mid }<\epsilon \mathrm{\mid }{F}_j\mathrm{\mid }$ for every $s\in S$ and hence the Folner condition.

Form the bipartite graph from [A2]. For every finite left subset $P\subseteq G\times \{1,2\}$ with projection $Q\subseteq G$, step 1.2 gives $\mathrm{\mid }N(P)\mathrm{\mid }=\mathrm{\mid }KQ\mathrm{\mid }\ge 2\mathrm{\mid }Q\mathrm{\mid }\ge \mathrm{\mid }P\mathrm{\mid }$, so [L2] produces a matching saturating $P$. By [A2], these finite matchings extend to an edge-respecting bijection $\mathrm{\Phi }:G\times \{1,2\}\to G$. For $r\in \{1,2\}$ and $k\in K$, put $C_{r,k}=\{\mathrm{\Phi }(g,r):\mathrm{\Phi }(g,r)=kg\}\mathrm{.}$ The finitely many $C_{r,k}$ are pairwise disjoint and partition $G$ because $\mathrm{\Phi }$ is bijective. For fixed $r$, the translated pieces $k^{-1}{C}_{r,k}$ partition $G$, because they are exactly the fibers in the domain copy $G\times \{r\}$. Thus the two subfamilies $(C_{1,k}{)}_{k\in K}$ and $(C_{2,k}{)}_{k\in K}$, with translators $k^{-1}$, form a paradoxical decomposition.

Steps 1.1 and 2.1 prove both directions, so $G$ is amenable exactly when it is not paradoxical.

By [L1], write $F_2=⟨a,b⟩$. For $x\in \{a,a^{-1},b,b^{-1}\}$, let $W(x)$ be the set of nonempty reduced words whose first letter is $x$. Put $P=\{a^n:n\ge 0\}$ and $P_{+}=\{a^n:n\ge 1\}$, and define $A_1=W(a)\setminus P_{+}$, $A_2=W(a^{-1})\cup P$, $B_1=W(b)$, and $B_2=W(b^{-1})$.

The four sets are pairwise disjoint and partition $F_2$: the usual five first-letter classes partition $F_2$, and $P_{+}$ has merely been moved from $W(a)$ into the piece containing the identity.

Left multiplication gives $aW(a^{-1})=F_2\setminus W(a)$ and $aP=P_{+}$, hence $F_2=A_1\bigsqcup a{A}_2$. Likewise $bW(b^{-1})=F_2\setminus W(b)$, hence $F_2=B_1\bigsqcup b{B}_2$. Thus the pieces of step 1.1, with translators $e,a,e,b$, satisfy both partition equalities in [L3].

Steps 1.1-3.1 give a paradoxical decomposition of $F_2$, so [L2] implies that $F_2$ admits no left-invariant mean and is therefore nonamenable.

If $G$ were amenable, then [L2] would make its subgroup $H$ amenable.

But $H\cong F_2$, and [L1] says $F_2$ is nonamenable. This contradiction proves that $G$ is nonamenable.

Choose word metrics on $G$ and $H$, quasi-inverse quasi-isometries $q:G\to H$ and $r:H\to G$, and constants $\lambda \ge 1,c,D_G,D_H\ge 0$ such that both maps satisfy the $(\lambda ,c)$ upper distance bound, $d_G(rq(x),x)\le D_G$, and $d_H(qr(y),y)\le D_H$. A fiber of $q$ has uniformly bounded cardinality: if $q(x)=q(x^{\mathrm{\prime }})$, the lower quasi-isometry inequality bounds $d_G(x,x^{\mathrm{\prime }})$, and a word-metric ball of fixed radius is finite. Let $M_q$ bound those fibers, and define $M_r$ similarly.

Assume $G$ is amenable. Let $E\subseteq H$ be finite and let $\epsilon >0$. Put $L=\max \{\mathrm{\mid }e\mathrm{\mid }:e\in E\}$, taking $L=0$ when $E=\mathrm{\varnothing }$, set $R=\lambda (L+D_H)+c+D_G$, and let $T$ be the finite radius-$R$ ball in $G$. Set $\delta =\epsilon \mathrm{/}(2M_q{M}_r(\mathrm{\mid }T\mathrm{\mid }+1))$. By [L2], choose a finite nonempty $A\subseteq G$ with $\mathrm{\mid }tA\mathrm{△}A\mathrm{\mid }<\delta \mathrm{\mid }A\mathrm{\mid }$ for every $t\in T$.

Put $B=N_{D_H}(q(A))$. It is finite and nonempty, and $\mathrm{\mid }B\mathrm{\mid }\ge \mathrm{\mid }q(A)\mathrm{\mid }\ge \mathrm{\mid }A\mathrm{\mid }\mathrm{/}{M}_q$. If $y\in N_L(B)\setminus B$, choose $b\in B$ and $a\in A$ with $d_H(y,b)\le L$ and $d_H(b,q(a))\le D_H$. Then $d_G(r(y),a)\le R$. Moreover $r(y)\mathrm{\notin }A$, since otherwise $d_H(y,q(r(y)))\le D_H$ would put $y$ in $B$. Hence $r(N_L(B)\setminus B)\subseteq N_R(A)\setminus A$, and the fiber bound for $r$ gives $\mathrm{\mid }{N}_L(B)\setminus B\mathrm{\mid }\le M_r\mathrm{\mid }{N}_R(A)\setminus A\mathrm{\mid }$.

Since $N_R(A)\setminus A\subseteq {\bigcup }_{t\in T}(tA\setminus A)$, step 2.1 gives $\mathrm{\mid }{N}_R(A)\setminus A\mathrm{\mid }<\mathrm{\mid }T\mathrm{\mid }\delta \mathrm{\mid }A\mathrm{\mid }$. For $e\in E$, one has $eB\setminus B\subseteq N_L(B)\setminus B$ and $\mathrm{\mid }eB\mathrm{△}B\mathrm{\mid }=2\mathrm{\mid }eB\setminus B\mathrm{\mid }$. Combining this with step 3.1 and $\mathrm{\mid }B\mathrm{\mid }\ge \mathrm{\mid }A\mathrm{\mid }\mathrm{/}{M}_q$ yields $\mathrm{\mid }eB\mathrm{△}B\mathrm{\mid }\mathrm{/}\mathrm{\mid }B\mathrm{\mid }<2M_q{M}_r\mathrm{\mid }T\mathrm{\mid }\delta <\epsilon$. Thus $B$ is an $(E,\epsilon )$-Folner set in $H$.

Since $E$ and $\epsilon$ were arbitrary, step 4.1 gives the Folner condition in $H$, so [L2] makes $H$ amenable. Applying the same argument to the quasi-inverse transfers amenability from $H$ to $G$. Therefore amenability depends only on quasi-isometry type, as asserted in [L1].

The additive group $(\mathbb{Z},+)$ is abelian and infinite.

By [L1], $(\mathbb{Z},+)$ is amenable. Since it is infinite, it refutes the statement.

The remark [L1] supplies a group that is nonamenable and has no nonabelian free subgroup at all.

Such a group cannot contain a free subgroup of rank $2$, so it refutes the statement.

Let $G=F_2$, which is nonamenable. The singleton $\{e\}$ satisfies $\mathrm{\mid }e\{e\}\mathrm{△}\{e\}\mathrm{\mid }=0$, so it is a $(\{e\},1)$-Folner set.

This single accidental Folner set does not make $G$ amenable, because [L1] requires the inequalities for all finite test sets and arbitrarily small tolerances. Therefore the statement is false.

Let $G=(\mathbb{R},+)$. It is abelian and therefore amenable by [L2], and it is uncountable by [L4]. Let $(F_n)$ be any sequence of finite nonempty subsets of $G$, and put $A={\bigcup }_n(F_n-F_n)$. Each finite subset of the ordered set $\mathbb{R}$ has a unique increasing enumeration, so the sets $F_n-F_n$ can be enumerated canonically and $A$ is at most countable. By [L4], choose $g\in \mathbb{R}\setminus A$.

For this $g$, one has $(g+F_n)\cap F_n=\mathrm{\varnothing }$ for every $n$, since an intersection would put $g$ in $F_n-F_n\subseteq A$. Hence $\mathrm{\mid }(g+F_n)\mathrm{△}{F}_n\mathrm{\mid }=2\mathrm{\mid }{F}_n\mathrm{\mid }$ and the ratio in [L1] is always $2$, never $0$. Thus $(F_n)$ is not a global Folner sequence. The amenability from step 1.1 does not force a contradiction, because [L3] is only a finite-test criterion and does not supply one countable family for all elements of an uncountable group. Since the sequence was arbitrary, the statement is false.