PTFA groups satisfy the finite Folner condition without a choice hypothesis

A torsion-free abelian group satisfies [F1]. Indeed the subgroup generated by a finite test set is isomorphic to ${\mathbb{Z}}^r$ by [F2]. For $r=0$ use the singleton identity set. For $r>0$ bound every test vector's coordinates in absolute value by $M\ge 0$ and take the integer cube $E_n=[-n,n{]}^r\cap {\mathbb{Z}}^r$, with $n\ge M$. A translate by any test vector has intersection with $E_n$ of cardinality at least $(2n+1-M{)}^r$, whereas both sets have cardinality $(2n+1{)}^r$. Thus its relative symmetric difference is at most $2[1-(1-M\mathrm{/}(2n+1){)}^r]$, which tends to zero. The transported cube witnesses the required condition in the original group. All choices concern a single finitely generated subgroup and a finite test set.

Suppose $1\to H\to G\stackrel{\pi }{\to }Q\to 1$ is any group extension and both $H,Q$ satisfy [F1]. Fix finite $S\subseteq G$ and $\epsilon >0$. Choose a finite nonempty $(\pi (S),\epsilon \mathrm{/}2)$-Folner set $E\subseteq Q$. For the finite set $E\cup \pi (S)E$, choose lifts $t_q\in G$. For each $s\in S$ and $q\in E$ with $\pi (s)q\in E$, put $h_{s,q}=t_{\pi (s)q}^{-1}s{t}_q\in H$. This is a finite collection of elements of $H$. Choose a finite nonempty set $B\subseteq H$ with $\mathrm{\mid }{h}_{s,q}B\mathrm{△}B\mathrm{\mid }<(\epsilon \mathrm{/}2)\mathrm{\mid }B\mathrm{\mid }$ for all of them, and let $F={\bigcup }_{q\in E}{t}_{q}B$. Its cosets are disjoint, so $\mathrm{\mid }F\mathrm{\mid }=\mathrm{\mid }E\mathrm{\mid }\mathrm{\mid }B\mathrm{\mid }$.

For $s\in S$, the cosets present in exactly one of $sF,F$ contribute $\mathrm{\mid }\pi (s)E\mathrm{△}E\mathrm{\mid }\mathrm{\mid }B\mathrm{\mid }$ elements to the symmetric difference. On a common coset indexed by $\pi (s)q$, the two subsets are $t_{\pi (s)q}{h}_{s,q}B$ and $t_{\pi (s)q}B$, whose difference has cardinality $\mathrm{\mid }{h}_{s,q}B\mathrm{△}B\mathrm{\mid }$. There are at most $\mathrm{\mid }E\mathrm{\mid }$ common cosets. Therefore $\mathrm{\mid }sF\mathrm{△}F\mathrm{\mid }\le \mathrm{\mid }\pi (s)E\mathrm{△}E\mathrm{\mid }\mathrm{\mid }B\mathrm{\mid }+{\sum }_{q\in E,\pi (s)q\in E}\mathrm{\mid }{h}_{s,q}B\mathrm{△}B\mathrm{\mid }<\epsilon \mathrm{\mid }E\mathrm{\mid }\mathrm{\mid }B\mathrm{\mid }=\epsilon \mathrm{\mid }F\mathrm{\mid }$. This proves the extension assertion for an arbitrary extension; the lifts were not required to form a homomorphism or a global transversal.

The trivial group satisfies [F1] using its singleton set. Induct up the given finite subnormal series. At each stage the kernel is the previous subgroup and the quotient is torsion-free abelian, so step 1.1 supplies the quotient condition and step 2.1 supplies the extension condition. This proves [F1] for every PTFA group in the full subnormal convention of [F3]. The construction uses only finitely many witnesses at each of finitely many stages. It constructs finite sets directly, without invariant means, ultrafilters or simultaneous choices over a family of tests, and hence adds no choice hypothesis.