Tarski proved that if a group $G$ is exponentially bounded, then for $a$, $b$ and $c$ in the associated (equidecomposability) type semigroup, we have $a+c=b+2c \Rightarrow a=b+c$. 


**Question:** Can this Tarski condition on the type semigroup be proved if one replaces "exponentially bounded" with "supramenable"? (I assume it's still not known if all supramenable groups are exponentially bounded.)


**Update:** Given Choice, this Tarski condition implies supramenability: If $(m+1)c\le mc$ for $m\in\mathbb Z^+$ and non-zero $c$ in the semigroup, then for some $d$, we have $d+(m+1)c=mc$ and iterating the Tarski condition we get $d+c=0$, which is impossible. Thus, $(m+1)c\not\le mc$, and if $c=[A]$ then that's what's needed for Tarski's Theorem to yield a $G$-invariant measure that assigns $1$ to $A$. (This gives an alternate proof of Rosenblatt's result that exponential boundedness implies supramenability.) So the question is basically whether the Tarski condition is equivalent to supramenability.


Mark Sapir asked for some definitions, which I originally put in the comments but now am moving here:


1. $G$ is supramenable iff for all $\varnothing\ne A\subseteq G$ there is a finitely-additive $\mu:2^G\to[0,\infty]$ with $\mu(A)=1$. Given Choice, this is true iff $G$ has no non-empty $G$-paradoxical subsets.
2. $G$ is exponentially bounded iff for all finite $F\subseteq G$ containing identity and closed under taking inverses we have $\lim |F^n|^{1/n} = 1$, where $F^n$ is the set of products of $n$ elements of $F$. ([Rosenblatt 1974](http://www.ams.org/journals/tran/1974-193-00/S0002-9947-1974-0342955-9/S0002-9947-1974-0342955-9.pdf) proved that exponentially bounded groups are supramenable.)
3. The type semigroup, due to Tarski, is defined on pages 164-5 of [this piece](http://rendiconti.dmi.units.it/volumi/23/04.pdf) by Laczkovich.



Does supramenability imply that $a+c=b+2c \Rightarrow a=b+c$ on the type semigroup?
The Tarski condition $a+c=b+2c\Rightarrow a=b+c$ is equivalent to supramenability, given AC.


Proof:


First, note that the Tarski condition on the type space $S=S(G)/G$ is easily equivalent to the condition that $a+b=2a\Rightarrow a=b$. (This is called strong separativity, I am told.) This is equivalent to the condition that $a=2a\Rightarrow a=0$, which is equivalent to supramenability. 


One direction of proof is trivial. 


Now suppose $x=2x\Rightarrow x=0$ for all $x$, and suppose that $a+b=2a$. If $a\le nb$ for some $n\in \mathbb N$, then by Lemma 7.6 of [Laczkovich](http://rendiconti.dmi.units.it/volumi/23/04.pdf), from $a+a=a+b$ we can conclude $a=b$ and we're done. 


So suppose that for no $n$ do we have $a\le nb$. Let $\phi:S\to[0,\infty]$ be the homorphism given by $\phi(x)=0$ if $x\le nb$ for some $n$ and $\phi(x)=\infty$ otherwise. Let $\mu:2^G\to[0,\infty]$ be the finitely-additive measure corresponding to $\phi$. 


By Prop 1.7 of [Armstrong](http://books.google.com/books?id=TXhb5wtS8D0C) (page 7; this uses AC), $\{ \mu \}$ can be extended to a dimensionally complete set $M$ of measures linearly ordered under Renyi order $\prec$, where $\nu\prec\rho$ iff $\nu(A)<\infty\Rightarrow \rho(A)=0$, and a set of measures on $G$ is dimensionally complete iff for each $\varnothing\ne A\subseteq G$ there is a unique $\nu$ in the set such that $\nu(A)\in(0,\infty)$.


Let $A$ and $B$ be subsets of $G$ such that $A=\pi[\alpha]$ and $B=\pi[\beta]$ where $a=[\alpha]$ and $b=[\beta]$ and $\pi$ is the projection of $G\times \mathbb N\to \mathbb N$ ($S$ is a set of equivalence classes in a subset of $g\times\mathbb N$). Observe that if $\psi:S\to[0,\infty]$ is a homomorphism and $\nu$ is the corresponding measure on $G$, then $\psi(a)\in(0,\infty)$ iff $\nu(A)\in(0,\infty)$ and the same for $b$ and $B$.


Let $\nu\in M$ be such that $\nu(A)\in(0,\infty)$. Note that $\mu(A)=\infty$ as $\phi(a)=\infty$. Thus, $\nu\prec \mu$ is impossible, and they are plainly not equal, so $\mu\prec\nu$. Now, $\mu(B)=0$. Thus, $\nu(B)=0$. Let $\psi:S\to[0,\infty]$ be the homomorphism corresonding to $\nu$. Then $\psi(a)\in(0,\infty)$ and $\psi(b)=0$. But $a+b=2a$ so $\psi(a)+\psi(b)=2\psi(a)$, which is impossible.


