Invariant means on commutative magmas It is a very standard fact that commutative semigroups admit an invariant mean and the proof basically relies on Markov-Kakutani fixed point theorem. Now, it seems to me that the proof of this theorem does not need the associativity (see for instance 
http://www.springerlink.com/content/g777570166731376/) and then the answer to the following question should be affirmative.

Question: Recall that a magma is a set equipped with a binary operation. Is it true that a commutative countable magma admits an invariant mean?

So, is that trivially true or am I missing some detail?
Thanks in advance,
Valerio
 A: Consider the free group $F$ generated by $a,b$ with the new operation: $u*v=$ the shortlex smallest word among $uv$ and $vu$ assumming $a>b>a^{-1}>b^{-1}$. This is a commutative magma. Consider the decomposition $F=F_a\cup F_b\cup F_{a^{-1}}\cup F_{b^{-1}}\cup \{1\}$ where $F_x$ is the set of (reduced) words starting with $x$. Would not this be a paradoxical decomposition of the magma?  This would show that the magma does not have an invariant mean. 
 Update Here is the proof. Suppose that there exists an invariant mean $m$. Note that for every reduced word $au$, $a^{-1}*au=u$ because $a$ is the biggest letter. Hence $a^{-1}F_a=F_b\cup F_{b^{-1}}\cup F_a\cup \{1\}$. Therefore $m(F_b)=m(F_{b^{-1}})=m(\{1\})=0$. Consider the set $F_{b,b'}$ of reduced words starting with $b$ and ending not in $b$. Then $b^{-1}F_{b,b'}$ contains all words starting in $a$ or $a^{-1}$ and ending not in $b$. Therefore the set of all words starting with $a$ or $a^{-1}$ and ending in $b$ has a full measure. Taking a word starting in $b^{-1}$ and ending in $b$ and multiplying it with $b$ on the left produces any word ending in $b$. Therefore the set of such words has measure 0, a contradiction.   
A: For the ones of you that are interested, here is a "finite" counterexample.
(How can I do the braces? I am going to use the round bracket, instead of brace)
Let $M=(-2,-1,0,1,2)$ equipped with the commutative operation $x\cdot y=\max(\min(x+y,2),-2)$. Observe that $1\cdot(-2)=(-1)$, and then the measure of the singleton $(-1)$ should be equal to the measure of the singleton $(-2)$. At the same way, one gets that the only possible invariant mean would be the uniform measure, but this is not invariant since $-2\cdot(-2,-1)=(-2)$.
