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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 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,


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mmm.. following the classical proof word by word, it seems that associativity is necessary to prove that the natural action of the magma $M$ on the set $\mathcal M$ of probability measures on $M$ is commutative. Anyway, let me give the explicit example: $M=[-k,k]\subseteq \mathbb Z$ and $$ x\cdot y=\max(\min(x+y,k),-k) $$ Maybe in this case the existence of an invariant mean follows just from the fact that $M$ is finite. – Valerio Capraro Jun 5 '11 at 13:28

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.

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Thank you very much! – Valerio Capraro Jun 5 '11 at 23:47

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 $-2\cdot(-2,-1)=(-2)$.

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I'm very confused, because it seems to me that this same idea shows that any finite 1-generated monoid would be a counterexample unless it's actually a group. Apparently not - what's wrong with my thinking? – Harry Altman Jun 6 '11 at 0:33
Why is the measure uniform? A finite semigroup is amenable iff it has unique minimal right ideal. But with your proof, the semigroup $0,a$ with zero product is not amenable. – Mark Sapir Jun 6 '11 at 18:10
Why is not amenable? The measure that gives measure $1$ to the element $0$ and measure $0$ to $a$ is invariant with respect to the zero-product. In my example, I have proved that one can translate the singleton $(−2)$ on the element $(−1)$ and so on, so that all singletons should have the same measure with respect to a possible invariant mean, but this is not the case because at the same time the set $(−2,−1)$ should have the same measure of the singleton $(−2)$. It seems to me this does not contradict your example, because it is not possible to translate $0$ over $a$. Am I missing something? – Valerio Capraro Jun 6 '11 at 23:15

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