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Does every proof that abelian groups are amenable rely on the axiom of choice?

So far, any proof I've seen that all, say countable discrete, abelian groups are amenable requires some sort of argument or technique that relies on choice, i.e. using Marcov-Kakutani or convergence w.r.t. ultrafilters, so I guess every proof needs choice. Is that the case?

To be clear, I don't have anything against choice, I'm just curious ;-)

Shaun
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j grk
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  • Even for $\mathbb{Z}$ the construction I know makes use of the Hahn-Banach theorem, i.e. the axiom choice is needed. Perhpas the reason is that the invariant mean is not unique. – Ryszard Szwarc Apr 01 '22 at 08:48
  • Yes, but what I wanted to know that the usage of choice is always required for any abelian group. – j grk Apr 01 '22 at 08:54
  • See also this MO-post, using Følner sequences and Terry Tao's answer including his blog (where he uses ultralimits for Proposition 6). So every proof so far uses it, it seems. – Dietrich Burde Apr 01 '22 at 09:11
  • In Tao's notes he makes use of the Hahn-Banach theorem as well, for finitely generated groups. Ultralimits are used for unions of such groups. Even for infinite group with one generator, i.e $\mathbb{Z}$, the axiom of choice is used. I do not know if it is possible to avoid that and indicate a particular mean. For some nonamenable groups, like free group with two generators, inexistence does not require axiom of choice. – Ryszard Szwarc Apr 01 '22 at 09:27
  • Perhaps you are asking whether it is true that the statement "every abelian group is amenable" implies the axiom of choice? Here is a related post regarding the Hahn-Banach theorem itself (although that post does not have any resolution). – Lee Mosher Apr 01 '22 at 13:33
  • And here is another related post regarding whether the axiom of choice is equivalent to the statement that every abelian group is injective. – Lee Mosher Apr 01 '22 at 13:37

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