A (discrete) group is amenable if it admits a finitely additive probability measure (on all its subsets), invariant under left translation. It is a basic fact that every abelian group is amenable. But the proof I know is surprisingly convoluted. I'd like to know if there's a more direct proof.

The proof I know runs as follows.

  1. Every finite group is amenable (in a unique way). This is trivial.

  2. $\mathbb{Z}$ is amenable. This is not trivial as far as I know; the proof I know involves choosing a non-principal ultrafilter on $\mathbb{N}$. This means that $\mathbb{Z}$ is amenable in many different ways, i.e. there are many measures on it, but apparently you can't write down any measure 'explicitly' (without using the Axiom of Choice).

  3. The direct product of two amenable groups is amenable. This isn't exactly trivial, but the measure on the product is at least constructed canonically from the two given measures.

  4. Every finitely generated abelian group is amenable. This follows from 1--3 and the classification theorem.

  5. The class of amenable groups is closed under direct limits (=colimits over a directed poset). This is like step 2: it seems that there's no canonical way of constructing a measure on the direct limit, given measures on each of the groups that you start with; and the proof involves choosing a non-principal ultrafilter on the poset.

  6. Every abelian group is amenable. This follows from 4 and 5, since every abelian group is the direct limit of its finitely generated subgroups.

Is there a more direct proof? Is there even a one-step proof?

Update Yemon Choi suggests an immediate simplification: replace 1 and 4 by

1'. Every quotient of an amenable group is amenable. This is simple: just push the measure forward.

4'. Every f.g. abelian group is amenable, by 1', 2 and 3.

This avoids using the classification theorem for f.g. abelian groups.

Tom Church mentions the possibility of skipping steps 1--3 and going straight to 4. If I understand correctly, this doesn't use the classification theorem either. The argument is similar to the one for $\mathbb{Z}$: one still has to choose an ultrafilter on $\mathbb{N}$. (One also constructs a Følner sequence on the group, a part of the argument which I didn't mention previously but was there all along).

Yemon, Tom and Mariano Suárez-Alvarez all suggest using one or other alternative formulations of amenability. I'm definitely interested in answers like that, but it also reminds me of the old joke:

Tourist: Excuse me, how do I get to Edinburgh Castle from here?

Local: I wouldn't start from here if I were you.

In other words, if a proof of the amenability of abelian groups uses a different definition of amenability than the one I gave, then I want to take the proof of equivalence into account when assessing the simplicity of the overall proof.

Jim Borger points out that if, as seems to be the case, even the proof that $\mathbb{Z}$ is amenable makes essential use of the Axiom of Choice, then life is bound to be hard. I take his point. However, one simplification to the 6-step proof that I'd like to see is a merging of steps 2 and 5. These are the two really substantial steps, but they're intriguingly similar. None of the answers so far seem to make this economy. That is, every proof suggested seems to involve two separate Følner-type arguments.

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    $\begingroup$ In #6, there appears to be a typo. Shouldn't the beginning of #6 read "every abelian group is amenable"? $\endgroup$ Jan 18, 2010 at 3:05
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    $\begingroup$ Thanks mathphysicist. Fixed. (I wrote "every abelian group is abelian", which even I can prove.) $\endgroup$ Jan 18, 2010 at 3:15
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    $\begingroup$ Can't one establish amenability of $\mathbb Z$ by using the second characterization of amenability given in wikipedia (Γ is amenable iff whenever Γ acts by isometries on a (separable) Banach space E, leaving a weakly closed convex subset C of the closed unit ball of E* invariant, then Γ has a fixed point in C) and the Ryll-Nardzewski fixed point theorem? This also needs Choice in order to prove Tychonof in order to prove the Banach–Alaoglu theorem, yet it seems more natural... $\endgroup$ Jan 18, 2010 at 3:36
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    $\begingroup$ I know nothing about these things, but here are some general remarks (which are probably not news to most people). It seems to me that if you need the axiom of choice for even the simplest of groups, like Z, then without refining your definitions, you shouldn't be surprised if every little result becomes a long, hard slog. An obvious attempt to repair this would be to find a version of amenability which (i) is equivalent assuming the axiom of choice and (ii) holds for Z without assuming the axiom of choice. $\endgroup$
    – JBorger
    Jan 18, 2010 at 4:09
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    $\begingroup$ "More natural" means mostly "I have trained myself to think it is natural that an arbitrarily humogous product of intervals is compact", I guess. $\endgroup$ Jan 18, 2010 at 4:09

5 Answers 5


Here is a simpler argument, combining 1--6 into one step.

Let $G$ be a countable abelian group generated by $x_1,x_2,\ldots$. Then a Følner sequence is given by taking $S_n$ to be the pyramid consisting of elements which can be written as

$a_1x_2+a_2x_2+\cdots+a_nx_n$ with $\lvert a_1\rvert\leq n,\lvert a_2\rvert\leq n-1,\ldots,\lvert a_n\rvert\leq 1$.

The invariant probability measure is then defined by $\mu(A)=\underset{\omega}{\lim}\lvert A\cap S_n\rvert / \lvert S_n\rvert$ as usual.

A more natural way to phrase this argument is:

  1. The countable group $\mathbb{Z}^\infty$ is amenable.
  2. All countable abelian groups are amenable, because amenability descends to quotients.

But I would like to emphasize that there is really only one step here, because the proof for $\mathbb{Z}^\infty$ automatically applies to any countable abelian group. This two-step approach is easier to remember, though. (The ideas here are the same as in my other answer, but I think this formulation is much cleaner.)

2016 Edit: Here is an argument to see that $S_n$ is a Følner sequence. It is quite pleasant to think about precisely where commutativity comes into play.

Fix $g\in G$ and any finite subset $S\subset G$. We first analyze the size of the symmetric difference $gS\bigtriangleup S$. Consider the equivalence relation on $S$ generated by the relation $x\sim y$ if $y=x+g$ (which is itself neither symmetric, reflexive, or transitive). We will call an equivalence class under this relation a "$g$-string". Every $g$-string consists of elements $x_1,\ldots,x_k\in S$ with $x_{j+1}=x_j+g$.

The first key observation is that $\lvert gS\bigtriangleup S\rvert$ is at most twice the number of $g$-strings. Indeed, if $z\in S$ belongs to $gS\bigtriangleup S$, then $z$ must be the "leftmost endpoint" of a $g$-string; if $z\notin S$ belongs to $gS\bigtriangleup S$, then $z-g$ must be the "rightmost endpoint" of a $g$-string; and each $g$-string has at most 2 such endpoints (it could have 1 if the endpoints coincide, or 0 if $g$ has finite order).

Our goal is to prove for all $g\in G$ that $\frac{\lvert gS_n \bigtriangleup S_n\rvert}{\lvert S_n\rvert}\to 0$ as $n\to \infty$. Since $\lvert abS\bigtriangleup S\rvert\leq\lvert abS\bigtriangleup bS\rvert+\lvert bS\bigtriangleup S\rvert= \lvert aS\bigtriangleup S\rvert+\lvert bS\bigtriangleup S\rvert$, it suffices to prove this for all $g_i$ in a generating set.

By the observation above, to prove that $\frac{\lvert g_iS_n \bigtriangleup S_n\rvert}{\lvert S_n\rvert}\to 0$, it suffices to prove that $\frac{\text{# of $g_i$-strings in $S_n$}}{\lvert S_n\rvert}\to 0$. Equivalently, we must prove that the reciprocal $\frac{\lvert S_n\rvert}{\#\text{ of $g_i$-strings in $S_n$}}$ diverges, or in other words that the average size of a $g_i$-string in $S_n$ diverges.

We now use the specific form of our sets $S_n=\{a_1g_1+\cdots+a_ng_n\,|\, \lvert a_i\rvert\leq n-i\}$. For any $i$ and any $n$, set $k=n-i$ (so that $\lvert a_i\rvert\leq k$ in $S_n$). The second key observation is that every $g_i$-string in $S_n$ has cardinality at least $2k+1$ unless $g_i$ has finite order. Indeed given $x\in S_n$, write it as $x=a_1g_1+\cdots+a_ig_i+\cdots+a_ng_n$; then the elements $a_1g_1+\cdots+bg_i+\cdots+a_ng_n\in S_n$ for $b=-k,\ldots,-1,0,1,\ldots,k$ belong to a single $g_i$-string containing $x$. If $g_i$ does not have finite order, these $2k+1$ elements must be distinct. This shows that the minimum size of a $g_i$-string in $S_n$ is $2n-2i+1$, so for fixed $g_i$ the average size diverges as $n\to \infty$.

When $g_i$ has finite order $N$ this argument does not work (a $g_i$-string has maximum size $N$, so the average size cannot diverge). However once $N<2k+1$, the subset containing the $2k+1$ elements above is closed under multiplication by $g_i$. In other words, once $n\geq i+N/2$ the set $S_n$ is $g_i$-invariant, so $\lvert g_iS_n\bigtriangleup S_n\rvert=0$.

I'm grateful to David Ullrich for pointing out that this claim is not obvious, since the quotient of a Følner sequence need not be a Følner sequence (Yves Cornulier gives an example here).

  • $\begingroup$ Thanks for the nice answer. I'd still like to get a one-step proof that works for all abelian groups, not just countable ones. That isn't because I'm particularly interested in uncountable abelian groups. Rather, it's because I find it intriguing that a fact about abelian groups is true in complete generality, but the known proofs all seem to go through multiple stages (well, at least two!), bootstrapping up through larger and larger classes of abelian group. Anyway, I've learned a lot from these answers, and yours is close enough to my ultimate goal that I'll click "accept". $\endgroup$ Jan 20, 2010 at 0:31
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    $\begingroup$ Probably a stupid question, but why is $S_n$ a Folner sequence? I don't see it. (I'll leave this comment short and to the point, and insert more confusion in a second comment...) $\endgroup$ Dec 31, 2015 at 15:07
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    $\begingroup$ I buy the second argument, but I don't see how it's a rephrasing of the first. Certainly amenability descends to quotients. So the existence of a Folner sequence descends to quotients. But the image of a Folner sequence under a quotient map need not be a Folner sequence - hence my confusion about your $S_n$. math.stackexchange.com/questions/1283824/… $\endgroup$ Dec 31, 2015 at 15:11
  • $\begingroup$ Maybe I should add that I tend to believe that $S_n$ actually is a Folner sequence; I just don't see how to get the estimates to prove it. Nor how it follows from the magic words "quotient map".. $\endgroup$ Dec 31, 2015 at 15:51
  • $\begingroup$ Tom: The elements $g_i$ are not free generators of a free abelian group, therefore, the elements of the $g_i$-string you listed need not be the same. Actually, my guess is that the sets you described is not a Folner sequence in general... $\endgroup$
    – Misha
    Jan 2, 2016 at 11:32

I believe that the neatest proof for amenability of abelian topological groups, is via the Markov-Kakutani fixed point theorem: see the half-page proof of this result as Theorem G.2.1 in: MR2415834 (2009i:22001) Bekka, Bachir; de la Harpe, Pierre; Valette, Alain Kazhdan's property (T). New Mathematical Monographs, 11. Cambridge University Press, Cambridge, 2008. xiv+472 pp. ISBN: 978-0-521-88720-5


There is a slightly quicker approach than that outlined by Mariano, using the Markov-Kakutani fixed point theorem. I first learned of this from Rudin's Functional Analysis -- more accurately, I remember skimming over that part as a struggling undergraduate, and then years later, once I'd heard of amenability, realizing the connection. (As far as I know, the word "amenable" is never mentioned in the book.)

The proof can be found via Theorems 5.23, 5.24 and 5.25 of the aforementioned book (2nd ed. if that makes any difference) and I don't think I can improve on the exposition there.

(Regarding the approach Tom outlines: it might be worth observing that amenability passes to quotient groups (this is particularly obvious using the invariant mean, but probably isn't too hard in most of the other formulations). Therefore, going from 3 to 4 doesn't need the classification theorem.)

A vague remark on possible alternative routes. From one point of view, the "invariant mean" formulation of amenability is just a convenient way to avoid epsilon-delta arguments with Folner nets. Now, Folner sequences can be easily constructed in ${\mathbb Z}^n$ for any finite $n$ -- just take cubes centred at zero of ever increasing width -- but at present I can't quite nail down how to transfer the construction over to arbitrary abelian groups. (Of course, things would be easier if we could pass to the formulation with an invariant mean; but then one might as well work with the invariant mean throughout.)

  • $\begingroup$ Thanks for fixing the links Michael -- I was attempting to use the interface's widget instead of using proper HTML, and something seemed to go down badly $\endgroup$
    – Yemon Choi
    Jan 18, 2010 at 4:16
  • $\begingroup$ Markdown provides a couple of other, less verbose notations for links. They are explained at mathoverflow.net/editing-help (I do not seem to be able to go to that page in any other way that by asking a question, though) $\endgroup$ Jan 18, 2010 at 4:22
  • $\begingroup$ @Leonid - quite right about being limited to the discrete case if you want Folner sequences. (If one believes that amenability is inherited by images of continuous group homomorphisms, then of course we only need the discrete case.) I wasn't sure about appealing to rates of growth -- how does your remark fit in with the existence of amenable groups of exponential or intermediate growth? $\endgroup$
    – Yemon Choi
    Jan 18, 2010 at 7:15
  • $\begingroup$ @Leonid: thanks, I really should have remembered that before I wrote my comment. (In defence, it was getting rather late and my brain was powering down, but still...) $\endgroup$
    – Yemon Choi
    Jan 18, 2010 at 20:38
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    $\begingroup$ I'm not sure what exactly the deleted comments you responded to said, but it is not true that Følner limits you to discrete groups. A locally compact group $G$ with Haar measure $\mu$ is amenable if and only if for every compact $K$ and every $\varepsilon \gt 0$ there is a compact set $F$ such that $\mu(F \mathbin{\Delta} kF) \leq \varepsilon \mu(F)$ for all $k \in K$. If $G$ happens to be $\sigma$-compact then you can arrange $F$ to be from an exhaustive sequence $F_n$. $\endgroup$ Jan 9, 2013 at 10:34

The simplest argument for me to keep in my head is (none of this is original, even to this thread):

  1. $\mathbb{Z}^n$ is amenable (because the cube of radius $k$ gives a Følner sequence).
  2. Thus finitely generated abelian groups are amenable (because amenability descends to quotients).
  3. Thus (discrete) abelian groups are amenable (because locally amenable groups are amenable).

Each of these properties is most easily proved using a different characterization of amenability, so this might not be pedagogically optimal, but since we already know the characterizations are equivalent we don't have to worry about that. Each of the parentheticals above is a property that everyone should have in mind when dealing with amenable groups, and so this argument also serves to remind me what's true in case I forget. Of course 1 and 2 could be combined (the ball of radius $k$ gives a Følner sequence in any f.g. abelian group, no need to be free) or deduced differently (you might prefer to justify 2 instead by "virtually amenable groups are amenable" or "amenable-by-amenable groups are amenable"), but I prefer this approach because it doesn't depend on the classification of f.g. abelian groups.

[Edit: upon rereading, I see that more of this than I realized was already present in Yemon's answer and the comments, so I'm making this community wiki.]

Second edit: in comments, Tom Leinster asked about seeing directly that f.g. abelian groups are amenable. Here are my thoughts: if you assume the classification and write $G=\mathbb{Z}^n\times T$ for $T$ finite, this is easy. Geometrically the most natural Følner sequence is the cube-of-radius-$k\times T$, but anything you try will work. If you don't want to use the structure theorem, I would note that if $G$ is abelian and has rank $n$, then it has polynomial growth of rank at most $n$; it follows that the ball of radius $k$ gives a Følner sequence. (Above I implicitly criticized this approach; that was because to bound the growth rate without knowing the structure of the group, you dominate it by the growth of the free abelian group. Thus this seemed to just be a hidden appeal to my step 2 above. It's perfectly valid though.)

Both approaches give a Følner sequence (because that's the definition of amenable I understand best). If you prefer the invariant mean definition, that's fine, but there is still value in sticking with Følner sequences as long as you can. For the simplest possible proof without relying on any equivalences, I would do the following:

  1. Finitely generated abelian groups admit Følner sequences (e.g. the ball of radius $k$).
  2. If $G$ is countable and every f.g. subgroup of $G$ admits a Følner sequence, then $G$ is amenable (admits an invariant f.a. probability measure).

To see 2, choose an increasing sequence of f.g. subgroups $G_i < G$ which exhausts $G$, and let $S_i^n$ be a Følner sequence for $G_i$; we can consider the $S_i^n$ as subsets of $G$. Then the measure we get is the "asymptotic density"

$\mu(A) = \underset{i\to \omega}{\lim}\underset{n\to\omega}{\lim}\ \ \vert A\cap S_i^n\vert / \vert S_i^n\vert$

where $\omega$ is a non-principal ultrafilter so that the limits exist. This is clearly a finitely additive probability measure, and to see that it's invariant note that every $\gamma\in G$ lies in all $G_N$ for $N$ sufficiently large.

  • $\begingroup$ It looks like your step 3 is the analogue of Tom's step 5; is it proven in the same way? $\endgroup$ Jan 18, 2010 at 19:46
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    $\begingroup$ Well, for what it's worth, Tom, I think you put very clearly what I was only groping towards mentioning: your three step argument is in fact what I had at the back of my mind when I write my own sleep-deprived answer. So +1, if only as a gesture :) $\endgroup$
    – Yemon Choi
    Jan 18, 2010 at 20:43
  • $\begingroup$ Tom, I like the idea of doing any f.g. abelian group instantly, with a single Folner sequence. But I'm unclear as to whether you're saying that needs the classification theorem (CT). I assumed "ball of radius r" referred to choosing a generating set then taking the usual length-of-word norm; and that strategy doesn't seem to depend on the CT. But then, if I understand correctly, you said that you preferred your original approach because it avoided the CT. $\endgroup$ Jan 18, 2010 at 22:08

The way I've seen it was also with the Markov-Kakutani fixed point theorem. The steps are these:

First define $$ K= \lbrace \Phi\in l^{\infty}(G)^\ast \mid \Phi(1) = 1 ,\Phi(F)\ge0\text{ whenever }F(g)\ge 0\text{ for all }g\in G\rbrace. $$

  1. Equip $l^{\infty}(G)^\ast $ with the weak$^\ast$-topology. Prove that $K$ is weak$^\ast$ compact and convex.

  2. Define for every $g\in G$ the map $$ T_g:K\to K:(T_g(\Phi)(F)=\Phi(F\cdot g). $$ Prove that every $T_g$, $g\in G$, is weak$^\ast$ continuous and affine.

  3. Since $G$ is a commutative group, we can apply the Markov-Kakutani fixed point theorem to the family $\lbrace T_g \mid g\in G\rbrace$ of affine maps from $K$ to $K$. So, we get $\Phi\in K$ such that $T_g\Phi=\Phi$ for all $g\in G$. Define $m(A)=\Phi(\chi_A)$ whenever $A\subset G$ and check that $m$ is an invariant mean on $G$.

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    $\begingroup$ In the definition of $K$ you should replace $\lVert \Phi\rVert \leq 1$ by $\Phi(\chi_G) = 1$ (which together with $\Phi(F) \geq 0$ implies $\lVert \Phi \rVert = 1$). The way you define it, $0 \in K$ is always a fixed-point, whether $G$ is amenable or not. $\endgroup$ Jan 9, 2013 at 11:37
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    $\begingroup$ @TheoBuehler I took the liberty of editing the post following your suggestion (it took only three years...) $\endgroup$ Jan 2, 2016 at 11:44

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