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I am using the following definition of amenable group: The at most countable group G is amenable iff there exists a sequence $\{F_n\}$ of finite subsets such that for every $g\in G$ we have $$\displaystyle\lim_{n\rightarrow\infty}\displaystyle\frac{|(gF_n)\triangle F_n|}{|F_n|}=0$$ and $G=\bigcup_n F_n$.

Now, I want to prove that a countable abelian group is amenable using just this definition. If $G=\{g_1, g_2,...\}$ in the following link an user says that the family $$F_n=\{a_1g_1+a_2g_2+\cdots+a_ng_n: |a_i|\leq n-i+1\} $$ works.

And I think that the following family it is useful as well $$F_n'=\{a_1g_1+a_2g_2+\cdots+a_ng_n: 0 \leq a_i \leq n\} $$

But I don't know how to prove this.

I could find these bounds $$|F_n|\leq (2n+1)(2n-1)\cdots 3 = (2n+1)!!$$ $$|F_n\cap gF_n|\leq (2n-1)(2n-1)!! $$ and $$|F_n'|\leq (n+1)^n$$ $$|F_n'\cap gF_n'|\leq n(n+1)^{n-1} $$

But I am not sure what step is next. Any help?

Thanks!

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I was the OP who originally gave the sequence $S_n=\{a_1g_1+\cdots+a_ng_n\,|\, \lvert a_i\rvert\leq n-i\}$ in question, and claimed that it forms a Følner sequence whenever $g_1,g_2,\ldots$ generate the abelian group $G$. 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 in this answer).

Neverthless, 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 $N$ the average size cannot diverge, since a $g_i$-string has maximum size $N$. 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 $\lvert g_iS_n\bigtriangleup S_n\rvert=0$.

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  • $\begingroup$ Very nice. When I see the observation that $S_n$ is a disjoint union of $g$-strings I say hmm, this seems promising, and I stop reading and start thinking (because I never understand anything unless I figure it out for myself). I come up with simple proof, look back and see it's exactly what you wrote. And there's a bonus - thinking about $g$-strings involves thinking about the string "$g$-string", and suddenly I burst out laughing. Doesn't matter what you say, will go to my grave uncertain whether you were trying to be funny... Thanks. $\endgroup$ – David C. Ullrich Jan 2 '16 at 14:45
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I can't tell you why that author's $F_n$ works. But I can show you how to prove that a countable abelian group is amenable, using nothing but that definition.

First, if $G$ is a finitely generated abelian group then $G$ is a direct sum of finitely many cyclic groups and one can easily give an explicit Folner sequence.

Now suppose that $G=(g_1,g_2,\dots)$ is abelian. Let $G_n$ be the subgroup generated by $g_1,\dots ,g_n$. Since $G_n$ is amenable there exists a finite set $F_n\subset G_n$ with $$\frac{|(g_jF_n)\triangle F_n|}{|F_n|}<\frac1n\quad(1\le j\le n).$$QED.

Edit: I see you included the condition $G=\bigcup F_n$ as part of the definition. That's not always included in the definition, as far as I've seen. Presumably because it's easy to see that it's irrelevant:

For the record I'm going to take the definition of "Folner sequence" as above except without that condition.

Note that if $(F_n)$ is a Folner sequence then it's easy to see that $\tilde F_n=t_nF_n$ defines another Folner sequence, for any $t_n\in G$. If $(F_n)$ is a Folner sequence for the group $G=(g_1,\dots)$ you can choose $t_n$ so that $g_n\in t_nF_n$, so $G=\bigcup \tilde F_n$.

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Edit: This was a comment to the effect that a statement labelled WK below would answer the OP's question. studiosus has pointed out that WK is false . I'm leaving this post here anyway; seems to me it at least clarifies some of the issues regarding the OP's question.

Original:

This is not an answer - it's a comment that doesn't fit in the comment box. I do have a thought that might help. Takes a little space to explain. All groups below are countable abelian.

Yes, he says that. In fact he says it's so if $\{g_1,g_2,\dots\}$ is just a set of generators. I can't imagine how one could prove that it's so. Which proves nothing.

I wonder whether what he says is actually so. Not that I'm qualified to speculate, knowing nothing about all this. But here's why I wonder: He goes on to say that another way to think of the same argument is [blah blah]. I actually do see how [blah blah] shows that a countable abelian group is amenable, but not using just the definition you say you want to use. I don't see how [blah blah] is another way to think of the same argument, quite.

What seems to me to be the "real" definition of amenability is this:

(1) There exists a finitely additive translation invariant probability measure (defined on the whole power set).

The definition you want to use is

(2) There exists a Folner sequence

(where the definition of "Folner sequence" is as in "your" definition).

I actually see how to show that (2) implies (1), although it's not quite trivial. I have no idea why (1) implies (2).

Anyway, [blah blah] amounts to this:

(i) If we let $(g_j)$ be the standard generators for $\Bbb Z^\omega$ (so $g_1=(1,0,0,\dots)$, etc.) then the $F_n$ above are a Folner sequence for $\Bbb Z^\omega$, hence $\Bbb Z^\omega$ is amenable.

(ii) Quotients of amenable groups are amenable.

I haven't worked it out, but (i) seem quite plausible, and it seems like the sort of thing one could establish by a simple counting argument.

And (ii) is more or less obvious, if we take (1) as the definition of amenability. Given the mysterious equivalence of (1) and (2), the following follows:

True Fact: The existence of a Folner sequence is preserved under quotients.

It is not clear to me whether the following stronger statement is true:

Who Knows (WK)? The image of a Folner sequence under a quotient is another Follner sequence.

And so we come to my point: If WK is actually true then you're set. It looks as though the author is using WK. If I wanted to settle this I'd try to prove WK. Or omg construct a counterexample; if WK is false then I have no idea why the author's claim that $(F_n)$ is a Folner sequence might hold, while if WK is true then it's all clear.

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