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How does one go about proving that the free group $<a,b,a^{-1},b^{-1}>$ lacks any Følner sequence?

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    $\begingroup$ I think it's easiest to show $F_2$ is not amenable by exhibiting a paradoxical decomposition, rather than working directly with Følner sets. $\endgroup$ – Cheerful Parsnip Nov 17 '14 at 23:10
  • $\begingroup$ What do $a^{-1}$ and $b^{-1}$ mean here? Usually one puts the generators first, and then the relations, if any, so are all of $a,b,a^{-1},b^{-1}$ considered to be free generators? [what confuses me is that in a free group one would automatically consider an inverse of a generator to be included without mentioning it] $\endgroup$ – coffeemath Nov 17 '14 at 23:10
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    $\begingroup$ @coffeemath, I agree. I just mean $<a,b>$ without any relations that are not imposed/guaranteed by the group axioms. My math professor wrote it as presented above though. It is just the set of all reduced words over alphabet $\{a,b\}$ endowed with a group operator. $\endgroup$ – user173897 Nov 17 '14 at 23:14
  • $\begingroup$ @GrumpyParsnip, I am not familiar with any of that terminology, so a more detailed explanation would be nice. My professor randomly assigned this question (it is off-topic to the class and we have not covered anything like/about this at all). It is not really homework, just a "you should look into this" sort of thing. $\endgroup$ – user173897 Nov 17 '14 at 23:16
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I'm going to sketch the standard proof, but it actually has a sticky technical point, so I'm not sure how satisfying you will find it.

First, a group $G$ is said to be amenable if it has a finitely additive, left-invariant probability measure. That is, there is a function $\mu\colon \mathcal P(G)\to [0,1]$ such that $\mu(G)=1$, $\mu(X\amalg Y)=\mu(X)+\mu(Y)$, and for all $g\in G$, $\mu(X)=\mu(gX)$. (Here $X \amalg Y$ is the union of $X,Y$ assuming they are disjoint.)

Now if a group has a Følner sequence $F_i$, you can define $$\mu(X)=\operatorname{ulim}_{i\to \infty} \frac{|X\cap F_i|}{|F_i|}.$$ The fancy $\operatorname{ulim}_{i\to\infty}$ is the ultrafilter limit, and is a way to force limits to exist even when they don't. (This is that sticky technical point.) You can verify that $\mu$ is a finitely additive probability measure. See Wikipedia's article on Følner sequences. So in other words, the existence of a Følner sequence implies amenability.

But now, we can also show that $\langle a,b\rangle$ has no such probabilty measure. Let $W(\alpha)$ be the set of reduced words beginning with $\alpha$. Then \begin{align*} \langle a,b\rangle &=\{1\}\amalg W(a)\amalg W(a^{-1})\amalg W(b)\amalg W(b^{-1})\\ &=W(a)\amalg aW(a^{-1})\\ &=W(b)\amalg bW(b^{-1}) \end{align*} Applying $\mu$ gives a contradiction, since the latter two equations give that $\mu(W(a)\amalg aW(a^{-1}))=\mu(W(b)\amalg bW(b^{-1}))=1$, and then the first equation gives $1=\mu(\{1\})+2$, which is a contradiction.

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  • $\begingroup$ Thank you. I am going to try to process all of that. I may come back occasionally with further questions, but I appreciate the work that you have done already. $\endgroup$ – user173897 Nov 18 '14 at 0:33
  • $\begingroup$ Sure, no problem. $\endgroup$ – Cheerful Parsnip Nov 18 '14 at 0:47

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