# Commensurability of finitely generated groups

Let’s begin with a definition.

Definition: Two groups $G,H$ are commensurable, which we’ll write $G\approx H$, if there exist subgroups $G_0 \leq G, H_0 \leq H$ such that $[G \colon G_0],[H \colon H_0]<\infty$ and $G_0 \cong H_0$.

This relation pops up in geometric group theory (among other places, I’m sure); for example the index in De La Harpe’s book lists about a dozen entries for “commensurable groups”.  Groups which are commensurable are easily seen to be quasi-isometric, and classifying groups up to quasi-isometry is a main focus in geometric group theory.

Commensurability of finitely generated groups is a countable Borel equivalence relation.  At some point, one begins to take the Borelness of these sorts of things for granted, so that part isn’t so surprising.  But the fact that the relation is countable is, I think.  It’s easy to see that a finitely generated group only has countably many finite-index subgroups; any finite-index subgroup is also finitely generated, and a finitely generated group only has countably many finitely generated subgroups.  But for this equivalence relation to be countable, one also needs to know that a given finitely generated group can only be a finite-index subgroup of countably many finitely generated subgroups.  This is not immediately obvious (at least, not to me).  As far as I can tell, the easiest way to see this is by appealing to the following theorem, due to Kaloujnine and Krasner:

Theorem (Kaloujnine-Krasner): Every extension of a group $N$ by a group $H$ is embeddable in the unrestricted wreath product $N \mathbin{Wr} H$.

So say that $N$ is a finite-index normal subgroup of a group $G$.  Then $G$ is the extension of $N$ by some finite group $H$, and so $G$ embeds into $N \mathbin{Wr} H = N \mathbin{wr} H$, since $H$ is finite.  Since $N \mathbin{wr} H$ is countable, it contains only countably many finitely generated subgroups, so there are only countably many possibilities for $G$ given $H$.  There are also only countably many possibilities for $H$, so there are only countably many groups which $N$ for which can be a finite-index normal subgroup.  What if $N$ is not a normal subgroup?  Then it contains a finite-index normal subgroup, so everything still works out.

Recall the terminology from the last post.  We’re going to look at the proof of the following result.

Theorem (Thomas): Commensurability of finitely generated groups is a universal countable Borel equivalence relation.

Note that it’s easy to see that commensurability is weakly universal, since it contains isomorphism of finitely generated groups as a subequivalence relation, and isomorphism is universal.  So this proof makes exactly the leap from weakly universal to universal that we’d like to make with the theorems I discussed last time.

The proof proceeds by creating a reduction from isomorphism to commensurability.  In fact, the reduction is easy to write down.  Let $S$ be your favorite infinite, finitely generated simple group, and let $A_5$ denote the alternating group on 5 elements.  Then the map

$G \mapsto W_G = (A_5 \mathbin{wr} G) \mathbin{wr} S$

is our reduction.  Clearly isomorphic groups will be sent to commensurable (in fact, isomorphic) groups.  The trick, of course, is to verify that this sends nonisomorphic groups to non-commensurable groups.

— Showing $W_A \approx W_B$ implies that particular subgroups are isomorphic —

Suppose that $W_A \approx W_B$.  We want to show that $A \cong B$.  It is enough to show that $(A_5 \mathbin{wr} A) \cong (A_5 \mathbin{wr} B)$, by a result of P.M. Neumann which states that $A \mathbin{wr} B \cong A' \mathbin{wr} B'$ iff $A\cong A'$ and $B\cong B'$.  The idea of the proof is to show that given a finite index subgroup of $W_A$ , one must be able to somehow recover $(A_5 \mathbin{wr} A)$.

A priori this could be quite difficult.  Suppose that you know two groups $G,H$ are isomorphic, say via $\pi$.  Say $K$ is a subgroup of $G$ that is somehow special or contains information you’ve coded into the construction of $G$, and $H$ also contains a special subgroup $L$.  There is no reason that $\pi(K)$ should have anything to do with $L$, and so there is no reason to expect the information you’ve coded in $K$ is the same as the information you’ve coded in $L$.

Let $K_A$ denote the base group of $W_A$.  Then other results of P.M. Neumann show that $W'_A=[S,K_A]S$.  This characterization, combined with a few other P.M. Neumann results (and here is where we use that $S$ is infinite and simple) tells us that if $H_A$ is a finite-index subgroup of $W_A$, then $W'_A=H'_A$.  So indeed, we can try and use the commutator subgroup of $W_A$ to recover $A$.  (All of these P.M. Neumann results are from the same paper, On the structure of standard wreath products of groups, which is a very thorough account of properties of wreath products.) This is great, because if $H_A \cong H_B$, then the isomorphism must send $H'_A$ to $H'_B$. So if we’ve coded information into $H'_A$ it must be the same as the information in $H'_B$. In this case, we’re hoping to be able to recover $A$ from $H'_A$.

— Finding $A$ in $H'_A$

So if $W_A \approx W_B$, we know that $[S,K_A]S\cong [S,K_B]S$.  We want to keep peeling away until we can see $A$ and $B$.  It seems intuitively obvious that we can get rid of the $S$‘s, but how?  Let $sf\in [S,K_A]S$ be such that $f\in[S,K_A]$, and $s\in S$, and let $\pi(sf)=s'f'\in[S,K_B]S$ with $f'\in[S,K_B]$ and $1\neq s'\in S$.  We’d like to show that $s\neq 1$ iff $s'\neq 1$, since this implies $[S,K_A]\cong [S,K_B]$.

We again look for certain subgroups that we know any isomorphism must preserve.  It would be nice to prove something like the following: $C_{W'_A}(sf)$ is cyclic iff $s\neq 1$.  Then since $C_{W'_B}(s'f')\cong C_{W'_A}(sf)$, we’d be done.  Unfortunately, things aren’t quite so easy.  Define $L_{sf}$ to be the normal closure of $sf$ in $W'_A$, and $R_{sf}=\langle (sf)^{L_{sf}} \rangle$ (i.e. $R_{sf}$ is the normal closure of $sf$ in $L_{sf}$).  Then one can show that $C_{L_{sf}}(R_{sf})=1$ iff $s=1$.  Since all of these groups are preserved under isomorphism, we now know $[S,K_A] \cong [S,K_B]$.

I don’t really have a good idea of how one decides to look at these groups beyond trial and error.  The one thing I’ll note is that centralizers are relatively easy to understand in wreath products $A \mathbin{wr} B$, since conjugating elements of the base group $K$ by elements of $B$ just shifts them around, and understanding conjugation by elements of $K$ is really only as difficult as understanding conjugation in $A$.  In particular, clearly $C_{A \mathbin{wr} B}(K)=\{e\}$, and so any subgroup containing $K$ will have trivial centralizer with respect to the whole group.  So it makes sense to look at centralizers.

For $X\subseteq S$, let $K^X_A=\{f\in K_A \mid \sigma(f)\subseteq X\}$, where $\sigma(f)$ denotes the support of $f$.  Now by a result of P.M. Neumann (again, I know), $[S,K_A]=W'_A\cap K_A$.  Obviously if $|X|=1$ we have $K^X_A\cong A_5 \mathbin{wr} A$, so if $K^X_A \leq W'_A \cap K_A$, we’d be well on our way to being done.  But in general this doesn’t happen.  We can do almost as well though.  Call $A_5 \mathbin{wr} A=\hat{A}$.  Using the P.M. Neumann result, it’s not too hard to check that if $s\neq t\in S$, $(W'_A\cap K_A^{\{s,t\}})/(W'_A\cap K_A^{\{s\}}) \cong \hat{A}$, and the analogous statements are true for $W_B$.  (Obviously $K_A^{\{s,t\}}/K_A^{\{s\}} \cong \hat{A}$, so this is just a matter of checking that enough of these groups are contained in $W'_A$ for such an isomorphism to still exist.)  So it is enough to show that there are some $a,b,c\in S$ with $a\neq b$ such that $\pi(W'_A \cap K_A^{\{c\}})=W'_B \cap K_B^{\{a\}}$ and $\pi(W'_A \cap K_A^{\{1,c\}})=W'_B \cap K_B^{\{a,b\}}$.

For $f\in [S,K_A]$, let $L_f=\langle f^{[S,K_A]} \rangle$, and define $R_f = C_{[S,K_A]}(L_f)$.  Note the similarity to the groups looked at couple of paragraphs ago.  A quick calculation shows that every element of $L_f$ has support contained in $\sigma(f)$, so $L_f \leq W'_A \cap K_A^{\sigma(f)}$.  We also note that since $A_5$ is simple and nonabelian, $\hat{A}$ has a unique nontrivial minimal normal subgroup, its base group $\hat{K}_A$.  Minimality is easy to check.  For uniqueness, since $C_{\hat{A}}(\hat{K}_A)=\{e\}$, we find that any normal subgroup of $\hat{A}$ must contain $\hat{K}_A$.  That second fact can be used to show that $L_f$ contains all the elements of $K_A$ with support contained in $\sigma(f)$ and range contained in $\hat{K}_A$.  It’s easy to see what happens to these functions under conjugation, and so after further calculation we find that $C_{[S,K_A]}(R_f)=W'_A \cap K_A^{\sigma(f)}$.  We now know that $|\sigma(f)|=1$ iff $C_{[S,K_A]}(R_f)$ is minimal among all groups of the form $C_{[S,K_A]}(R_h)$ with $h\in [S,K_A]$, and since similar results hold in $[S,K_B]$, we get our desired result.

To sum up, I think the main insight is that some subgroups don’t really change under isomorphism. (There is a little work to get from asking about commensurability to asking about isomorphism, but this is taken care of by P.M. Neumann’s results.) Derived subgroups map to derived subgroups; centralizers map to centralizers. This means that we can try and use those subgroups to encode the information we’re interested in preserving. Perhaps it’s an obvious idea, but until I actually wrote out what was happening, I hadn’t thought of things in those terms.