Let $F$ be a non-abelian free group, say finitely generated. Suppose $N$ is a commensurated subgroup of $F$. That is $N \cap N^g$ is of finite index in both $N$ and in $N^g$ for any $g \in F$.

What is known about $N$?

I guess $N$ does not have to be normal as it can be normal in a subgroup of finite index or of finite index in a normal subgroup. Are there other examples?

  • 2
    $\begingroup$ To complement Uri's answer, the answer is yes if $N$ is finitely generated: then $N=1$ or $N$ has finite index. This follows from Marshall Hall's theorem. $\endgroup$
    – YCor
    May 3, 2018 at 22:19
  • $\begingroup$ Let me also remark that for some groups it is known that every commensurated subgroup is virtually normal. As follows from my answer below, this is not the case for $\text{SL}_2(\mathbb{Z})$, which is virtually free. However this is the case for $\text{SL}_n(\mathbb{Z})$, $n\geq 3$. This result is due to Venkataramana. Further results in this direction were also proved by Shalom-Willis. In general, the Margulis-Zimmer conjecture, stating that in higher rank lattices every commensurated subgroup is virtually normal, is still open. Problem being the cocompact lattices. $\endgroup$
    – Uri Bader
    May 4, 2018 at 7:49

2 Answers 2


There are many commensurated subgroups in a free groups which are not of the kind you described above.

We all know that normal subgroups $N\lhd \Gamma$ could be thought of as kernels of surjective maps $\Gamma\to G$. Here is an analogous way of thinking of commensurated subgroups $N<\Gamma$.

Fix a group $\Gamma$. For every totally disconnected locally compact (tdlc) group $G$, a compact open subgroup $K<G$ and a homomorphism $\rho:\Gamma\to G$, $N=\rho^{-1}(K)$ is commensurable in $\Gamma$ (as $K$ is commensurtaed in $G$). In fact, every commensurated $N<\Gamma$ is obtained this way (a construction of a group $G$ and a map $\rho$ as above is called sometimes the "Schlichting completion" of $\Gamma$ wrt $N$).

If $\rho(\Gamma)$ is dense in $G$ then (a finite index subgroup of) $K$ is normal in (a finite index subgroup of) $G$ iff (a finite index subgroup of) $N$ is normal in (a finite index subgroup of) $\Gamma$. This is easy to see.

Now, to give an example answering the question just fix a tdlc group that you know well enough (say $\text{PGL}_2(\mathbb{Q}_p)$) and a compact open subgroup which is far from being normal (say $\text{PGL}_2(\mathbb{Z}_p)$) and find an injective dense homomorphism $\rho:F_n\to G$ by fixing any generic enough $n$-tuple in $G$ (which could be made very concrete).


As Uri says, there are many commensurated subgroups that are not closely related to normal subgroups. I don't know anything special to say about free groups. However, there are still interesting connections between commensurated subgroups and 'virtually normal' subgroups, where $N$ is virtually normal if there is some subgroup $M$ of $N$ of finite index that is normal in $\Gamma$. The main result I know about this is joint work with Caprace, Kropholler and Wesolek: https://arxiv.org/abs/1706.06853

Given a group $\Gamma$ and a subgroup $N$, define the residual closure of $N$ in $\Gamma$ to be $\widetilde{N} = \bigcap_{K \in \mathcal{K}} K$, where $\mathcal{K}$ is the class of virtually normal subgroups that contain $K$ (alternatively, take the class of subgroups $LN$ where $L$ is normal in $\Gamma$ and $LN/L$ is finite).

It turns out that if $N$ is commensurated and $\Gamma$ is generated by finitely many cosets of $N$, then $\widetilde{N}$ is virtually normal. In particular, if $\Gamma$ is a finitely generated group, any commensurated subgroup $N$ that arises as an intersection of virtually normal subgroups must itself be virtually normal.


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