# Proving that commensurability is transitive

We have that two groups $\Gamma$ and $\Gamma'$ are commensurable if there exist finite index subgroups $G \leq \Gamma$ and $G' \leq \Gamma'$ such that $G \cong G'$. We denote this $\Gamma \approx \Gamma'$.

I am trying to prove that this gives a transitive relation, but I just don't see why it needs to be. Clearly if $\Gamma \approx \Gamma'$ and $\Gamma' \approx \Gamma''$ we have finite index subgroups $G \leq \Gamma$, $G' \leq \Gamma'$, $H' \leq \Gamma'$ and $H'' \leq \Gamma''$ with $G \cong G'$ and $H' \cong H''$ but this doesn't mean that $G \cong H''$...

I feel that I must be missing something obvious, but I can't see why we have to have finite index subgroups of $\Gamma$ and $\Gamma''$ which are isomorphic to each other.

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An idea:

$$G\cong G'\implies \,\forall\,H'\le \Gamma'\;\exists\, H\le \Gamma\;\;s.t.\;\;G\cap H\cong G'\cap H'$$

because isomorphic groups have isomorphic subgroups. But then in fact

$$G'\cap H'\cong K'\cap H''\;,\;\;\text{for some finite index}\;\ K'\le\Gamma'\;\text{(same argument as above)}$$

and now remember that $\,[G:H]\,,\,[G:K]<\infty\implies [G:H\cap K]<\infty$

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Thank you, I think I understand why that would work. – user79474 May 26 '13 at 14:15
But the $K'$ you mention is a subgroup of $\Gamma''$ I think? – user79474 May 26 '13 at 17:02
To make this idea work, you need to pick actual isomorphisms $g : G \to G'$ and $h : H' \to H''$, to represent all those vague $\cong$ symbols. – Lee Mosher Jun 21 '13 at 17:23