Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 1 down vote accepted

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 $\,[G:H]\,,\,[G:K]<\infty\implies [G:H\cap K]<\infty$

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.