# Maximal embedding of the intersection of commensurable subgroups

Let $G$ and $H$ be infinite subgroups of some ambient group, and let $\Gamma$ denote their intersection. If $$\Gamma :G$$ and $$\Gamma: H$$ are finite, so that $G$ and $H$ are commensurable, when does it hold that $\gcd($\Gamma :G$,$\Gamma: H$)=1$? Does anybody have any ideas about how one might approach a problem like this?

This is always the case if $G$ (thus also $H$) is an infinite cyclic group because $\Gamma$ is generated by $g^N=h^M$ for some pair of co-prime integers $M,N$. If this is a silly question please excuse me- I am not a group theorist and I don't even have any intuition for how hard this question is!

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First, you reversed the order of the groups in the indices, i.e. should be $[G:\Gamma]$ rather than $[\Gamma:G]$. Second, I cannot understand your argument about an infinte cyclic group. Could you please clarify this? –  Yiftach Barnea Mar 20 '11 at 23:34
Here's some intuition for how to approach it: try the next easiest example after $\mathbb{Z}$, namely $\mathbb{Z}^2$. –  HJRW Mar 21 '11 at 0:03
Or even $\mathbb{Z}/2\times\mathbb{Z}/2$. –  HJRW Mar 21 '11 at 2:42
You could perhaps try to understand $G \ast_\Gamma H$? I don't understand the kind of statement you are hoping to prove. –  Colin Reid Mar 21 '11 at 15:38
Also, do you require $G \cong H$? –  Colin Reid Mar 21 '11 at 15:40
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