Commensurability of Subgroups Let $G$ be some group, and let $S_1$ and $S_2$ be some subgroups. Then $S_1$ and $S_2$ are said to be commensurable iff $|S_1 : S_1 \cap S_2|$ and $|S_2 : S_1 \cap S_2|$ are finite. I am trying to show that this is an equivalence relation on the collection of all subgroups of $G$. Reflexivity and symmetry are trivial to verify; however, transitivity has been giving me trouble for a few days now. I found this, but I don't understand the following excerpt from the linked page: > $$[S_1:S_1∩ S_3] \le [S_1:S_1∩S_2∩S_3]$$ $$ = [S_1:S_1 ∩ S_2][S_1 ∩ S_2:S_1 ∩ S_2∩S_3]$$ $$\le [S_1:S_1∩S_2][S_2:S_2∩S_3]$$ I cannot figure out how to get the third line from the second line. Why is $$[S_1 ∩ S_2:S_1 ∩ S_2∩S_3] \le [S_2:S_2∩S_3]?$$

To simplify notation, let me write $S_{12}=S_1\cap S_2$, $S_{123}=S_1\cap S_2\cap S_3$, etc. Define $f:S_{12}/S_{123}\to S_2/S_{23}$ by $f(aS_{123})=aS_{23}$. To prove this is well-defined, we must show that if $aS_{123}=bS_{123}$ for $a,b\in S_{12}$ then $aS_{23}=bS_{23}$. But if $aS_{123}=bS_{123}$ then $b^{-1}a\in S_{123}$ which implies $b^{-1}a\in S_{23}$ so $aS_{23}=bS_{23}$.

Now I claim that $f$ is injective, and thus $[S_{12}:S_{123}]\leq[S_2:S_{23}]$. To prove this, suppose $a,b\in S_{12}$ are such that $aS_{23}=bS_{23}$. This means $b^{-1}a\in S_{23}$. Since $a,b\in S_{12}$, we also have $b^{-1}a\in S_1$, so $b^{-1}a\in S_{123}$. Thus $aS_{123}=bS_{123}$. That is, $f(aS_{123})=f(bS_{123})$ implies $aS_{123}=bS_{123}$, so $f$ is injective.

More generally, the same argument shows that if $A$ is a group with subgroups $B$ and $C$, then $[C:B\cap C]\leq [A:B]$. (In your case, we have $A=S_2$, $B=S_{23}$, and $C=S_{12}$.)