TITLE: Commensurability of Subgroups
QUESTION [5 upvotes]: 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 
$$[S_1 ∩ S_2:S_1 ∩ S_2∩S_3] \le [S_2:S_2∩S_3]?$$

REPLY [2 votes]: Just to complement Eric's answer:
for two subgroups $H,K$ of a group $G$, define $d(H,K)=[H:H\cap K][K:H\cap K]$. Here both indices are viewed as cardinals: if both are finite one takes the product, and if at least one is infinite, the product is meant to be the max (in accordance to obvious cardinal arithmetic).
Then $d$ is submuliplicative: $d(H,L)\subset d(H,K)d(K,L)$ for all $L$. In particular, the condition $H\sim K \Leftarrow d(H,K)<\omega$ is an equivalence relation (called being commensurate), and on commensuration classes, $\log d$ is a distance.
As a number of authors, I call groups $G_1$, $G_2$ commensurable if they have isomorphic finite index subgroups. This is a distinct notion (of course, any two commensurate subgroups of a given group are commensurable, and the converse fails). Authors also call it "abstractly commensurable" when they use "commensurable" in the other meaning.
"-able" suggests the possibility of something, here it is the existence of an isomorphism. It is not appropriate in the case of commensuration.