Commensurability of Subgroups

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 $C=S_{12}$.)

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.