Q:

Orbit-Stabilizer Theorem to prove a quotient set inequality

Is there a way to apply the orbit-stabilizer formula to conclude that for $H,K
\leq G$, then $[G: H\cap K]\leq [G:H][G:K]$?
The inequality isn't too hard to see, just by taking the map from $g/(H\cap K)\to G/H\times G/K$ defined as
$$
g(H\cap K)\mapsto (gH,gK)
$$
which is well defined and injective. I want to know if there is perchance a kind of combinatorial argument using the orbits and stabilizers under a group action of some sort. Cheers.

A:

It suffices to show that $[H:H\cap K]\leq [G:K]$. For this, let $H$ act on the set of left cosets of $K$ in $G$ by left multiplication, and consider the orbit of $K$. The stabilizer of $K$ under this action is $H\cap K$; therefore, the number of elements in the orbit is $[H:H\cap K]$. Since the orbit of $K$ under this action is contained in the orbit of $K$ under the action of $G$ (which is transitive), we conclude that $[H:H\cap K]$ (the size of the orbit of $K$ under the action of $H$) is less than or equal to $[G:K]$ (the size of the orbit of $K$ under the action of $G$).
Now you have $[G:H\cap K] = [G:H][H:H\cap K] \leq [G:H][G:K]$, as desired. 
Note that this inequality holds in cardinality, so we don't even need the indices to be finite.

