How to prove this index is finite? Let $H,K<G$, $[G:H]$ is finite, show that $[K: H\cap K]$ is finite. 
Here "$[G:H]$" denote the left-coset index. 
My approach: 
I wrote out the partition of $G$ explicitly as
$$G=H+a_1H+\cdots+a_nH$$
I want to proceed to the following 
$$K=G\cap K=H\cap K+(a_1H)\cap K+\cdots+(a_nH)\cap K$$
I intended to show that
$$(aH)\cap K=a(H\cap K)$$
which seemed impossible to do. 
Any neat way to do this?
 A: I think you can do this by defining a proper map.  
Let $k (H \cap K)$ be a coset in $ K / H \cap K$.   Define a map 
$$k ( H \cap K) \mapsto k H $$
Prove this map is well defined.  Given two equal cosets
$k(H\cap K)=m(H \cap K)$, this implies $m^{-1} k  \in H\cap K \leq H$.
Thus $m^{-1} k H = H$ if and only if $k H = mH$.  So by the glory of shrek, the map is well defined. 
Now if the map is injective, each coset $k ( H \cap K)$ corresponds to one and only one coset in $G/H$.   That is, no two different cosets in $K / H \cap K$ cosets map to the same coset in $G/H$.  Thus if we show injectivity, we can prove 
$$| K/ H \cap K | \leq |G/ H| $$
I think this is pretty easy to see. 
and this proves our problem.
Suppose $kH = mH$ for $k, m \in K$.  Then $m^{-1} k H = H$.  This implies 
$m^{-1} k \in H$.  Thus $m^{-1} k H \cap K = H \cap K$ and thus $k (H \cap K) = m (H \cap K)$. So that the map is indeed injective and we proved that 
$$[K : H \cap K] \leq [G : H] < \infty $$
Just to make sure, in general given a function  $f: A \to B$ that is injective, then $|A| \leq |B|$.
$f: A \to f(A)$ is already a bijection and so 
$$|A|  = |f(A)| \leq |B|$$
