Let K<=H<=G (subgroups). Prove that if index (G:K) is finite, then indices (H:K) and (G:H) are also finite.
Let B be a group and A<B be a subgroup. Say that two elements x,y from B are equivalent x ~ yif x = ay for some a from A. Each equivalency class is called adjacent class, and theset of all such classes (i.e. factor space) is called the set of adjacency classes and denoted by B/A. Then the index (B:A) is the number of elements in B/A.
Thus each adjacency class of B/A has the form A*y for some y from B.
Suppose now that K < H < G.
1) Let K*x, (xfrom H), be any adjacency class of H/K. Since x belogns to G as well, we see that K8x is also anadjacency class of G/K, whence number (H:K) of classes in H/K is not greater than the number (G:K) of classes in G/K, and so (H:K) <= (G:K) < infinity, so (H:K) is finite.
2) Let H*y, (y from G), be any adjacency class of G/H. Since H is a dijoint union of (H:K) adjacency classes H*x_1, H*x_2, ... we see that Hy is also a disjoint union of (H:K) classes H*x_1*y, H*x_2*y, ...
As (G:K), and (H:K) are finite we obtain that (G:H) = (G:K) / (H:K) and this number is also finite.