Let $G$ be a finite group with $H,K$ subgroups. Suppose that there is a series of normal subgroups: $$ G \ \trianglerighteq G_1 \ \trianglerighteq G_2 \ \trianglerighteq \dots \ \trianglerighteq G_r = H. $$
How can I prove that $(H:H\cap K)$ divides $(G:K)$? I've tried to use some very common identities, but nothing works, because I can't write $(G:H)$.
Any help would be very appreciated! Thanks in advance!
Edit: After some digging, I think I was able to answer the question. Well, is true that $$ (G:H \cap K) = (G:H)(H:H\cap K) $$ and $$ (G:H \cap K) = (G:K)(K:H\cap K). $$ So, $$ (G:H)(H:H\cap K) = (G:K)(K:H\cap K) \implies \frac{(G:H)}{(K:H\cap K)}(H:H\cap K)=(G:K). $$ But, since $H$ is normal in $G$, we have $$\frac{(G:H)}{(K:H\cap K)} = \frac{\frac{|G|}{|H|}}{\frac{|K|}{|H\cap K|}} = \frac{|G|}{\frac{|H||K|}{|H\cap K|}} = \frac{|G|}{|HK|} = (G:HK) \in \Bbb{N}. $$
So, $(H:H\cap K)$ divides $(G:K)$.