Homomorphism and indexes of subgroups Let $G$ be a finite group, $f:G \to M$ group homomorphism and $H \leq G$. Show that $[f(G):f(H)]$ divides $[G:H]$.
I have already shown that $f(H) \leq f(G) \leq M$. I guess I should use Lagrange's theorem somehow, but I don't have any other ideas on how to proceed. Any help would be appreciated!

With $K=\ker f$ and $L=H\cap K$ (which is a subgroup of $K$)
$$G/K\cong f(G)$$
$$H/L\cong f(H)$$
Thus with $|G|=g$ etcetera we get that $h/l$ divides $g/k$ and the index $$[f(G):f(H)]=\dfrac{g/k}{h/l}$$ divides $[G:H]=g/h$. Indeed, with $k=lm$ we get that it is $[G:H]/{m}$.
