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!

Sketch: Let $K$ be the kernel of $f$.


*

*Since $K$ is normal, $HK$ is a subgroup of $G$.

*There is a bijection between cosets of $HK$ in $G$ and cosets of $f(H)$ in $f(G)$.  The bijection is given by $gHK$ corresponds to $f(g)f(H)$.  To see where this comes from, observe that $f^{-1}(f(H))=HK$.  
One direction of the proof: Suppose that the cosets $f(g_1)f(H)$ and $f(g_2)f(H)$ are equal.  Then, there exists some $h\in H$ such that $f(g_1)f(h)=f(g_2)$.  Therefore, $f(g_2^{-1}g_1h)=e_M$.  Therefore, $g_2^{-1}g_1h\in K$ so there is some $k\in K$ such that $g_2^{-1}g_1h=k$ or that $g_2^{-1}g_1=kh^{-1}\in KH$.  This implies that $g_2HK=g_1HK$.
Starting with $g_2^{-1}g_1\in HK$, write $g_1=g_2hk$ and use this to show that $f(g_1)f(H)=f(g_2)f(H)$ by a similar argument.


*The result follows since $[G:H]=[G:HK][HK:H]$ and $[G:HK]=[f(G):f(H)]$.

