I am supposed to proof the following theorem:
Let $A$ be an abelian group and $B\leq A$. In addition let $f: A \to A'$ be an homomorphism of two abelian groups. Show that: $[A:B] = [\text{im}(A) : \text{im}(B)][\text{ker}(A):\text{ker}(b)]$
So using the theorem of Lagrance and an isomorphism from $\frac{G}{\text{ker}(f)} \to \text{im}(f)$ I've already proofen the theorem for finite groups. However we only introduced the theorem of Lagrance for finite groups and I didn't even use the fact that $A,B$ are abelian groups so I assume I am missing something. I really would appreciate some help ^^
Edit: I just had a look into our script and we also wrote the last line with this notation: [A:B] = [$A^f$:$B^f$] : [$A_f$:$B_f$] is this a more wide spread notation?