Prove $[G:A] = [G:B]\cdot [B:A]$ if $A\leqslant B\leqslant G$ If $G$ is a group and $A\leqslant B\leqslant G$ then prove
  $$[G:A] = [G:B]\cdot [B:A]$$


For finite groups the theorem is obivious and simple to prove. But for infinte groups it seems a lot harder.

I've tried the following:

Let $L_G(A)$ be the set of left cosets of $A$ in $G$, then we need to prove
$$
|L_G(A)| = |L_G(B)|\cdot |L_B(A)|
$$
I was thinking of a trying to find a bijection $\alpha$ between $L_G(A)$ and $L_G(B) \times L_B(A)$. It seems logical to choose something like this:
$$
\begin{align*}
\alpha: L_G(A) &amp; \to L_G(B) \times L_B(A)\\
gA &amp; \mapsto (gB, b_gA) 
\end{align*}$$
But how can I choose this $b_g$? 

I know that $L_B(A)$ partitions $B$, this means that for any $ga \in gB$ (since $A\leqslant B)$ then $\exists b_i$ such that $ga \in g(b_iA)$ and then using this $b_i$. But is this useful?
Each coset $xB$ of $B$ in $G$ is the disjoint union of $|B:A|$ cosets
of $A$ in $B$. In detail, is $B=\bigcup_{i\in I} y_i A$ is a disjoint
coset decomposition, so is $xB=\bigcup_{i\in I} xy_i A$. Therefore
$G$ is the disjoint union $\bigcup_{j\in J} x_j B$
of $|G:B|$ $B$-cosets, and so a disjoint union 
$\bigcup_{j\in J}\bigcup_{i\in I} x_j y_i A$
of $|G:B||B:A|$
$A$-cosets.
we can use the isomorphism theorem  $G/ B/B/A \simeq G/A$