Prove that $|G:A|=|G:B||B:A|$ for $A\subseteq B\subseteq G$ Given $G$ a group and $A,B$ subgroups of $G$ such that $A\subseteq B\subseteq G$, I want to prove $|G:A|=|G:B||B:A|$ where these are the indexes of the respective groups.
I have been presented with a proof as follows but get stuck near to the end;
Let $Bx_i$ and $Ab_j$ be the set of distinct right cosets in $G$, $B$ respectively for some $i\in I, j\in J$.
Then $|G:B|=|I|$ and $|B:A|=|J|$.
With this set up, the proof claims that $Ab_jx_i$ are all the distinct right cosets of $A$ in $G$.
I understand how the proof has shown that these are the distinct right cosets of $A$, but do not understand the deduction that these are distinct.
This is the given argument;
Suppose $Ab_jx_i=Ab_kx_l$. As $Ab_j,Ab_k \subseteq B$, this implies $Bx_i =Bx_l$ and hence $i=l$. A similar argument is used to show $Ab_j=Ab_k$ implies $j=k$.
Surely we can have two cosets being equal to each other with distinct elements representing each coset and so why would $i=l$ here?
eg. if $G=(\mathbb Q^*, \times)$ and $H=\{-1,1\}$ then the right cosets $H(1)=H(-1)=H$.
 A: Essential is: if two cosets have a non-empty intersection then they are equal.
$Bx_i\supseteq Ab_jx_i=Ab_kx_l\subseteq Bx_l$ tells us that the cosets $Bx_i$ and $Bx_l$ have a non-empty intersection hence are equal. 
Then $x_i$ and $x_l$ both represent the same coset, but among all representatives $x_k$ there is only one that represents the coset $Bx_i=Bx_l$. 
We could rephrase that by saying that among all indices $k$ there is only one such that $x_k$ represents coset $Bx_i=Bx_l$. 
This justifies the conclusion that $i=l$
Then $i=l$ and setting $x_i=x=x_j$ now from $Ab_jx=Ab_kx$ we can conclude that also $Ab_j=Ab_k$.
Then cosets $Ab_j$ and $Ab_k$ have non-empty intersection hence are equal, so that also $j=k$.
A: I think what the author of the proof has done is that he is implicitly (or explicitly somewhere) using transversals. Given a group $G$ and a subgroup $H$ of $G$, a transversal is a set $S\subseteq G$ such that for all cosets can be written as $gH$ for some unique $g\in S$. 
Example: $G=\Bbb{Z}$ and $H=4\Bbb{Z}$. A transversal for $H$ in $G$ could be $\{0,1,2,3\}$ or $\{4, 11,-3, 10\}$.
You're right that, in general, there are quite a few elements in $G$ with $g_1H=g_2H$. All that the author has done is that they are assuming that there is an already chosen set of transversals for $A$ in $B$ (the $b_i$'s) and $B$ in $G$ (the $x_i$'s).
A: I will give you an alternative proof, which is also the way Lagrange's theorem is usually proven. 
Define for $x,y \in G$ the relation $\equiv$
$x \equiv y \iff x^{-1}y \in H$ 
It is easy to verify this is an equivalence relation $G$. In particular, this implies that the set $\{[g]\mid g \in G\}$ forms a partition of the group. So, all equivalence classes are disjunct and their union is the entire group $G$.
Now, it is easy to verify that $[g] = gH$
Hence, the cosets $\{gH\mid g \in G\}$ form a partition of the group!
It follows that:
$$G = \bigcup_{g \in G} gH$$
Let $A$ be the set of representants of different equivalence classes
and hence, because the cosets are disjunct:
$$|G| = \left|\bigcup_{g \in G} gH\right| = \sum_{g \in A}|gH| = \sum_ {g \in A} |H| = |A||H| =  [G:H] |H|$$
We conclude:
$$[G:H] = \frac{|G|}{|H|}$$
Now, if $A \subset B \subset G$ and $A,B$ are groups
then $A$ is a subgroup of $B$ and $B$ is a subgroup of $G$.
Hence, we have:
$$[G:B] = \frac{|G|}{|B|}$$ 
and $$|B:A| = \frac{|B|}{|A|}$$
and we can conclude:
$$[G:B]|B:A| = \frac{|G|}{|B|} \frac{|B|}{|A|} = \frac{|G|}{|A|} = [G:A]$$
