A bijection between between the orbits $G$ on $G \backslash B \times G \backslash B$ and orbits of $B$ on $G \backslash B$?

Question

Let $G$ be a group and $B$ a subgroup. Can somebody show me how that orbits of $G$ on $G \backslash B \times G \backslash B$ are in bijection with the orbit of $B$ on $G \backslash B$?

Here is what I've come up with so far:

Let $(aB,cB) \in G \backslash B \times G \backslash B$.

Then $G.(aB,cB) = \{(gaB,gcB) : g \in G\}$

By setting $ga = h$ we can see that this is in bijection with:

$=\{(hB,ha^{-1}cB) : h \in G\}$

Now, if we have $(hB,ha^{-1}cB) = (h'B,h'a^{-1}cB)$ then $(h'^{-1}hB,h'^{-1}ha^{-1}cB)=(B,a^{-1}cB)$ and in particular $h'^{1}h \in B$. I felt like this might be going in the write direction to showing the desired bijection.

Help appreciated!

Answer 1

Two step hint:

step 1

show that every orbit of $G$ on $G/B\times G/B$ contains an element of the form $(B,gB)$.

step 2

Consider $\{(B,gB):g\in G\}$. What is the subgroup of $G$ that fixes this? Notice that orbits of $G$ are in bijection with the orbits of this stabiliser and up to a few explanatory words, you're there.