Originally Posted by

**EntryLevel** They are the stabilizers in G.

Actually, I decided to ignore the advice about the 2nd isomorphism theorem and I thought of the following map that I think is a bijection.

$\displaystyle \phi :G / HG_a \rightarrow \{O_1,...,O_r\}$

$\displaystyle \phi (g)=$ the orbit containing g(a), where a is given in the problem.

The map is definitely surjective. It is well defined because if you have two cosets $\displaystyle gHG_a$ and $\displaystyle g'HG_a$ then it must be true that $\displaystyle g=g'hg_a$ for elements $\displaystyle h\in H$ and $\displaystyle g_a\in G_a$

$\displaystyle g(a)=g'hg_a(a)=g'h(a)=g'(a)$

where the 2nd equality is because g_a is a stabilizer and the last equality because in part (a), it is shown that any element of G permutes the different orbits, taking all elements in one orbit to all elements in another...

Finally it is injective because if $\displaystyle g(a)=g'(a)$ then $\displaystyle g^{-1}g'\in G_a\subset HG_a$

This seems to work... perhaps the hint was a red-herring (or perhaps I made a mistake).

Seems to work from what I see. I was going to just note the following. Since $\displaystyle G$ acts transitively on $\displaystyle A$ you have that $\displaystyle \text{card}(A)=[G:G_a]$ for any $\displaystyle a$, right (this is just the orbit stabilizer theorem and using the fact that the orbit must be the whole set). But, from the orbit decomposition theorem and the fact that all the orbits of $\displaystyle A$ under $\displaystyle H$ are equipotent we have that $\displaystyle \text{card}(A)=r\text{card}(O_1)$. So, we have that

$\displaystyle \displaystyle \begin{aligned}\frac{|G|}{|HG_a|} &=\frac{|G|}{\displaystyle \frac{|H||G_a|}{|H\cap G_a|}}\\ &=\frac{|G|}{\displaystyle \frac{|H|}{|H\cap G_a|}|G_a|}\\ &=\frac{\displaystyle \frac{|G|}{|G_a|}}{\text{card}(O_1)}\\ &=\frac{\text{card}(A)}{\text{card}(O_1)}\\ &=\frac{r\text{card}(O_1)}{\text{card}(O_1)}\\ &=r\end{aligned}$