I found this question, about finding the order of the rotational symmetry group in two ways.
It's by using the orbit-stabilizer theorem on a triangle, and by using it on a square.
I know that the orbit stabilizer theorem is the one below, but I don't get how we get a different order even though it's all the same group in the end.
$$\left|{\operatorname{Orb} \left({x}\right)}\right| = \left[{G : \operatorname{Stab} \left({x}\right)}\right] = \dfrac {\left|{G}\right|} {\left|{\operatorname{Stab} \left({x}\right)}\right|}$$
Why are the results that we get different for picking a triangle and picking a square?
There is a wonderful explanation by gowers on this website which explains the intuition behind the orbit stabiliser theorem in the first few paragraphs. He also goes in depth into multiple proof strategies of the theorem. If you have any questions about his explanation comment on this post and I can help.