0
$\begingroup$

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.

enter image description here

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Huh, it seems pretty simple. As I see it, the white triangles are different from the black triangles. For the squares, I guess it would be 6 * 5, since there are 6 squares and five triangles. If I picked a white triangle, would that mean it's 24 * 2 since there are 24 white triangles and two paths? Thanks for the link honestly, I'm going to look into it some more :) $\endgroup$ – Andrew Raleigh Nov 25 '16 at 1:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.