# Quick question about colouring the vertices of a fixed square using three colours.

If a square remains fixed in the plane, how many different ways can the corners of the square be colored if three colors are used?

Why does the answer use $$D_{4}$$ when the square cannot move? I don't understand. The answer is $$21$$ ways. Can someone please explain why they choose $$D_{4}$$ when it says the square is fixed? Is it because the square does not change if reflected and rotated, only the vertices change?

• If the square remains fixed in the plane, then I say the answer is clearly $81$. Coloring the top left corner green and the other three red is as I read this problem distinct from coloring the bottom right corner green and the rest red. If they want us to use Burnside's lemma and $D_4$, then in my opinion the problem author has chosen their words unwisely. – Arthur Dec 5 '18 at 21:39
• @Arthur I agree, this problem is worded very vague! They definitely use Burnside's theorem, where it says $\frac{3^{4}+3+3^{2}+3+3^{2}+3^{2}+3^{3}+3^{3}}{8}$. – numericalorange Dec 5 '18 at 22:00
• If this were on a test, my solution, which any grader (who is reasonable according to my personal standards) would respect, is to start by stating "I interpret 'fixed in the plane' to mean that coloring the top left corner green and the other three red is distinct from coloring the bottom right corner green and the rest red" and then proceed to calculate the answer of $81$. Unless it was abundantly clear from context (e.g. a test specifically on the section of Burnside's lemma), that ought to do the trick. – Arthur Dec 5 '18 at 22:19