# Dihedral groups non-commutavity

Here is a result about dihedral groups. $$rs = sr ^{-1}$$, where $$r$$ is a rotation of $$\frac{2 \pi}{n}$$ radians and $$s$$ is a reflection about the line of symmetry from vertex $$i$$ and the origin. This is quite understable, and I can basically prove this using my intuition, but I cannot reach a rigorous proof. I also thought that as this result is familiar, so I searched all the questions in MSE but I did not find any. So I posted this question. Can you please help?

• Don't ever think you can vandalize your questions and get away with it. – user21820 Apr 3 at 3:03

Any reflection is its own inverse. Since $$rs$$ is a reflection we have $$(rs)^{-1}=rs$$. On the other hand we also know that $$(rs)^{-1}=s^{-1}r^{-1}=sr^{-1}$$. ($$s=s^{-1}$$ because $$s$$ is a reflection as well). Since the inverse is unique we conclude that $$rs=sr^{-1}$$.