Let G be the group of 6th roots of unity. What are the generators of this group? What are the primitive 6th roots of unity?
I know that the primitive roots are $\frac{1}{2} \pm i\frac{\sqrt3}{2}$. Would these also be the generators of the cyclic group? I don't think they are but how would I determine the generator?
Best Answer
If you're familiar with writing complex numbers in polar form, you can list the six sixth roots of unity as \begin{equation} e^0, e^{i\frac{\pi}{3}}, e^{i\frac{2\pi}{3}}, e^{i\pi}, e^{i\frac{4\pi}{3}}, e^{i\frac{5\pi}{3}}. \end{equation} It is then easy to compute the subgroup generated by each of these. For example, \begin{equation} \langle e^{i\frac{4\pi}{3}} \rangle = \lbrace e^{i\frac{4\pi}{3}}, e^{i\frac{2\pi}{3}}, 1 \rbrace, \end{equation} so $e^{i\frac{4\pi}{3}}$ is not a generator of $G$. You can compute the subgroups generated by the other elements in the same way, and determine which roots give you all of $G$. (Notice the similarity with modular arithmetic.)