Determine the generator for the cyclic group formed by the solutions of $x^9 = 1$.

Question

Find the solutions to $x^9 = 1$ and determine the generator for the cyclic group formed by the solutions.

The equation can be factored as $(x^3 - 1)(x^6 + x^3 + 1) = 0$ and the solutions are $$\begin{align*} x &= 1\\ x &= -\sqrt[\leftroot{1}\uproot{3}9]{-1} \\ x &= (-1)^{2/9} \\ x &= -\sqrt[\leftroot{1}\uproot{3}3]{-1} \\ x &= (-1)^{4/9} \\ x &= -(-1)^{5/9} \\ x &= (-1)^{2/3} \\ x &= -(-1)^{7/9} \\ x &= (-1)^{8/9} \end{align*}$$ but I'm unsure how to determine the generator. How are these found generally and what would be my options to start determining it?

Answer 1

Your elements are basically $(e^{ik\pi/9})_{k=0}^8$. They form the cyclic group of $9$ elements under ordinary multiplication. The most obvious generator is $r=e^{i\pi/9}$ (with $k=1$).

However, any such element with $k$ relatively prime to $9$ (e.g. $4,5,8$) would generate the group as well. For example, starting with $r^4$, you get a cycle $$ \left< r^4,r^8,r^{12} = r^3,r^7,r^{11}=r^2,r^6,r^{10} = r,r^5, r^9 = 1 \right> $$

Answer 2

Note that

$x^9=1=\cos 2n\pi +\sin 2n\pi=e^{i2n\pi}$

$\implies x=e^{i2n\pi/9}; n=0,1,2,...,8$ are roots.

Assume $e^{i2\pi/9}=a$. Then the powers of it coprime to $9$ viz. $a,a^2,a^4,a^5,a^7 $ and $a^8$ are generators.

Answer 3

The multiplicative group of $9$-th roots is isomorphic to the additive group $\mathbf Z/9\mathbf Z$, under the isomorphism: $$\mathbf Z/9\mathbf Z\longrightarrow U_9,\quad k\bmod9\longmapsto\mathrm e^{\tfrac{2ik\pi}9}.$$ Under this isomorphism, generators of $\mathbf Z/9\mathbf Z$ map onto generators of $U_9$, i.e. primitive $9$-th roots of unity. Therefore it is enough to know the generators of $\mathbf Z/9\mathbf Z$, which is standard.