Q) Show that $x$ or $2x$ is a generator of the cyclic group $(\mathbb{Z}_3[x]/\langle f(x)\rangle)^*$ where $f(x)$ is a cubic irreducible polynomial over $\mathbb{Z}_3$.
My attempt: Let $F= \mathbb{Z}_3[x]/<f(x)>$, $[F:\mathbb{Z}_3] = 3\implies |F|=3^3=27\implies F\simeq GF(27) \implies F^* \simeq (GF(27))^*\simeq \mathbb{Z}_{26}$
If $x\in F ^* \implies |x|$ divides $|F^*|$ by Lagrange's theorem. $\implies |x|\in \{1,2,13,26\}$.
I can't understand why $|x|\neq 1,2$? If $|x|=1$, $x$ is not a generator, but why can't $2x$ be a generator? $|x|=1\implies x=1\implies 2x=2$. Is $2x$ not a generator because $\nexists n$ s.t. $2^n = 1 $ (mod 26). I am also not sure if I should take modulo $26$. I did mod $26$ because $F^*\simeq \mathbb{Z}_{26}$.
Similarly, $|x|=2\implies x^2 = 1 \implies (2x)^2 = 4 \neq 1; (2x)^{13} = 2^{13}x = 2x $ (mod 26) $\neq 1; (2x)^{26} = 2^{26} \neq 1$ (mod 26). Am I thinking right? Thanks.