I need to find a generator of the multiplicative group of $\mathbb{Z}/23\mathbb{Z}$ as a cyclic group.
Since $\mathbb{Z}/23\mathbb{Z}$ only has $23$ elements and ord$(x)$ where $x$ is a generator must divide $23$, then does this mean the generator can only be $1$ or $23$? Or have I got the wrong idea?
It would help if someone can provide the solution to this problem, I don't find hints very useful.
Best Answer
As has been pointed out in the comments, $\mathbb{F}_{23} = \mathbb{Z}/23$ has $22$ elements, and is a cyclic group. Since the order of any element in that group must divide $22$, all orders must be $1$, $2$, $11$, or $22$. You are looking for an element of order $22$. To find them, you need only compute the second and $11^{\mathrm{th}}$ powers of each element modulo 23; when neither is $1$, you have found an element of order $22$.