Referring to the group of units.
My first thought is yes, since $1$ would be the generator.
Although I think I'm getting confused between the generator and the identity element in this case. $1$ is definitely the latter, is it also a generator?
I believe I'm doing the operation incorrectly. For example, $3 \cdot 3 =1$ correct? $3$ couldn't be the generator because if I kept doing this operation I would start repeating the same elements, but not all the elements ($1,3,5,7)$
Best Answer
I guess you are mixing $\mathbb{Z_8}$ and $U(8)$. $1$ is identity in $U(8)$ as operation is multiplication here. But in $\mathbb{Z_8}$ operation is addition, so $0$ is the identity and $1$ is the generator in $\mathbb{Z_8}$ as you can have any elements less than $8$ by adding $1$. No $U(8)$ is not cyclic. To see it pick an element of $U(8)$ and keep multiplying it with itself. If you find one with order 4, that is your generator. but every element here is of order 2.
$\textbf{Theorem-}$ $U(n)$ is cyclic iff $n$= $2,4, p^k,2p^k $ where $p$ is an odd prime. Try to prove it.