Prove that $(\mathbb{Z}/13\mathbb{Z})^∗$ is a cyclic group by finding a generator

Question

Prove that $(\mathbb{Z}/13\mathbb{Z})^∗$ is a cyclic group by finding a generator.

I am a bit confused about this question. I know that the answer is $g=2$, but I am confused because then that gives an element $0$ which is not in the group. When we take $[2]^n$, we get all the elements $[1,2,..., 12]$ but we also get $[0]$ for $2^{13}$, right? Can someone clarify what I am doing wrong here? Or if I have some definition wrong.

Answer 1

Why?

$[2]^{13}=[2^{13}]=[8192]=[630\times13+2]=[2]$!

Afterall, $[2]^2=[4]\ne[1], [2]^3=[8]\ne[1], [2]^4=[3]\ne[1], [2]^6=[12]\ne[1]$. And seeing it is enough to proove $(\Bbb Z/13\Bbb Z)^*=<[2]>$.

Because by Fermat's Little Theorem, $[2]^{12}=[1]$, so order of $[2]$ must be divisors of $12$.

Answer 2

The multiplicative groups don't include zero. $2^{12} = 1$ mod $13$, hence $2^{13}=2$ mod $13$. This follows from basic rules of modular arithmetic.