Prove, that the multiplicative group of a finite field is cyclic.

Question

I have almost completed the proof, I have shown, that there exists an element of order e, where e is min{n:$g^n=1$} and $g\in G$ where $G$ is the multiplicative group of a finite field. Now I noticed, that the polynomial $x^e-1$ is solved by all $g\in G$. I am finished if I can show, that e=q-1 where q=$|G|$. However I only know that q-1$\leq$e. My idea is, that if q-1 <e, e >min{n:$g^n=1$}, so e has to be q-1 wich means $G$ is cyclic. Is the reasoning for q-1=e sufficient?

Answer 1

Hint: Suppose $q-1<e$:

By division with remainder, $e = p(q-1) + r$ with remainder $0\leq r<q-1$. Then $1=g^e = (g^{q-1})^p g^r = g^r$ for all $g$.