Primitive elements proof

Question

I am looking for the proof of this:

Let $p$ be prime. Then there exists a primitive element $g$ modulo $p$.

I have done the following:

By Fermat's Little theorem, we have that $g^{p-1} \equiv 1 \mod p$. Since $\phi(p)=p-1$, then the multiplicative order of $g$ modulo $p$ is equal to $\phi(p)$, and hence $g$ is a primitive element modulo $p$.

Does anybody know if this is true?

Not every $g$ is a primitive element. For instance it's true that $1^{p-1}\equiv 1\pmod{p}$, but $1$ is certainly not primitive. However, it's true, because the group of units of a finite field is cyclic. — egreg, Apr 21 '15 at 17:50
possible duplicate of Proving that the multiplicative group mod p (p is prime) is cyclic — Jack D'Aurizio, Apr 21 '15 at 18:07

score 0 · Answer 1 · answered Apr 21 '15 at 17:59

Consider $2 \bmod 7$. You will find that $2^3\equiv 1 \bmod 7$ so that $2$ is not a primitive element (nor is $1$ for that matter). You need to show that there is an element of multiplicative order $6$, and you can't assume that every element has order $6$.

It turns out (and you will need to prove) that the multiplicative group of non-zero elements modulo $p$ is a cyclic group of order $p-1$. When $p-1=6$ the cyclic group of order $6$ has just two elements of order $6$. Finding an actual primitive element for any prime is tricky - there is no general formula.

score 0 · Answer 2 · answered Apr 21 '15 at 19:07

From $g^{p-1}\equiv 1\mod p$, you only can conclude the order of $g$ is a divisor of $p-1$.

Actually, since $\mathbf F_p^{\times}$ is a cyclic group of order $p-1$, it has $\,\varphi(p-1)$ generators. More precisely, if $\zeta\,$ is a generator, the other generators are $\zeta^k$, where $1\le k< p-1,\enspace k\wedge(p-1)=1$.

Primitive elements proof

2 Answers2