# Any finite field of $q$ elements has exactly $\Phi(q-1)$ primitive roots

Is the following prove of the above statement correct?

$$\bullet\$$Any finite field of $$q$$ elements is isomorphic to $$\mathbb{F}_q$$ and we know that $$\mathbb{F}_q^*$$ is a cyclic group of $$q-1$$ elements. Let $$\mathbb{F}_q^*=\langle \alpha \rangle$$ with $$\alpha\in \mathbb{F}_q^*$$. If we take another $$\beta\in\mathbb{F}_q^*$$, we can therefore write it as a power of $$\alpha$$: $$\beta=\alpha^k$$ for some $$k$$.

Now we have that $$\mathbb{F}_q^*=\langle \beta\rangle \iff \gcd(q-1,k)=1$$ as we know that $$\mathbb{F}_q^*$$ is cyclic of order $$q-1$$, thus isomorphic to $$\mathbb{Z}/(q-1)\mathbb{Z}$$.

This in turn gives us $$\gcd(q-1,k)=1 \iff k\in (\mathbb{Z}/(q-1)\mathbb{Z})^*$$ and the latter group has $$\Phi(q-1)$$ elements, where $$\Phi$$ denotes the Euler totient function.

Therefore, there are $$\Phi(q-1)$$ choices for $$k$$, so $$\Phi(q-1)$$ choices for $$\beta$$ such that $$\mathbb{F}_q^*=\langle \beta \rangle$$. In other words: there are $$\Phi(q-1)$$ primitive roots in each field with $$q$$ elements. $$\bullet\bullet$$

I am particularly unsure of the step where we conclude that $$\gcd(q-1,k)=1$$. It seems slightly logical, in a simpler group setting of $$\mathbb{Z}/n\mathbb{Z}$$, but I can't quite understand or proof exactly why that particular statement is true.

• A cyclic group of order $m$ has $\varphi(m)$ generators by a simple argument. Now use the fact that $\mathbb{F}_q^*$ is cyclic. It's easier to think about this as $\mathbb{Z}_{q-1}$ additively, which it is up to (unknown) iso. (This is basically what you're doing.) – Randall Mar 22 at 15:53