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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.

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  • $\begingroup$ 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.) $\endgroup$ – Randall Mar 22 at 15:53

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