I want to prove the following theorem: Given a finite field $\langle F, +, \cdot \rangle$, the multiplicative group $\langle F^*, \cdot \rangle$ of all nonzero elements of F is cyclic.
I am trying to follow this asnwer: https://math.stackexchange.com/a/59911/811720.
I don't understand why given a $y \in G_d$, the cyclic group $\langle y \rangle = \{ y^k | k \in \mathbb{Z}\}$ would be a subset or would be equal to $G_d$.
Could you provide more details on why this holds?
As far as I understand, since $ y \in G_d$ and $G_d = \{x \in G | x^d = 1 \}$, we have that $ \langle y \rangle = \{ y^k | k \in \mathbb{Z}\} = \{ x^k | k \in \mathbb{Z} \land x \in G \land x^d = 1 \}$.
How can one see that this is the same as $G_d$ from here?