0

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?

cbakos
  • 153
  • 1
    The argument does not claim that $\langle y\rangle$ would be equal to $G_d$. All it says is that $x^d=1$ for every $x\in\langle y\rangle$. – Jyrki Lahtonen Mar 21 '22 at 18:20
  • It says: "But the subgroup ⟨y⟩ has cardinality d, so from the hypothesis we have that ⟨y⟩={x∈G∣x^d=1}, which by definition is G_d. Doesn't this mean they are claimed to be equal? – cbakos Mar 21 '22 at 18:30
  • You need to look again at the definition of $G_d$ – Matthew Towers Mar 21 '22 at 18:39
  • According to the linked andwer, G_d is made up of elements of G (x∈G) with order d (x^d=1), hence my definition (as the group is multiplicative). What's wrong with my interpretation? What is the correct formulation then? – cbakos Mar 21 '22 at 18:46
  • 1
    The answer defines $G_d$ as the set of elements of order $d$ (which is not a subgroup). When $G$ is abelian (as is the case here), the subset ${x\mid x^d=1}$, on the other hand is a subgroup. Not all the elements in that subgroup have order $d$. For example $x=1$ is in there, and it has order $1$. – Jyrki Lahtonen Mar 21 '22 at 20:16

0 Answers0