"Given a Galois Field $(\mathbb{F}, +, \cdot)$ of order 8. With an element $x \in \mathbb{F}$ we create a group $(\{x^m | m \in \mathbb{Z}\}, \cdot)$. ($x^m$ is calculated via the second operator $\cdot$ in the field, for example: $x^3 = x\cdot x\cdot x$). What is $\max_{x\in\mathbb{Z}}|\{ x^m|m \in \mathbb{Z}) \}|$?"
I don't really understand the question or know how to answer... Help please?
On
As $F$ is a field of order $8$, the multiplicative group $F^\times$ is of order $7$. Unless you start with $x=0$ (which is pointless), you generate a subgroup of this group of order $7$. You should knwo that a group of order $7$ must be cyclic. What does that tell you about the answer to the porblem? (It turns out that $F^\times$ is always cyclic for finite fields, but in the problem at hand we may shortcut the argument by noting that $7$ is prime - the law of small numbers is working for us here).
Hint: Since the field has order $8$, it's $\mathbb{F_{2^3}}$. Now consider a fixed $x\in\mathbb{F_{2^3}}$ and look at the sequence $1,x,x^2,x^3,\cdots$. As this is a finite field, this sequences is guaranteed to be periodic. The question is: how long can this sequence be? Try looking at the Frobenius Endomorphism and see if you can learn anything about the length of the sequence based on the action of the Endomorphism on the element $x$.