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Let $F$ be a field of $8$ elements and $A =\{x\in F:x^7=1\text{ and }x^k\neq 1\}$ for all natural numbers $k<7$. Then the number of elements in $A$ is
1. $1$
2. $2$
3. $3$
4. $6$

How can I solve this problem, I am completely stuck on it. Can anyone help me? Thanks

2 Answers 2

3

The problem is asking for the number of elements of $F$ that have multiplicative order $7$. In other words, how many generators does $F^{\times}$ have? The answer should follow at once from the fact that $F^{\times}$ is a cyclic group of order $7$.

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Do you know how to construct a field of order 8? Once you have one example you should just be able to compute the 7th powers of the elements.