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Let $F$ be a field of $8$ elements. Let $A$ be a subset of $F$ and

$A = \{x \in F \mid x^7=1$ and $x^k \neq 1$ for $k<7\}$. Then find the number of elements in $A$.

Argument: Since $F$ is a field of $8$ elements, we know that in a field, set of non-zero elements form a group with respect to multiplication (say $G$), hence there exists a group with $7$ elements(leave the zero element from the field) since the order is $7$(prime number) $G$ is cyclic. Therefore the number of generators(i.e multiplicative order is $7$) of $G$ is $6$, which is nothing but the set $A$...Hence there is $6$ elements in the set $A$..

I don't know whether my argument is right or not..please let me know where i am wrong...

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    $\begingroup$ Well it's all about $\phi(7)$ (Euler phi function)(primitive roots) and your arguments seems all right to me $\endgroup$ – Kushal Bhuyan May 9 '16 at 13:54
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    $\begingroup$ Yes - the argument is correct as you stated it (even if you never heard of Euler phi function or primitive roots - you don't need them or need to know about them for your proof to be perfectly fine as you stated it). $\endgroup$ – mathguy May 9 '16 at 21:43

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