I'd like to know if my solution to the following problem is correct

"Let F be a field of 8 elements and A = {x $\displaystyle \epsilon $ F such that x^{7}=1 and

x^{k}$\displaystyle \neq $ 1 for all natural numbers k<7}"

Find the no. of elements of A.

MY SOLUTION

I used the following result.

The nth roots of unity form a group which is cyclic and it'll have 'd' elements

where d = g.c.d (n, |F|-1).

Here n = 7 and |F|-1 =7 which means there are d = g.c.d(7,7) = 7

But since the question wants to simply find numbers such that

i) x^{7}= 1 and

ii) x^{k}$\displaystyle \neq $1 for any k<7,

the actual answer should be one less than 7, as we gotta remove the

multiplicative identity 1 from this list of elements.

I guess the answer has to be 6.

Am I right in my conclusion?

Thanks in advance.