the number of generators of a cyclic group of order d is φ(d), the euler totient of d. if d = p, a prime number, then φ(p) = p-1 (so yes, you are correct).

(assumed as given: the group of units of a finite field is cyclic).

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let's make a finite field of order 8. to do this, we need an irreducible cubic in Z_{2}[x]. let's try f(x) = x^{3}+x+1:

f(0) = 1

f(1) = 1, so f has no roots in Z_{2}, so f has no linear factors (and thus is irreducible) in Z_{2}[x].

so let's form F = Z_{2}(u), where u is a root of f. by definition, we have: u^{3}= -u-1 = u+1 (since in any field of characteristic 2, -a = a for all a).

let's go ahead and list all possible elements of F = {0,1,u,u+1,u^{2},u^{2}+u,u^{2}+1,u^{2}+u+1}.

(viewing F as the vector space Z_{2}xZ_{2}xZ_{2}, with (ordered) basis {u^{2},u,1} these would be:

F = {(0,0,0),(0,0,1),(0,1,0),(0,1,1),(1,0,0),(1,1,0),( 1,0,1),(1,1,1)} if you prefer).

let's explicitly find a generator for F* (which will prove it's cyclic of order 7). we'll try u first:

u^{2}≠ 1 (u is not of order 2, trivial).

u^{3}= u+1 (not of order 3)

u^{4}= u(u^{3}) = u(u+1) = u^{2}+u (not of order 4)

u^{5}= (u^{2})(u^{3}) = u^{2}(u+1) = u^{3}+u^{2}= u^{2}+u+1 (not of order 5)

u^{6}= (u^{3})^{2}= (u+1)^{2}= u^{2}+1 (using the frobenius morphism, and also: not of order 6)

u^{7}= u(u^{6}) = u(u^{2}+1) = u^{3}+u = (u+1)+u = 1 <---u is of order 7.

since the order of u^{k}is 7/(gcd(k,7)) we see that |u^{k}| = 7 for k = 1,2,3,4,5,6, and only u^{7}= 1 has order 7/7 = 1.