a) We say g is a generator for G iff every element in G can be written as gk for some integer k. Supposing G is a finite group with n elements, show that g is a generator for G iff the order of g is n.

b) We say a group G is cyclic iff it contains a generator g. Show that if G contains a prime number of elements, then G is cyclic.

c) Show that a finite cyclic group with n elements contains exactly phi(n) generators.

Now here is what I have done so far:

a) not sure how to proceed

b) not sure how to proceed

c) let G = <g>. Then <g> = {gk | k in Z}.

Suppose that G = <a>. Then we claim <ai> = G if and only if gcd(i,n)=1. Now suppose that gcd(i,n)=1 and also suppose that e=(ai)k=aik. It follows that n|ik. But if gcd(i,n) = 1, then we also have n|k. In particular, the order of (ai) >= n. But the order of any element of G is <= n and so the order of (ai) is exactly n.

Now suppose conversely that G = <ai>. In particular, this means that the order of ai is exactly n. But now suppose that k > 0 divides both i and n and we will obtain a contradiction. Then (ai)n/k = ei/k = e. In particular, the order of ai is less than n/k < n. This is a contradiction.