Group generators and order

a) We say g is a generator for G iff every element in G can be written as g^{k} 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> = {g^{k} | k in Z}.

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

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