I see how I can prove (a):
I can say,
Let g be an element in G which does not have finite order. Since (g)^k is in G for each value of k = 0, 1, 2, ..., we can conclude that we may find integers m and n such that (g)^m = (g)^n. If m > n, then g^(m-n) = e. Thus, m-n is the order of G and is...