In any group G, if x has order n, then x^k has order n/gcd(n,k). (If you need help with this, let me know.) In particular, x^k has order n iff (n,k) = 1. So for your problem, the group G is the multiplicative group of integers mod p, of order p-1. Then assuming g has order p-1, g^k has order p-1 iff (k,p-1)=1. So since p-1 is even, k must be odd.