Thread: Primitive Roots - Show that k is odd?

If g and h are primitive roots of an odd prime p, then g is congruent to h^k (mod p) for some integer k. Show that k is odd.

Suppose g and h are primitive roots of an odd prime p. This means that gⁱ is congruent to a (mod p) and h^j is congruent to b (mod p) with gcd(i,p) = 1 and gcd(j,p)=1.
We also know that g is congruent to h^k (mod p).

I am not sure if I need to manipulate this to show that k is of the form 2m+1 or what. Any advice would be helpful.

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.