If gcd(y,p-1)=1 then {1,2,3,...,p-1}={y,2y,3y,...,(p-1)y} mod p.
Given a=g^b, then a=(g^y)^(bc) where c=y^(-1) mod(p-1)
hi, i'm stuck on this assignment question. my teacher said it's really easy so i think i'm just missing something obvious? thanks
Let p be an odd prime and g be a primitive root modulo p. Let y be an integer such
that gcd(y, p-1) = 1. Show that g^y is a primitive root modulo p.