QN: Prove that if there exists a primitive root modulo n, then there exists a primitive root modulo every divisor of n.
If $a$ is a primitive root mod n, then $a^{x}\equiv 1$ mod n has least integer solution $x= ord_n(a)= \phi(n)$. If p is any prime divisor of n, then $a^{\phi(n)}\equiv a^{\phi(p)} \equiv 1$ mod p. From this how do I extend it to modulo a general divisor d of n?
$a^{\phi(d)} \equiv 1$ mod d seems to hold but how do I show $ord_d(a)= \phi(d)$?
Thanks in advance!