# Math Help - show the inverse?

1. ## show the inverse?

show that if r is a primitive root modulo the positive integer m then the r^-1 is also a primitive root modulo m, where r^-1 is an inverse of r modulo m

2. $r^{-1} \equiv r^{\phi(m)-1}(\bmod m)$. Now use the fact that $r^k$ is a primitive root if and only if $\gcd(k,\phi(m)) = 1$ where $k$ is a positive integer.

3. Hello,

A precision (just in case) :

$\phi(m)$ stands for the Euler Totient function. It also designs the cardinal of the subgroup containing the invertible elements taken from Z/mZ.
This is why $a^{\phi(m)}$ is always $\equiv 1 [m]$.