TITLE: Show that r is a primitive root?
QUESTION [5 upvotes]: Show that if $r$ is a primitive root modulo the positive integer $m$, then $ {\overline r }$ is also a primitive root modulo m if $ {\overline r }$ is an inverse of $r$ modulo $m$.
My TA did not go over any proofs in my self study class. We just learned the topic and did some examples. I was wondering what this proof might look like. It seems it might help because it looks similar to the problems with examples. 
Can someone please show me?

REPLY [4 votes]: A primitive root of $m$ is a number $g$ such that $g,g^2,g^3, \dots, g^{\varphi(m)}$ are all incongruent modulo $m$. So $g$ is a primitive root of $m$ precisely if $g$ has order $\varphi(m)$ modulo $m$.
So it is enough to show that for any $x$, if $xy\equiv 1\pmod{m}$, then $x^k\equiv 1\pmod m$ if and only if $y^k\equiv 1\pmod{m}$. This is immediate, since $(xy)^k=x^ky^k$.