must a primitive root be invertible?

Question

I think so, consider $g$, a primitive root. We know that $g^{\phi(m)} \equiv 1 \pmod {m}$.

Then, $g^{-1} = g^{\phi(m) - 1}$ since:

$$ gg^{-1} = g\cdot g^{\phi (m) - 1} = g^{\phi(m)} = 1$$

As an implication, $\gcd(g, m) = 1$.

I think this is all really elementary numbers theory but I just wanted to make sure :)

Thanks

Answer 1

Sure. But you are not explaing yourself clearly. You are aiming at proving that $g^{-1}=g^{\phi(m)-1}$. Therefore, the proof cannot begin with “$gg^{-1}=$”. Just delete that and it will be fine.