How to prove that $\sum_\limits{k=1\\(k,p-1)=1}^{p-1}g^k \equiv \mu(p-1)$ (mod p) for prime p and primitive root g

Question

p is a prime and g is a primitive root modules p, and I want ot prove that:

$\sum_\limits{k=1\\(k,p-1)=1}^{p-1}g^k \equiv \mu(p-1)$ (mod p)

$\mu(x)$ is the Möbius function

I know how to deal with $\sum_\limits{k=1}^{p-1}g^k $ but there is a limit $(k,p-1)=1$ so I don't know how to keep going.

Answer 1

Here is my answer inspired by @reuns:

$\sum_{k=1,(k,p-1)=1}^{p-1}g^k = \sum_{k=1}^{p-1} g^k \sum_{d| (k,p-1)} \mu(d) = \sum_{d| p-1} \mu(d) \sum_{m=1}^{(p-1)/d}g^{md}$

And only when $d = p-1$, $\sum_{m=1}^{(p-1)/d}g^{md}\not\equiv0$ (mod p) , otherwise $\sum_{m=1}^{(p-1)/d}g^{md}\equiv0$ (mod p)

So we can get the answer