Prove $a \equiv b \pmod{p-1}$ if $g^a \equiv g^b \pmod{p}$

Question

Let $g$ be a primitive root for $\mathbb{F}_p$.

Suppose that $x = a$ and $x = b$ are both integer solutions to the congruence $g^x \equiv h \pmod{p}$. Prove that $a \equiv b \pmod{p-1}$.

I'm at a loss for where to even start here.

Answer 1

Note that $g$ is a primitive root means that if $g^x\equiv 1 \mod p$ then $p-1|x$. From your condition, one has $g^a\equiv g^b \mod p$, or $g^{a-b}\equiv 1\mod p$. Thus $(p-1)|(a-b)$, or $a\equiv b \mod (p-1)$.