The first paragraph of the proof of the Theorem 10.4. of the book Introduction to Analytic Number Theory by Tom M. Apostol. says :
The numbers $1, 2,..., p - 1$ are distributed into disjoint sets $A(d)$, each set corresponding to a divisor $d$ of $p - 1$. Here we define $$ A(d) = {\{x \ : \ 1 \le x \le p - 1 \ \ \text{and}\ \ \text{exp}_p(x) = d}\}.$$
How $A(d_1) \cap A(d_2)= \emptyset$ if $ d_1 \ne d_2$ ?
Foe example consider $p=17$. So the divisors of $p-1$ are $1,2,4,8,16.$ For any number $x$ if $x^4 \equiv 1 \ (\text{mod } 17)$ then also $x^8 \equiv 1 \ (\text{mod } 17)$ then $x \in A(4) \cap A(8)$, meaning that it is nonempty, just opposite to the book's claim!
PS Should I upload full text of proof of the Theorem 10.4.?
$\exp_p(x)=d$ means $d$ is the smallest positive integer such that $x^d \equiv 1 \mod p$.
The keyword here is the smallest.
It is impossible to have two smallest different number.
if $x^2 \not\equiv 1 \mod p$ and $x^4 \equiv 1 \mod p$, then $x \in A(4)$.
Even though $x^8 \equiv 1 \mod p$, since $8$ is bigger than $4$, $x \not\in A(8).$