Let $PSL(3,q)$, $q$ odd. Then for $p \mid q(q^2 - 1)$ and $s \mid q-1$ ($p,s$ prime) we have a $p$-subgroup $X\ne 1$ such that $s$ divides $|N_G(X)|$

Question

Let $G = PSL(3, q)$ with $q$ odd, the projective special linear group over a finite field of order $q$. Let $p, s$ be prime numbers. If $p$ divides $q(q^2 - 1)$ and $s$ divides $q-1$, then there exists a nontrivial $p$-subgroup $X \le G$ (i.e. a subgroup whose order is a power of $p$) such that $s$ divides $|N_G(X)|$.

Why is this so and where does these subgroups come from?

This occurs in an argument of a proof, but I do not see it. Just as a note, we have $$ |G| = \frac{(q^3 - 1)(q^3-q)q^2}{\gcd(q-1,3)}. $$