Fifth Sylow Theorem

Theorem
The number of Sylow $p$-subgroups of a finite group is a divisor of their common index.

Proof
By the Orbit-Stabilizer Theorem, the number of conjugates of $$P$$ is equal to the index of the normalizer $$N_G \left({P}\right)$$.

Thus by Lagrange's Theorem, the number of Sylow $p$-subgroups divides $$\left|{G}\right|$$.

Let $$m$$ be the number of Sylow $p$-subgroups, and let $$\left|{G}\right| = k p^n$$. From the Fourth Sylow Theorem, $$m \equiv 1 \left({\bmod\, p}\right)$$.

So it follows that $$m \nmid p \implies m \nmid p^n$$.

Thus $$m \backslash k$$ which is the index of the Sylow $p$-subgroups in $$G$$.