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 $G$.

Also known as
Some sources call this the fourth Sylow theorem and merge it with what we call the Fourth Sylow Theorem.

Others merge this result with what we call the Fourth Sylow Theorem and call it the third Sylow theorem.

Others merge this with what we call the Third Sylow Theorem and call it the third Sylow theorem.