Fifth Sylow Theorem/Proof 1

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 $\map {N_G} P$.

Thus by Lagrange's Theorem, the number of Sylow $p$-subgroups divides $\order G$.

Let $m$ be the number of Sylow $p$-subgroups, and let $\order G = k p^n$.

From the Fourth Sylow Theorem, $m \equiv 1 \pmod p$.

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

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

$\blacksquare$

Sources

• 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $11$: The Sylow Theorems: Corollary $11.11$