Intersection of Normal Subgroup with Sylow P-Subgroup

Theorem

Let $P$ be a Sylow $p$-subgroup of a finite group $G$.

Let $N$ be a normal subgroup of $G$.

Then $P \cap N$ is a Sylow $p$-subgroup of $N$.

Proof

Since $N \triangleleft G$, we see that:

$\left \langle {P, N} \right \rangle = P N$

from Subgroup Product with Normal Subgroup as Generator.

Since $P \cap N \le P$, it follows that $\left|{P \cap N}\right| = p^k$ where $k > 0$.

By Order of Subgroup Product:

$\left|{P N}\right| \left|{P \cap N}\right| = \left|{P}\right| \left|{N}\right|$

Hence from Lagrange's Theorem:

$\left[{N : P \cap N}\right] = \left[{P N : P}\right]$

By Index of Subgroup of Subgroup:

$\left[{G : P}\right] = \left[{G : P N}\right] \left[{P N : P}\right]$

Thus:

$\left[{P N : P}\right] \backslash \left[{G : P}\right]$

where $\backslash$ denotes divisibility.

Thus:

$p \nmid \left[{P N : P}\right]$

so:

$p \nmid \left[{N : P \cap N}\right]$

Thus $P \cap N$ is a Sylow $p$-subgroup of $N$.

$\blacksquare$

Sources

• John F. Humphreys: A Course in Group Theory (1996): $\S 11$: Proposition $11.14 \ \text{(i)}$