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 $N \lhd G$, we see that:

$\gen {P, N} = P N$

from Subset Product with Normal Subgroup as Generator.

Since $P \cap N \le P$, it follows that:

$\order {P \cap N} = p^k$

where $k > 0$.

By Order of Subgroup Product:

$\order {P N} \order {P \cap N} = \order P \order N$

Hence from Lagrange's Theorem:

$\index N {P \cap N} = \index {P N} P$

By Tower Law for Subgroups:

$\index G P = \index G {P N} \index {P N} P$

Thus:

$\index {P N} P \divides \index G P$

where $\divides$ denotes divisibility.

Thus:

$p \nmid \index {P N} P$

so:

$p \nmid \index N {P \cap N}$

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

$\blacksquare$

Sources

• 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $11$: The Sylow Theorems: Proposition $11.14 \ \text{(i)}$