Intersection of Normal Subgroup with Sylow P-Subgroup
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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 $k > 0$.
- $\order {P N} \order {P \cap N} = \order P \order N$
Hence from Lagrange's Theorem:
- $\index N {P \cap N} = \index {P N} P$
- $\index G P = \index G {P N} \index {P N} P$
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 $11.14 \ \text{(i)}$