Sylow Subgroup is Hall Subgroup
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Theorem
Let $G$ be a group.
Let $H$ be a Sylow $p$-subgroup of $G$.
Then $H$ is a Hall subgroup of $G$.
Proof
Let $p$ be prime.
Let $G$ be a finite group such that $\order G = k p^n$ where $p \nmid k$.
By definition, a Sylow $p$-subgroup $H$ of $G$ is a subgroup of $G$ of order $p^n$.
By Lagrange's Theorem, the index of $H$ in $G$ is given by:
- $\index G H = \dfrac {\order G} {\order H}$
So in this case:
- $\index G H = \dfrac {k p^n} {p^n} = k$
As $p \nmid k$ it follows from Prime not Divisor implies Coprime that $k \perp p$.
The result follows from the definition of Hall subgroup.
$\blacksquare$