Sylow Subgroup is Hall Subgroup

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 $\left|{G}\right| = k p^n$ 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 $\left[{G : H}\right] = \dfrac {\left|{G}\right|} {\left|{H}\right|}$.

So in this case $\left[{G : H}\right] = \dfrac {k p^n} {p^n} = k$.

As $p \nmid k$ it follows from Prime Not Divisor then Coprime that $k \perp p$.

The result follows from the definition of Hall subgroup.

$\blacksquare$