Sylow Subgroup is Hall Subgroup

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Let $G$ be a group.

Let $H$ be a Sylow $p$-subgroup of $G$.

Then $H$ is a Hall subgroup of $G$.


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.