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 $$\left[{G : H}\right] = \frac {\left|{G}\right|} {\left|{H}\right|}$$.

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

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.