If a group has a normal subgroup of order $d$ for every divisor $d$ of the order of the group then it is nilpotent

Question

How do I prove this?

Let $G$ be a finite group. For every $d$ dividing $|G|$, there is a normal subgroup of order $d$. Then $G$ is nilpotent.

The proof in my notes says that clearly every Sylow subgroup is normal, and we should "apply" this fact. However I don't see why this implies $G$ is nilpotent.

Answer 1

If every Sylow subgroup is normal, then $G$ is the direct product of its Sylow subgroups. This is because the intersection of Sylow subgroups of different $p$ is trivial, because $p$ divides the order of any non trivial element of a $p$-group.

The direct product of nilpotent groups is nilpotent, and $p$-groups are nilpotent. So $G$ is nilpotent.

Answer 2

A finite group $G$ is nilpotent if and only all it’s Sylow subgroups are normal. Choose $d=$ highest prime power for each of the primes dividing $|G|$.