A finite group is nilpotent iff it is the product of its Sylow subgroups

Statement

Let $G$ be a finite group. The following conditions are equivalent:

1. $G$ is nilpotent.
2. For each prime number $p$, $G$ has exactly one Sylow p-subgroup.
3. $G$ is the internal direct product of its Sylow subgroups, one for each prime divisor of $|G|$.

Remarks

The direction from 3 to 1 holds because finite p-groups are nilpotent and a finite direct product of nilpotent groups is nilpotent. The direction from 1 to 2 uses the Frattini argument. In a nilpotent group, a proper subgroup is strictly smaller than its normaliser. For example, $\mathbb{Z}/12\mathbb{Z} \cong \mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}$ is nilpotent, but $S_3$ has three Sylow $2$-subgroups, so it is not nilpotent.

Explanations

L1 human

A finite nilpotent group splits into its $p$-parts, one for each prime. The parts do not interact, so the group is their direct product. Conversely, a direct product of $p$-groups is nilpotent.