---
tag: 00Y9N80VSS
kind: statement
role: theorem
name: A finite group is nilpotent iff it is the product of its Sylow subgroups
msc: [20F18, 20D15]
uses: [0DZPD2MG33, 0JWZX6FKSH, 0Q2MVQ7JMT, 081WTKYX6C, 0QTZDKJEDV]
sources:
  - url: https://en.wikipedia.org/wiki/Nilpotent_group
    locator: Properties
  - url: https://en.wikipedia.org/wiki/Nilpotent_group
    locator: Examples
---

## Statement

Let $G$ be a finite [[group|G5S1M9]].
The following conditions are equivalent:

1. $G$ is [[nilpotent|0Q7MMQHC1P]].
2. For each prime number $p$, $G$ has exactly one [[Sylow p-subgroup|0053CQE9Y2]].
3. $G$ is the [[internal direct product|0VBZH90R1A]] 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|081WTKYX6C]] and a finite direct product of nilpotent groups is nilpotent.
The direction from 1 to 2 uses the [[Frattini argument|0QTZDKJEDV]]. 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.