---
tag: DA515YH1J0
kind: definition
name: Residually finite group
msc: ["20E26"]
defines: HEE922EHQ8
uses: [G5S1M9, 0CB0DX5BVV, 0DTXBQPA2S, 0JYTG5K9Y4]
sources:
  - url: https://en.wikipedia.org/wiki/Residually_finite_group
---

## Definition

Let $G$ be a [[group|G5S1M9]].
$G$ is *residually finite* if for every $g \in G$ with $g \ne 1$ there is a [[normal subgroup|0CB0DX5BVV]] $N$ of $G$ of finite [[index|0DTXBQPA2S]] with $g \notin N$.

## Remarks

Equivalently, for every $g \ne 1$ there are a finite group $Q$ and a [[homomorphism|0JYTG5K9Y4]] $\varphi \colon G \to Q$ with $\varphi(g) \ne 1$.
Equivalently, the intersection of all normal subgroups of finite index of $G$ is trivial.

Example: $\Z$ is residually finite.
For $n \ne 0$, the subgroup $N = (\abs{n} + 1)\Z$ is normal of index $\abs{n} + 1$, and $n \notin N$ since $0 < \abs{n} < \abs{n} + 1$.

Non-example: the additive group $\Q$ is not residually finite.
Each quotient of $\Q$ is divisible, and a finite divisible abelian group is trivial, since multiplication by its order is surjective and also zero.
So $\Q$ has no proper normal subgroup of finite index, and the condition fails for the element $1 \in \Q$, which is not the identity element $0$.