Residually finite group

Remarks

Equivalently, for every $g \ne 1$ there are a finite group $Q$ and a homomorphism $\varphi \colon G \to Q$ 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 $0$.