Profinite group

Remarks

A topological group is profinite if and only if it is compact, Hausdorff and totally disconnected. Then $G$ is the inverse limit of its finite quotients $G/N$, where $N$ runs over the open normal subgroups.

Example: the p-adic integers $\Z_p = \varprojlim_n \Z/p^n\Z$, with the reduction maps $\Z/p^{n+1}\Z \to \Z/p^n\Z$. Each $\Z/p^n\Z$ is finite and discrete, so $\Z_p$ is profinite.

Non-example: $\Z$ with the discrete topology is Hausdorff and totally disconnected but not compact, so it is not profinite. Its profinite completion $\hat\Z = \varprojlim_n \Z/n\Z$ is profinite and contains $\Z$ as a dense subgroup.

Every finite group with the discrete topology is profinite. Closed subgroups, quotients by closed normal subgroups, and arbitrary products of profinite groups are profinite.