profinite topology
Definition
Let $G$ be a group.
- The profinite topology on $G$ is the one with base $$\mathcal{B}={\text{cosets of finite index normal subgroups}}$$
- A
topological group $G$ is called residually finite if it its profinite topology is Hausdorff.
- Equivalently, $$\bigcap_{H\in\mathcal{B}}H={1_G}$$ (Property 1).
Examples
1 Galois groups are residually finite. In direct limit, we describe that a Galois extension $K/k$ is the direct limit of its finite Galois subfields $L/k$: $$K/k\cong\lim_\to L/k.$$ The basis on $G=\text{Gal}(K/k)$ is the set of $\text{Gal}(K/L)$. Suppose $\sigma\in\bigcap_{H\in\mathcal{B}}H$. Then $\sigma$ fixes every $L$, hence fixes $K$, hence $\sigma=1_G$ and by Property (1) we are done.
In fact, Galois groups are even stronger in their topology: they are profinite groups.
2 (additive $\mathbb{Z}$ and $\mathbb{R}$) The profinite topology on $\mathbb{Z}$ is not compact: consider the open cover $${U_p}\cup {1+5\mathbb{Z}}\cup {1-5\mathbb{Z}},$$ where $U_p=p\mathbb{Z}$ for $p$ prime. There is no finite subcover, as this would imply that there are only finitely many primes.
Properties
- $G$ is residually finite iff $\bigcap_{H\in\mathcal{B}}H={1_G}$.
- An equivalent bases is the one consisting of all left cosets of all subgroups of finite index.
- Hence in the profinite topology, every finite index subgroup is open. But this is not generally true for profinite groups. Namely, the topologies may not agree.
Proofs 2 The proposed basis contains the one in the definition, so it suffices to consider the other inclusion. Any subgroup contains a nonempty normal subgroup ( core) $N\subset H\subset G$. The profinite topology is translation invariant, so for any $g\in H$ we can translate $N$ to an open set $gN$ containing $g$. Since we can do this for every $g\in H$, $H$ is the union of all such translated open sets, hence in the topology.