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The profinite completion is initial among continuous homomorphisms from G to profinite groups

Statement

The profinite completion is initial among continuous homomorphisms from $G$ to profinite groups.

Facts & Assumptions

Given: An abstract group $G$ , a profinite group $P$ , and a homomorphism $f : G \to P$ that is continuous for the profinite topology on $G$ .

[L1]

By definition, choose a topological group isomorphism $θ : P \to {\underset{\leftarrow}{l i m}}_{i} P_{i}$ with every $P_{i}$ finite discrete (A profinite group is a topological group isomorphic to an inverse limit of finite discrete groups).

[L2]

The canonical map $ι_{G} : G \to \hat{G}$ has dense image (The canonical map to the profinite completion has kernel equal to the finite residual and has dense image).

[F1]

The compatible-tuples construction satisfies the inverse-limit universal property (The compatible-tuple construction satisfies the inverse-limit universal property in groups).

[F2]

The completion $\hat{G}$ is the inverse limit of the finite quotients $G / N$ with its inverse-limit topology, and the $N$ th coordinate of $ι_{G} (g)$ is $g N$ (The profinite completion is the inverse limit of the finite quotients G over N, The canonical map sends g to its coherent system of residue classes).

[L3]

A map into an inverse limit is continuous exactly when all coordinate composites are continuous (A map into an inverse limit is continuous exactly when all coordinate composites are continuous).

[L4]

An inverse limit of finite discrete groups is Hausdorff (Inverse limits of finite discrete groups are Hausdorff and totally disconnected, and compact assuming Choice).

Proof

technique · direct

1.1

For each coordinate $q_{i} : P \to P_{i}$ of [L1], put $N_{i} : = \ker (q_{i} \circ f)$ . This is normal, and $G / N_{i}$ is isomorphic to the image of $q_{i} \circ f$ in the finite group $P_{i}$ , so $N_{i}$ has finite index. Let ${\overline{f}}_{i} : G / N_{i} \to P_{i}$ be the induced homomorphism.

L1givenconstructalgebra

2.1

Let $π_{N_{i}} : \hat{G} \to G / N_{i}$ be the completion coordinate and define ${\hat{f}}_{i} : = {\overline{f}}_{i} \circ π_{N_{i}}$ . It is continuous because both finite quotients are discrete and $π_{N_{i}}$ is a coordinate projection for the topology in [F2]. Moreover, [F2] gives ${\hat{f}}_{i} \circ ι_{G} = q_{i} \circ f .$

F2L1step 1.1construct

3.1

If $i \leq j$ and $φ_{i j} : P_{j} \to P_{i}$ is the transition map, then $q_{i} = φ_{i j} \circ q_{j}$ , so $N_{j} \subseteq N_{i}$ . Let $ψ_{i j} : G / N_{j} \to G / N_{i}$ be the natural quotient map. The two induced maps satisfy $φ_{i j} \circ {\overline{f}}_{j} = {\overline{f}}_{i} \circ ψ_{i j},$ and the completion coordinates satisfy $ψ_{i j} \circ π_{N_{j}} = π_{N_{i}}$ . Hence $φ_{i j} \circ {\hat{f}}_{j} = {\hat{f}}_{i}$ , so $({\hat{f}}_{i})_{i}$ is a compatible cone.

L1step 1.1step 2.1algebra

4.1

By [F1], the compatible cone from step 3.1 induces a homomorphism $h : \hat{G} \to {\underset{\leftarrow}{l i m}}_{i} P_{i}$ with coordinate maps ${\hat{f}}_{i}$ . The coordinate identities in step 2.1 give $h \circ ι_{G} = θ \circ f$ . Define $\hat{f} : = θ^{- 1} \circ h$ ; then $\hat{f} \circ ι_{G} = f$ .

F1L1step 2.1step 3.1construct

5.1

Every coordinate composite of $h$ is the continuous map ${\hat{f}}_{i}$ , so [L3] makes $h$ continuous. The inverse $θ^{- 1}$ is continuous because $θ$ is a topological group isomorphism, hence $\hat{f}$ is continuous.

L1L3step 2.1step 4.1

5.2

For uniqueness, let $u, v : \hat{G} \to P$ be continuous homomorphisms with $u \circ ι_{G} = v \circ ι_{G} = f$ . By [L1] and [L4], $P$ is Hausdorff. The equalizer of $u$ and $v$ is therefore closed, while [L2] says it contains the dense subset $ι_{G} [G]$ . Thus the equalizer is all of $\hat{G}$ , and $u = v$ .

L1L2L4step 4.1algebra

6.1

Steps 4.1, 5.1, and 5.2 give the unique continuous homomorphism $\hat{f} : \hat{G} \to P$ extending $f$ . Therefore $\hat{G}$ is initial among continuous homomorphisms from $G$ to profinite groups.

step 4.1step 5.1step 5.2∎

Depends on

Used by

Dependency tree · two levels

24 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Brian Osserman, Math 6112 notes on inverse limits and profinite groups (standard reference, not scraped)
H. W. Lenstra, Profinite groups and Galois groups (standard reference, not scraped)