Equivalence of definitions of profinite group

This article gives a proof/explanation of the equivalence of multiple definitions for the term profinite group
View a complete list of pages giving proofs of equivalence of definitions

Statement

The following are equivalent for a topological group:

1. It is the inverse limit of an inverse system of finite groups, each equipped with the discrete topology.
2. It is a compact totally disconnected T0 topological group.

A topological group satisfying both equivalent conditions is termed a profinite group.

Facts used

1. Hausdorffness is product-closed
2. Hausdorffness is hereditary
3. Tychonoff's theorem
4. Closed subset of compact space implies compact
5. Connectedness is continuous image-closed
6. Compact and totally disconnected implies every open neighborhood of identity contains an open normal subgroup
7. Compact implies every open subgroup has finite index
8. Compact to Hausdorff implies closed

Proof

(1) implies (2)

Given: An inverse system $G_{i},i\in I$ of finite groups with discrete topologies. $G$ is the inverse limit of the system.

To prove: $G$ is compact, $T_{0}$, and totally disconnected.

Proof

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	$G$ can be identified with a closed subgroup of the product $\prod _{i\in I}G_{i}$, equipped with the product topology and the direct product group structure.	definition of inverse limit	$G$ is an inverse limit of the $G_{i}$s
2	$G$ is Hausdorff (hence $T_{0}$)	Facts (1), (2)	all the $G_{i}$s are finite and discrete, hence Hausdorff	Step (1)	By Fact (1), the product $\prod _{i\in I}G_{i}$ is Hausdorff, hence by Fact (2), so is any subgroup of that. By Step (1), we thus get that $G$ is Hausdorff.
3	$G$ is compact	Facts (3), (4)	all the $G_{i}$s are finite, hence compact	Step (1)	By Fact (3), the product $\prod _{i\in I}G_{i}$ is compact, hence by Fact (4), so is any closed subgroup of that. By Step (1), we thus get that $G$ is compact.
4	$G$ is totally disconnected	Fact (5)	all the $G_{i}$s are discrete, hence totally disconnected	Step (1)	[SHOW MORE] Suppose $C$ is a connected subset of $G$. For each projection map from $G$ to $G_{i}$, the image of $C$ in $G_{i}$ is connected, hence a point in $G_{i}$. Since each of the coordinate projections of the set is a single point, the whole set $C$ must be a single point in the product $\prod _{i\in I}G_{i}$. This proves that $G$ is totally disconnected.

(2) implies (1)

Given: A compact $T_{0}$ totally disconnected group $G$.

To prove: $G$ is the inverse limit of an inverse system of finite groups.

Proof:

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	The open normal subgroups of $G$ form a basis at the identity element for $G$.	Fact (6)	$G$ is compact and totally disconnected		Follows directly from Fact (6).
2	If $N$ is an open normal subgroup of $G$, then $G/N$ is a finite group with the discrete topology.	Fact (7)	$G$ is compact		By Fact (7), $G/N$ is finite. Further, $N$ is open, and so are all its cosets, so the coset space gets a discrete quotient topology.
3	Consider the homomorphism from $G$ to the inverse limit of the inverse system of quotients of $G$ by open normal subgroups. This is continuous and has dense image.
4	The homomorphism of Step (3) is a closed map.	Fact (8)
5	The homomorphism of Step (3) is injective.
6	The homomorphism of Step (3) is an isomorphism of topological groups.