characterization of Alexandroff groups

Topological groupMathworldPlanetmath G is called Alexandroff if G is an Alexandroff space as a topological spaceMathworldPlanetmath. For example every finite topological group is Alexandroff. We wish to characterize them. First recall, that if A is a subset of a topological space, then Ao denotes an intersectionMathworldPlanetmath of all open neighbourhoods of A.

Lemma. Let X be an Alexandroff space, f:X××XX be a continuous map and xX such that f(x,,x)=x. Then f(A××A)A, where A={x}o.

Proof. Let A={x}o. Of course A is open (because X is Alexandroff). Therefore f-1(A) is open in X××X. Thus (from the definition of product topology and continuous map), there are open subsetes V1,,VnX such that each Vi is an open neighbourhood of x and


Now let Ui=ViA. Of course xUi, so Ui is nonempty and Ui is open. Furthermore UiVi and thus


On the other hand UiA and Ui is open neighbourhood of x. Thus Ui=A, because A is minimalPlanetmathPlanetmath open neighbourhood of x. Therefore


which completesPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath the proof.

PropositionPlanetmathPlanetmath. Let G be an Alexandroff group. Then there exists open, normal subgroupMathworldPlanetmath H of G such that for every open subset UG there exist {gi}iIG such that


Proof. Let H={e}o be an intersection of all open neighbourhoods of the identityPlanetmathPlanetmathPlanetmath eG. Let U be an open subset of G. If gU, then g-1U is an open neighbourhood of e. Thus Hg-1U and therefore gHU. Thus


To complete the proof we need to show that H is normal subgroup of G. Consider the following mappings:

M:G×GG is such that M(x,y)=xy;
ψ:GG is such that ψ(x)=x-1;
φg:GG is such that φg(x)=gxg-1 for any gG.

Of course each of them is continuous (because G is a topological group). Furthermore each of them satisfies Lemma’s assumptionsPlanetmathPlanetmath (for x=e). Thus we have:

gHg-1=φg(H)H for any gG.

This shows that H is a normal subgroup, which completes the proof.

Corollary. Let G be a topological group such that G is finite and simple. Then G is either discrete or antidiscrete.

Proof. Of course finite topological groups are Alexandroff. Since G is simple, then there are only two normal subgroups of G, namely the trivial group and entire G. Therfore (due to proposition) the topology on G is ,,generated” by either the trivial group or entire G. In the first case we gain the discrete topology and in the second the antidiscrete topology.

Title characterization of Alexandroff groups
Canonical name CharacterizationOfAlexandroffGroups
Date of creation 2013-03-22 18:45:43
Last modified on 2013-03-22 18:45:43
Owner joking (16130)
Last modified by joking (16130)
Numerical id 4
Author joking (16130)
Entry type Theorem
Classification msc 22A05