quotient group of a topological group by its identity component is totally disconnected

Assume that G is a topological groupMathworldPlanetmath and Ge is the identity component. It is well known, that Ge is a normal subgroupMathworldPlanetmath of G, thus we may speak about quotient groupMathworldPlanetmath.

PropositionPlanetmathPlanetmathPlanetmath. The quotient group G/Ge is totally disconnected.

Proof. First of all note that connected componentsMathworldPlanetmathPlanetmathPlanetmathPlanetmath of G are of the form gGe. Indeed, Ge is a connected component of eG and for any gG we have a homeomorphismPlanetmathPlanetmath fg:GG such that fg(x)=gx. Thus fg(Ge)=gGe is a connected component of gG (please, see this entry (http://planetmath.org/HomeomorphismsPreserveConnectedComponents) for more details).

Now let π:GG/Ge be the quotient map (which is open and onto) and AG/Ge be an arbitrary, connected subset of G/Ge. Assume that there are at least two points in A. Consider the subset π-1(A)G (which is the union of some cosets). Since A has at least two points, then π-1(A) contains at least two cosets, which are connected components of G. Thus π-1(A) is not connected. Therefore there exist U,Vπ-1(A) such that U,V are open (in π-1(A)), disjoint and UV=π-1(A).

Note that if xU, then the connected component of x (which is equal to xGe) is contained in U. Indeed, assume that xGeU. Then there is hxGe such that hU. Then, since UV=π-1(A) we have that hV. But then UxGe and VxGe are nonempty open and disjoint subsets of xGe such that (UxGe)(VxGe)=xGe. ContradictionMathworldPlanetmathPlanetmath, because xGe is connected. Analogusly, whenever xV, then xGeV.

Therefore both U and V are unions of cosets. Thus π(U) and π(V) are disjoint. Furthermore π(U)π(V)=A and both π(U), π(V) are open in A (because π is an open map). This means that A is not connected. Contradiction. Thus A has at most one element, which completesPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath the proof.

Remark. This proposition can be easily generalized as follows: assume that X is a topological spaceMathworldPlanetmath, X=Xi is a decomposition of X into connected components and R is an equivalence relationMathworldPlanetmath associated to this decomposition (i.e. xRy if and only if there exists i such that x,yXi). Then, if the quotient map π:XX/R is open, then X/R is totally disconnected.

Title quotient group of a topological group by its identity component is totally disconnected
Canonical name QuotientGroupOfATopologicalGroupByItsIdentityComponentIsTotallyDisconnected
Date of creation 2013-03-22 18:45:35
Last modified on 2013-03-22 18:45:35
Owner joking (16130)
Last modified by joking (16130)
Numerical id 7
Author joking (16130)
Entry type Theorem
Classification msc 22A05