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

Assume that $G$ is a topological group^{} and ${G}_{e}$ is the identity component. It is well known, that ${G}_{e}$ is a normal subgroup^{} of $G$, thus we may speak about quotient group^{}.

Proposition^{}. The quotient group $G/{G}_{e}$ is totally disconnected.

Proof. First of all note that connected components^{} of $G$ are of the form $g{G}_{e}$. Indeed, ${G}_{e}$ is a connected component of $e\in G$ and for any $g\in G$ we have a homeomorphism^{} ${f}_{g}:G\to G$ such that ${f}_{g}(x)=gx$. Thus ${f}_{g}({G}_{e})=g{G}_{e}$ is a connected component of $g\in G$ (please, see this entry (http://planetmath.org/HomeomorphismsPreserveConnectedComponents) for more details).

Now let $\pi :G\to G/{G}_{e}$ be the quotient map (which is open and onto) and $A\subseteq G/{G}_{e}$ be an arbitrary, connected subset of $G/{G}_{e}$. Assume that there are at least two points in $A$. Consider the subset ${\pi}^{-1}(A)\subseteq G$ (which is the union of some cosets). Since $A$ has at least two points, then ${\pi}^{-1}(A)$ contains at least two cosets, which are connected components of $G$. Thus ${\pi}^{-1}(A)$ is not connected. Therefore there exist $U,V\subseteq {\pi}^{-1}(A)$ such that $U,V$ are open (in ${\pi}^{-1}(A)$), disjoint and $U\cup V={\pi}^{-1}(A)$.

Note that if $x\in U$, then the connected component of $x$ (which is equal to $x{G}_{e}$) is contained in $U$. Indeed, assume that $x{G}_{e}\u2288U$. Then there is $h\in x{G}_{e}$ such that $h\notin U$. Then, since $U\cup V={\pi}^{-1}(A)$ we have that $h\in V$. But then $U\cap x{G}_{e}$ and $V\cap x{G}_{e}$ are nonempty open and disjoint subsets of $x{G}_{e}$ such that $(U\cap x{G}_{e})\cup (V\cap x{G}_{e})=x{G}_{e}$. Contradiction^{}, because $x{G}_{e}$ is connected. Analogusly, whenever $x\in V$, then $x{G}_{e}\subseteq V$.

Therefore both $U$ and $V$ are unions of cosets. Thus $\pi (U)$ and $\pi (V)$ are disjoint. Furthermore $\pi (U)\cup \pi (V)=A$ and both $\pi (U)$, $\pi (V)$ are open in $A$ (because $\pi $ is an open map). This means that $A$ is not connected. Contradiction. Thus $A$ has at most one element, which completes^{} the proof. $\mathrm{\square}$

Remark. This proposition can be easily generalized as follows: assume that $X$ is a topological space^{}, $X=\bigcup {X}_{i}$ is a decomposition of $X$ into connected components and $R$ is an equivalence relation^{} associated to this decomposition (i.e. $xRy$ if and only if there exists $i$ such that $x,y\in {X}_{i}$). Then, if the quotient map $\pi :X\to X/R$ 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 |