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

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

∈

G

and for any

g

∈

G

we have a

homeomorphism

f

g

:

G

→

G

such that

f

g

⁢

(

x

)

=

g

⁢

x

. Thus

f

g

⁢

(

G

e

)

=

g

⁢

G

e

is a connected component of

g

∈

G

(please, see this entry (

http://planetmath.org/HomeomorphismsPreserveConnectedComponents

) for more details).

Now let

π

:

G

→

G

/

G

e

be the

quotient map

(which is open and onto) and

A

⊆

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

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

,

V

⊆

π

-

1

⁢

(

A

)

such that

U

,

V

are open (in

π

-

1

⁢

(

A

)

),

disjoint

and

U

∪

V

=

π

-

1

⁢

(

A

)

.

Note that if

x

∈

U

, then the connected component of

x

(which is equal to

x

⁢

G

e

) is

contained

in

U

. Indeed, assume that

x

⁢

G

e

⊈

U

. Then there is

h

∈

x

⁢

G

e

such that

h

∉

U

. Then, since

U

∪

V

=

π

-

1

⁢

(

A

)

we have that

h

∈

V

. But then

U

∩

x

⁢

G

e

and

V

∩

x

⁢

G

e

are nonempty open and disjoint subsets of

x

⁢

G

e

such that

(

U

∩

x

⁢

G

e

)

∪

(

V

∩

x

⁢

G

e

)

=

x

⁢

G

e

.

Contradiction

, because

x

⁢

G

e

is connected. Analogusly, whenever

x

∈

V

, then

x

⁢

G

e

⊆

V

.

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

completes

the proof.

□

Remark.

This proposition can be easily generalized as follows: assume that

X

is a

topological space

,

X

=

⋃

X

i

is a decomposition of

X

into connected components and

R

is an

equivalence relation

associated to this decomposition (i.e.

x

⁢

R

⁢

y

if and only if there exists

i

such that

x

,

y

∈

X

i

). Then, if the quotient map

π

:

X

→

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

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