characterization of Alexandroff groups

characterization of Alexandroff groups

Topological group

G

is called

Alexandroff

if

G

is an Alexandroff space as a

topological space

. 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

A

o

denotes an

intersection

of all open

neighbourhoods

of

A

.

Lemma.

Let

X

be an Alexandroff space,

f

:

X

×

⋯

×

X

→

X

be a

continuous map

and

x

∈

X

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

V

1

,

…

,

V

n

⊆

X

such that each

V

i

is an open neighbourhood of

x

and

f

⁢

(

V

1

×

⋯

×

V

n

)

⊆

A

.

Now let

U

i

=

V

i

∩

A

. Of course

x

∈

U

i

, so

U

i

is nonempty and

U

i

is open. Furthermore

U

i

⊆

V

i

and thus

f

⁢

(

U

1

×

⋯

×

U

n

)

⊆

A

.

On the other hand

U

i

⊆

A

and

U

i

is open neighbourhood of

x

. Thus

U

i

=

A

, because

A

is

minimal

open neighbourhood of

x

. Therefore

f

⁢

(

A

×

⋯

×

A

)

=

f

⁢

(

U

1

×

⋯

×

U

n

)

⊆

A

,

which

completes

the proof.

□

Proposition

.

Let

G

be an Alexandroff group. Then there exists open,

normal subgroup

H

of

G

such that for every

open subset

U

⊆

G

there exist

{

g

i

}

i

∈

I

⊆

G

such that

U

=

⋃

i

∈

I

g

i

⁢

H

.

Proof.

Let

H

=

{

e

}

o

be an intersection of all open neighbourhoods of the

identity

e

∈

G

. Let

U

be an open subset of

G

. If

g

∈

U

, then

g

-

1

⁢

U

is an open neighbourhood of

e

. Thus

H

⊆

g

-

1

⁢

U

and therefore

g

⁢

H

⊆

U

. Thus

U

=

⋃

g

∈

U

g

⁢

H

.

To complete the proof we need to show that

H

is normal subgroup of

G

. Consider the following

mappings

:

M

:

G

×

G

→

G

⁢

is such that

⁢

M

⁢

(

x

,

y

)

=

x

⁢

y

;

ψ

:

G

→

G

⁢

is such that

⁢

ψ

⁢

(

x

)

=

x

-

1

;

φ

g

:

G

→

G

⁢

is such that

⁢

φ

g

⁢

(

x

)

=

g

⁢

x

⁢

g

-

1

⁢

for any

⁢

g

∈

G

.

Of course each of them is

continuous

(because

G

is a topological group). Furthermore each of them

satisfies

Lemma’s

assumptions

(for

x

=

e

). Thus we have:

H

⁢

H

=

M

⁢

(

H

×

H

)

⊆

H

;

H

-

1

=

ψ

⁢

(

H

)

⊆

H

;

g

⁢

H

⁢

g

-

1

=

φ

g

⁢

(

H

)

⊆

H

⁢

for any

⁢

g

∈

G

.

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

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