local finiteness is closed under extension, proof that

local finiteness is closed under extension, proof that

Let

G

be a group and

N

a

normal subgroup

of

G

such that

N

and

G

/

N

are both

locally finite

.
We aim to show that

G

is locally finite.
Let

F

be a finite subset of

G

.
It suffices to show that

F

is

contained

in a

finite subgroup

of

G

.

Let

R

be a set of coset representatives of

N

in

G

,
chosen so that

1

∈

R

.
Let

r

:

G

/

N

→

R

be the

function

mapping

cosets to their representatives,
and let

s

:

G

→

N

be defined by

s

⁢

(

x

)

=

r

⁢

(

x

⁢

N

)

-

1

⁢

x

for all

x

∈

G

.
Let

π

:

G

→

G

/

N

be the

canonical projection

.
Note that for any

x

∈

G

we have

x

=

r

⁢

(

x

⁢

N

)

⁢

s

⁢

(

x

)

.

Put

A

=

r

⁢

(

⟨

π

⁢

(

F

)

⟩

)

, which is finite as

G

/

N

is locally finite.
Let

B

=

s

⁢

(

F

∪

A

⁢

A

∪

A

-

1

)

, let

C

=

B

∪

B

-

1

and let

D

=

{

a

-

1

⁢

c

⁢

a

∣

a

∈

A

⁢

and

⁢

c

∈

C

}

⊆

N

.

Put

H

=

⟨

D

⟩

, which is finite as

N

is locally finite.
Note that

1

∈

A

⊆

R

and

1

∈

B

⊆

C

⊆

D

⊆

H

≤

N

.

For any

a

1

,

a

2

∈

A

we have

a

1

⁢

a

2

=

r

⁢

(

a

1

⁢

a

2

⁢

N

)

⁢

s

⁢

(

a

1

⁢

a

2

)

∈

A

⁢

B

.
Note that

D

-

1

=

D

,
and so every element of

H

is a

product

of elements of

D

.
So any element of the form

a

-

1

⁢

h

⁢

a

, where

a

∈

A

and

h

∈

H

,
is a product of elements of the form

a

-

1

⁢

a

1

-

1

⁢

c

⁢

a

1

⁢

a

for

a

1

∈

A

and

c

∈

C

;
but

a

1

⁢

a

=

a

2

⁢

b

for some

a

2

∈

A

and

b

∈

B

,
so

a

-

1

⁢

h

⁢

a

is a product of elements of the form

b

-

1

⁢

a

2

-

1

⁢

c

⁢

a

2

⁢

b

=

b

-

1

⁢

(

a

2

-

1

⁢

c

⁢

a

2

)

⁢

b

∈

C

⁢

D

⁢

B

⊆

H

,
and therefore

a

-

1

⁢

h

⁢

a

∈

H

.

We claim that

A

⁢

H

≤

G

.
Let

a

1

,

a

2

∈

A

and

h

1

,

h

2

∈

H

.
We have

(

a

1

⁢

h

1

)

⁢

(

a

2

⁢

h

2

)

=

a

1

⁢

a

2

⁢

(

a

2

-

1

⁢

h

1

⁢

a

2

)

⁢

h

2

.
But, by the previous paragraph,

a

1

⁢

a

2

∈

A

⁢

B

and

a

2

-

1

⁢

h

1

⁢

a

2

∈

H

,
so

a

1

⁢

a

2

⁢

(

a

2

-

1

⁢

h

1

⁢

a

2

)

⁢

h

2

∈

A

⁢

B

⁢

H

⁢

H

⊆

A

⁢

H

.
Thus

A

⁢

H

⁢

A

⁢

H

⊆

A

⁢

H

.
Also,

(

a

1

⁢

h

1

)

-

1

=

h

1

-

1

⁢

a

1

-

1

∈

H

⁢

a

1

-

1

.
But

a

1

-

1

=

r

⁢

(

a

1

-

1

⁢

N

)

⁢

s

⁢

(

a

1

-

1

)

∈

A

⁢

B

,
so

H

⁢

a

1

-

1

⊆

H

⁢

A

⁢

B

⊆

A

⁢

H

⁢

A

⁢

H

⊆

A

⁢

H

.
Thus

(

A

⁢

H

)

-

1

⊆

A

⁢

H

.
It follows that

A

⁢

H

is a

subgroup

of

G

, and it is clearly finite.

For any

x

∈

F

we have

x

=

r

⁢

(

x

⁢

N

)

⁢

s

⁢

(

x

)

∈

A

⁢

B

.
So

F

⊆

A

⁢

H

, which

completes

the proof.

Title

local finiteness is

closed under

extension

, proof that

Canonical name

LocalFinitenessIsClosedUnderExtensionProofThat

Date of creation

2013-03-22 15:36:53

Last modified on

2013-03-22 15:36:53

Owner

yark (2760)

Last modified by

yark (2760)

Numerical id

6

Author

yark (2760)

Entry type

Proof

Classification

msc 20F50

Related topic

LocallyFiniteGroup

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