examples of infinite simple groups

examples of infinite simple groups

Let

X

be a set and let

f

:

X

→

X

be a

function

. Define

C

⁢

(

f

)

=

{

x

∈

X

|

f

⁢

(

x

)

≠

x

}

.

Throughout, we will say that

f

:

X

→

X

is a

permutation

on

X

iff

f

is a

bijection

and

C

⁢

(

f

)

is a

finite set

.

For permutation

f

:

X

→

X

, the set

C

⁢

(

f

)

will play the role of a ,,bridge” between the

infinite

world and the finite world.

Let

S

⁢

(

X

)

denote the group of all permutations on

X

(with

composition

as a

multiplication

). For

f

∈

S

⁢

(

X

)

, subset

A

⊂

X

will be called

f

-finite

iff

A

is finite and

C

⁢

(

f

)

⊆

A

. This is

equivalent

to the fact, that

A

is finite and if

f

⁢

(

x

)

≠

x

, then

x

∈

A

.

It is easy to see, that if

f

∈

S

⁢

(

X

)

and

A

is

f

-finite, then

f

⁢

(

A

)

=

A

. Thus, we have well defined permutation (on a finite set)

f

A

:

A

→

A

by the

formula

f

A

⁢

(

x

)

=

f

⁢

(

x

)

.

Lemma.

For any subset

A

⊆

X

and any

f

,

g

∈

S

⁢

(

X

)

such that

A

is

f

-finite and

g

-finite we have that

A

is

f

∘

g

-finite and

(

f

∘

g

)

A

=

f

A

∘

g

A

.

Proof.

Assume, that

A

is

f

-finite and

g

-finite. Let

x

∈

X

be such that

(

f

∘

g

)

⁢

(

x

)

≠

x

. Assume, that

x

∉

A

. Then

f

⁢

(

x

)

=

g

⁢

(

x

)

=

x

and thus

(

f

∘

g

)

⁢

(

x

)

=

x

.

Contradiction

. Thus

x

∈

A

, so

C

⁢

(

f

∘

g

)

⊆

A

and since

A

is finite, then

A

is

f

∘

g

-finite. Finally, the

equality

(

f

∘

g

)

A

=

f

A

∘

g

A

holds, because

(

f

∘

g

)

A

is well definied (since

A

is

f

∘

g

-finite) and the

operation

(

⋅

)

A

does not change the formulas of functions.

□

Now we can talk about the sign of a permutation. For

f

∈

S

⁢

(

X

)

define

sgn

⁢

(

f

)

=

sgn

⁢

(

f

A

)

.

It can be easily checked, that

sgn

is well defined (indeed, sign depends only on those

x

∈

X

for which

f

⁢

(

x

)

≠

x

). Furthermore, it follows directly from the definition, that

sgn

:

S

⁢

(

X

)

→

{

-

1

,

1

}

is a

group homomorphism

(in

{

-

1

,

1

}

we have standard multiplication). Define

A

⁢

(

X

)

=

ker

⁢

(

sgn

)

.

Briefly speaking,

A

⁢

(

X

)

is the

subgroup

of

even permutations

on a set

X

(a.k.a. the

alternating group

for the set

X

).

Now, we shall prove the following

proposition

, using the fact, that for any finite set

X

with at least

5

elements

, the group

A

⁢

(

X

)

is simple (this is well known fact).

Proposition.

If

X

is an

infinite set

, then

A

⁢

(

X

)

is a

simple group

.

Proof.

Assume, that

A

⁢

(

X

)

is not simple and let

N

⊆

A

⁢

(

X

)

be a proper, nontrivial,

normal subgroup

. For a subset

Y

⊆

X

define

N

Y

=

{

f

Y

|

f

∈

N

⁢

and

⁢

C

⁢

(

f

)

⊆

Y

}

.

Note, that

A

⁢

(

Y

)

=

{

f

Y

|

f

∈

A

⁢

(

X

)

⁢

and

⁢

C

⁢

(

f

)

⊆

Y

}

.

Obviously

N

Y

⊆

A

⁢

(

Y

)

is a subgroup (due to lemma) of

A

⁢

(

Y

)

. We will show, that it is normal. Let

f

Y

∈

N

Y

and

g

Y

∈

A

⁢

(

Y

)

. We have to show, that

g

Y

∘

f

Y

∘

g

Y

-

1

∈

N

Y

. Of course

g

∘

f

∘

g

-

1

∈

N

,

because

N

is normal (here

f

,

g

correspond to

f

Y

,

g

Y

). It follows from lemma (note, that

Y

is

g

∘

f

∘

g

-

1

-finite), that

g

Y

∘

f

Y

∘

g

Y

-

1

=

(

g

∘

f

∘

g

-

1

)

Y

∈

N

Y

,

which shows, that

N

Y

is normal. To obtain the contradiction, we need to show, that there exists

Y

⊆

X

with at least

5

elements, such that

N

Y

is nontrivial and proper (because in this case

A

⁢

(

Y

)

is simple).

Let

f

∈

N

be such that

f

≠

id

X

and let

g

∈

A

⁢

(

X

)

be such that

g

∉

N

. Let

Y

be any

f

-finite and

g

-finite subset of

X

with at least

5

elements (such subset exists). Then

N

Y

is nontrivial, because

f

Y

∈

N

Y

is nontrivial.

Now assume, that

g

Y

∈

N

Y

, i.e. assume, that there exists

h

∈

N

with

C

⁢

(

h

)

⊆

Y

, such that

g

Y

=

h

Y

. Then (due to lemma)

Y

is

h

∘

g

-

1

-finite, and since

g

Y

=

h

Y

we have that for any

x

∈

Y

the following holds:

(

h

∘

g

-

1

)

⁢

(

x

)

=

x

.

On the other hand, for

x

∈

X

\

Y

we have

g

⁢

(

x

)

=

h

⁢

(

x

)

=

x

. This shows, that

h

=

g

, but

h

∈

N

and

g

∉

N

. Contradiction. Thus

g

Y

∉

N

Y

, so

N

Y

is proper.

This

completes

the proof.

□

Remark.

This proposition shows, that the class of simple groups is actually a

proper class

, i.e. it is not a set. Therefore studying infinite simple groups can be very difficult.

Title

examples of infinite simple groups

Canonical name

ExamplesOfInfiniteSimpleGroups

Date of creation

2013-03-22 19:09:17

Last modified on

2013-03-22 19:09:17

Owner

joking (16130)

Last modified by

joking (16130)

Numerical id

5

Author

joking (16130)

Entry type

Example

Classification

msc 20E32

Generated on Fri Feb 9 19:27:44 2018 by

LaTeXML