two isomorphic groups

two isomorphic groups

The set of

3

⁢

x

⁡

3

permutation matrices

form a group under

matrix multiplication

. This example demonstrates that fact and develops the multiplication table and compares it to

S

3

. Although there are alternative ways to fill in the table, this example serves to help the beginner. Here we will see that the two groups have the
same

structure

. We begin by defining the elements of our group.

I

=

(

1

0

0

0

1

0

0

0

1

)

R

=

(

1

0

0

0

0

1

0

1

0

)

A

=

(

0

1

0

0

0

1

1

0

0

)

S

=

(

0

0

1

0

1

0

1

0

0

)

B

=

(

0

0

1

1

0

0

0

1

0

)

T

=

(

0

1

0

1

0

0

0

0

1

)

Here, our group is just

P

3

=

{

I

,

A

,

B

,

R

,

S

,

T

}

. Now, we can start to multiply and then fill in the table.
First, we calculate the square of each

elements

.

A

2

=

(

0

1

0

0

0

1

1

0

0

)

⁢

(

0

1

0

0

0

1

1

0

0

)

=

(

0

0

1

1

0

0

0

1

0

)

=

B

B

2

=

(

0

0

1

1

0

0

0

1

0

)

⁢

(

0

0

1

1

0

0

0

1

0

)

=

(

0

1

0

0

0

1

1

0

0

)

=

A

R

2

=

(

1

0

0

0

0

1

0

1

0

)

⁢

(

1

0

0

0

0

1

0

1

0

)

=

(

1

0

0

0

1

0

0

0

1

)

=

I

S

2

=

(

0

0

1

0

1

0

1

0

0

)

⁢

(

0

0

1

0

1

0

1

0

0

)

=

(

1

0

0

0

1

0

0

0

1

)

=

I

T

2

=

(

0

1

0

1

0

0

0

0

1

)

⁢

(

0

1

0

1

0

0

0

0

1

)

=

(

1

0

0

0

1

0

0

0

1

)

=

I

Now starting with the upper left

3

⁢

x

⁡

3

block, we go through the table.

A

⁢

B

=

(

0

1

0

0

0

1

1

0

0

)

⁢

(

0

0

1

1

0

0

0

1

0

)

=

(

1

0

0

0

1

0

0

0

1

)

=

I

B

⁢

A

=

(

0

0

1

1

0

0

0

1

0

)

⁢

(

0

1

0

0

0

1

1

0

0

)

=

(

1

0

0

0

1

0

0

0

1

)

=

I

We can

complete

the upper left

3

⁢

x

⁡

3

block of the table and complete

diagonal

using the above values. We note that no row or column can have a repeated elements which follows from the of a group. Next, we work on the upper right

3

⁢

x

⁡

3

block of the table.

A

⁢

R

=

(

0

1

0

0

0

1

1

0

0

)

⁢

(

1

0

0

0

0

1

0

1

0

)

=

(

0

0

1

0

1

0

1

0

0

)

=

S

A

⁢

S

=

(

0

1

0

0

0

1

1

0

0

)

⁢

(

0

0

1

0

1

0

1

0

0

)

=

(

0

1

0

1

0

0

0

0

1

)

=

T

Now we can complete the upper right 3 x 3 block of the table. Next, we work on the lower left

3

⁢

x

⁡

3

block of the table.

R

⁢

A

=

(

1

0

0

0

0

1

0

1

0

)

⁢

(

0

1

0

0

0

1

1

0

0

)

=

(

0

1

0

1

0

0

0

0

1

)

=

T

R

⁢

B

=

(

1

0

0

0

0

1

0

1

0

)

⁢

(

0

0

1

1

0

0

0

1

0

)

=

(

0

0

1

0

1

0

1

0

0

)

=

S

S

⁢

A

=

(

0

0

1

0

1

0

1

0

0

)

⁢

(

0

1

0

0

0

1

1

0

0

)

=

(

1

0

0

0

0

1

0

1

0

)

=

R

Now we can complete the lower left

3

⁢

x

⁡

3

block of the table. Finally, we work on the lower right

3

⁢

x

⁡

3

block of the table.

R

⁢

S

=

(

1

0

0

0

0

1

0

1

0

)

⁢

(

0

0

1

0

1

0

1

0

0

)

=

(

0

0

1

1

0

0

0

1

0

)

=

B

S

⁢

R

=

(

0

0

1

0

1

0

1

0

0

)

⁢

(

1

0

0

0

0

1

0

1

0

)

=

(

0

1

0

1

0

0

0

0

1

)

=

A

This completes the

multiplication

and the table is given below.

I

A

B

R

S

T

I

I

A

B

R

S

T

A

A

B

I

S

T

R

B

B

I

A

T

R

S

R

R

T

S

I

B

A

S

S

R

T

A

I

B

T

T

S

R

B

A

I

Next, we want to compare this table to the

symmetric group

S

3

. We begin as before by defining the elements
as follows.

e

=

(

1 2 3

1 2 3

)

r

=

(

1 2 3

2 1 3

)

a

=

(

1 2 3

2 3 1

)

s

=

(

1 2 3

3 2 1

)

b

=

(

1 2 3

3 1 2

)

t

=

(

1 2 3

1 3 2

)

The multiplication table for this group is obtained within the entry

symmetric group on three letters

. The table is:

e

a

b

r

s

t

e

e

a

b

r

s

t

a

a

b

e

s

t

r

b

b

e

a

t

r

s

r

r

t

s

e

b

a

s

s

r

t

a

e

b

t

t

s

r

b

a

e

Define the following

homomorphism

φ

:

S

3

→

P

3

by the following:

φ

⁢

(

e

)

=

I

;

φ

⁢

(

a

)

=

A

;

φ

⁢

(

b

)

=

B

;

φ

⁢

(

r

)

=

R

;

φ

⁢

(

s

)

=

S

;

φ

⁢

(

t

)

=

T

.

Since

φ

is a

bijection

, we conclude that

P

3

and

S

3

are

isomorphic

.

Title

two

isomorphic groups

Canonical name

TwoIsomorphicGroups

Date of creation

2013-03-22 15:52:28

Last modified on

2013-03-22 15:52:28

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

10

Author

Wkbj79 (1863)

Entry type

Example

Classification

msc 20B30

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