G⁢L_2⁢(ℤ)

G

⁢

L

2

⁢

(

ℤ

)

Let

M

2

⁢

(

ℤ

)

be the ring of

2

⁢

x

⁢

2

matrices with

integer

entries, and define

G

⁢

L

2

⁢

(

ℤ

)

to be the

subring

of matrices

invertible

over

ℤ

. Thus for

M

∈

M

2

⁢

(

ℤ

)

,

M

∈

G

⁢

L

2

⁢

(

ℤ

)

⇔

det

⁡

M

=

±

1

Let

A

⁢

u

⁢

t

ℤ

⁢

(

ℤ

⊕

ℤ

)

be the ring of

automorphisms

of

ℤ

⊕

ℤ

as a

ℤ

-module. Then

G

⁢

L

2

⁢

(

ℤ

)

≅

A

⁢

u

⁢

t

ℤ

⁢

(

ℤ

⊕

ℤ

)

as rings, under the obvious

operations

.

To see this, we demonstrate a natural correspondence between endomorphisms of

ℤ

⊕

ℤ

and

M

2

⁢

(

ℤ

)

and show that invertible endomorphisms correspond to

invertible matrices

. Let

φ

:

ℤ

⊕

ℤ

→

ℤ

⊕

ℤ

be any

ring homomorphism

. It is clear that

φ

is determined by its action on

(

1

,

0

)

and

(

0

,

1

)

, since

φ

⁢

(

x

,

y

)

=

φ

⁢

(

x

⁢

(

1

,

0

)

+

y

⁢

(

0

,

1

)

)

=

x

⁢

φ

⁢

(

1

,

0

)

+

y

⁢

φ

⁢

(

0

,

1

)

Suppose then that

φ

⁢

(

1

,

0

)

=

(

a

,

b

)

and

φ

⁢

(

0

,

1

)

=

(

c

,

d

)

. Then

φ

⁢

(

x

,

y

)

=

(

a

⁢

x

,

b

⁢

x

)

+

(

c

⁢

y

,

d

⁢

y

)

=

(

a

⁢

x

+

c

⁢

y

,

b

⁢

x

+

d

⁢

y

)

=

(

a

c

b

d

)

⁢

(

x

y

)

Now,

φ

is

surjective

if both

(

1

,

0

)

and

(

0

,

1

)

are in its image. But

(

1

,

0

)

∈

im

⁢

φ

if and only if there is some

(

x

,

y

)

such that

a

⁢

x

+

c

⁢

y

=

1

b

⁢

x

+

d

⁢

y

=

0

Solving this pair of

equations

for

y

we see that we must have

y

⁢

(

b

⁢

c

-

a

⁢

d

)

=

1

and thus

b

⁢

c

-

a

⁢

d

=

±

1

. Similarly,

(

0

,

1

)

∈

im

⁢

φ

if and only if

y

⁢

(

a

⁢

d

-

b

⁢

c

)

=

±

1

. Thus

φ

is surjective precisely when

a

⁢

d

-

b

⁢

c

=

±

1

, i.e. precisely when the

matrix representation

of

φ

has determininant

±

1

. This then gives a map from

A

⁢

u

⁢

t

ℤ

⁢

(

ℤ

⊕

ℤ

)

to

G

⁢

L

2

⁢

(

ℤ

)

that is obviously a ring isomorphism. This concludes the proof.

A

=

A

⁢

u

⁢

t

ℤ

⁢

(

ℤ

⊕

ℤ

)

has a simple and well-known set of

generators

as a group:

r

=

(

x

,

y

)

↦

(

y

,

x

)

s

=

(

x

,

y

)

↦

(

x

,

x

+

y

)

Note that

s

m

=

(

x

,

y

)

↦

(

x

,

m

⁢

x

+

y

)

for any integer

m

. We now prove this fact.

Define the

subgroup

A

′

⊂

A

by

A

′

=

<

r

,

s

>

, the subgroup of

A

generated by

r

and

s

. If

φ

1

,

φ

2

∈

A

, define

φ

1

∼

φ

2

if

φ

1

and

φ

2

are in the same

A

′

-coset.

Our objective is to show that

A

′

=

A

, which we can do by showing that each

φ

∼

e

, where

e

is the

identity transformation

of

A

. This demonstration is essentially an application of the

Euclidean algorithm

. For suppose

φ

⁢

(

x

,

y

)

=

(

a

⁢

x

+

c

⁢

y

,

b

⁢

x

+

d

⁢

y

)

Assume, by applying

r

if necessary, that

a

≤

b

, and choose

m

such that

b

=

a

⁢

m

+

q

,

0

≤

q

<

a

. Then

r

⁢

s

-

m

⁢

φ

⁢

(

x

,

y

)

=

r

⁢

(

a

⁢

x

+

c

⁢

y

,

(

b

-

a

⁢

m

)

⁢

x

+

(

d

-

c

⁢

m

)

⁢

y

)

=

r

⁢

(

a

⁢

x

+

c

⁢

y

,

q

⁢

x

+

d

⁢

y

)

=

(

q

⁢

x

+

d

⁢

y

,

a

⁢

x

+

c

⁢

y

)

, so that

φ

∼

(

x

,

y

)

↦

(

q

⁢

x

+

d

⁢

y

,

a

⁢

x

+

c

⁢

y

)

Continuing this process, we

eventually

see that

φ

∼

(

x

,

y

)

↦

(

c

⁢

y

,

b

⁢

x

+

d

⁢

y

)

But

a

⁢

d

-

b

⁢

c

=

±

1

, so we have

b

⁢

c

=

±

1

. Applying either

s

d

or

s

-

d

as appropriate, we get

φ

∼

(

x

,

y

)

↦

(

c

⁢

y

,

b

⁢

x

)

∼

(

x

,

y

)

↦

(

b

⁢

x

,

c

⁢

y

)

Thus, we are done if we show that all such forms

(

b

⁢

x

,

c

⁢

y

)

with

b

,

c

=

±

1

are in the same

A

′

-coset as

e

. The case where

b

=

c

=

1

is obvious. For the other cases, note that

(

x

,

y

)

↦

(

x

,

-

y

)

=

s

-

1

⁢

r

⁢

s

⁢

r

⁢

s

-

1

⁢

r

(

x

,

y

)

↦

(

-

x

,

y

)

=

r

⁢

s

-

1

⁢

r

⁢

s

⁢

r

⁢

s

-

1

⁢

r

and

(

x

,

y

)

↦

(

-

x

,

-

y

)

is obviously the

composition

of these two.

This result is often phrased by saying that the matrices

(

1

0

1

1

)

,

(

0

1

1

0

)

generate

G

⁢

L

2

⁢

(

ℤ

)

as a

multiplicative group

.

Title

G

⁢

L

2

⁢

(

ℤ

)

Canonical name

GL2mathbbZ

Date of creation

2013-03-22 16:31:38

Last modified on

2013-03-22 16:31:38

Owner

rm50 (10146)

Last modified by

rm50 (10146)

Numerical id

7

Author

rm50 (10146)

Entry type

Application

Classification

msc 20G15

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