3Modules

IB Groups, Rings and Modules

3.5 Conjugacy of matrices*

We are now going to do some fun computations of conjugacy classes of matrices,

using what we have got so far.

Lemma.

Let

α, β

:

V → V

be two linear maps. Then

V

α

∼

=

V

β

as

F

[

X

]-modules

if and only if

α

and

β

are conjugate as linear maps, i.e. there is some

γ

:

V → V

such that α = γ

−1

βγ = α.

This is not a deep theorem. This is in some sense just some tautology. All

we have to do is to unwrap what these statements say.

Proof.

Let

γ

:

V

β

→ V

α

be an

F

[

X

]-module isomorphism. Then for

v ∈ V

, we

notice that if β(v) is just X · v in V

β

, and α(v) is just X · v in V

α

. So we get

β ◦ γ(v) = X · (γ(v)) = γ(X · v) = γ ◦ α(v),

using the definition of an F[X]-module homomorphism.

So we know

βγ = γα.

So

α = γ

−1

βγ.

Conversely, let

γ

:

V → V

be a linear isomorphism such that

γ

−1

βγ

=

α

. We

now claim that

γ

:

V

α

→ V

β

is an

F

[

X

]-module map. We just have to check that

γ(f · v) = γ(f(α)(v))

= γ(a

0

+ a

1

α + ···+ a

r

α

r

)(v)

= γ(a

0

v) + γ(a

1

α(v)) + γ(a

2

α

2

(v)) + ··· + γ(a

n

α

n

(v))

= (a

0

+ a

1

β + a

2

β

2

+ ··· + a

n

β

n

)(γ(v))

= f · γ(v)

So classifying linear maps up to conjugation is the same as classifying modules.

We can reinterpret this a little bit, using our classification of finitely-generated

modules.

Corollary.

There is a bijection between conjugacy classes of

n × n

matrices

over

F

and sequences of monic polynomials

d

1

, ··· , d

r

such that

d

1

| d

2

| ··· | d

r

and deg(d

1

, ··· , d

r

) = n.

Example.

Let’s classify conjugacy classes in

GL

2

(

F

), i.e. we need to classify

F[X] modules of the form

F[X]

(d

1

)

⊕

F[X]

(d

2

)

⊕ ··· ⊕

F[X]

(d

r

)

which are two-dimensional as

F

-modules. As we must have

deg

(

d

1

d

2

···d

r

) = 2,

we either have a quadratic thing or two linear things, i.e. either

(i) r = 1 and deg(d

1

) = 2,

(ii) r

= 2 and

deg

(

d

1

) =

deg

(

d

2

) = 1. In this case, since we have

d

1

| d

2

, and

they are both monic linear, we must have d

1

= d

2

= X − λ for some λ.

In the first case, the module is

F[X]

(d

1

)

,

where, say,

d

1

= X

2

+ a

1

X + a

2

.

In the second case, we get

F[X]

(X − λ)

⊕

F[X]

(X − λ)

.

What does this say? In the first case, we use the basis 1

, X

, and the linear map

has matrix

0 −a

2

1 −a

1

In the second case, this is

λ 0

0 λ

.

Do these cases overlap? Suppose the two of them are conjugate. Then they have

the same determinant and same trace. So we know

−a

2

= 2λ

a

2

= λ

2

So in fact our polynomial is

X

2

+ a

1

X + a

2

= X

2

− 2λ + λ

2

= (X − λ)

2

.

This is just the polynomial of a Jordan block. So the matrix

0 −a

2

1 −a

1

is conjugate to the Jordan block

λ 0

1 λ

,

but this is not conjugate to

λI

, e.g. by looking at eigenspaces. So these cases

are disjoint.

Note that we have done more work that we really needed, since

λI

is invariant

under conjugation.

But the first case is not too satisfactory. We can further classify it as follows.

If X

2

+ a

1

X + a

2

is reducible, then it is

(X − λ)(X − µ)

for some µ, λ ∈ F. If λ = µ, then the matrix is conjugate to

λ 0

1 λ

Otherwise, it is conjugate to

λ 0

0 µ

.

In the case where

X

2

+

a

1

X

+

a

2

is irreducible, there is nothing we can do

in general. However, we can look at some special scenarios and see if there is

anything we can do.

Example.

Consider

GL

2

(

Z/

3). We want to classify its conjugacy classes. By

the general theory, we know everything is conjugate to

λ 0

0 µ

,

λ 0

1 λ

,

0 −a

2

1 −a

1

,

with

X

2

+ a

1

X + a

2

irreducible. So we need to figure out what the irreducibles are.

A reasonable strategy is to guess. Given any quadratic, it is easy to see if it

is irreducible, since we can try to see if it has any roots, and there are just three

things to try. However, we can be a bit slightly more clever. We first count how

many irreducibles we are expecting, and then find that many of them.

There are 9 monic quadratic polynomials in total, since

a

1

, a

2

∈ Z/

3. The

reducibles are (

X − λ

)

2

or (

X − λ

)(

X − µ

) with

λ 6

=

µ

. There are three of each

kind. So we have 6 reducible polynomials, and so 3 irreducible ones.

We can then check that

X

2

+ 1, X

2

+ X + 2, X

2

+ 2X + 2

are the irreducible polynomials. So every matrix in

GL

2

(

Z/

3) is either congruent

to

0 −1

1 0

,

0 −2

1 −1

,

0 −2

1 −2

,

λ 0

0 µ

,

λ 0

1 λ

,

where

λ 6

=

µ

and

λ, µ ∈

(

Z/

3)

×

(since the matrix has to be invertible). The

number of conjugacy classes of each type are 1

,

1

,

1

,

3

,

2. So there are 8 conjugacy

classes. The first three classes have elements of order 4

,

8

,

8 respectively, by

trying. We notice that the identity matrix has order 1, and

λ 0

0 µ

has order 2 otherwise. Finally, for the last type, we have

ord

1 0

1 1

= 3, ord

2 0

1 2

= 6

Note that we also have

|GL

2

(Z/3)| = 48 = 2

4

· 3.

Since there is no element of order 16, the Sylow 2-subgroup of

GL

2

(

Z/

3) is not

cyclic.

To construct the Sylow 2-subgroup, we might start with an element of order

8, say

B =

0 1

1 2

.

To make a subgroup of order 6, a sensible guess would be to take an element of

order 2, but that doesn’t work, since

B

4

will give you the element of order 2.

Instead, we pick

A =

0 2

1 0

.

We notice

A

−1

BA =

0 1

2 0

0 1

1 2

0 2

1 0

=

1 2

0 2

0 2

1 0

=

2 2

2 0

= B

3

.

So this is a bit like the dihedral group.

We know that

hBi C hA, Bi.

Also, we know

|hBi|

= 8. So if we can show that

hBi

has index 2 in

hA, Bi

, then

this is the Sylow 2-subgroup. By the second isomorphism theorem, something

we have never used in our life, we know

hA, Bi

hBi

∼

=

hAi

hAi ∩ hBi

.

We can list things out, and then find

hAi ∩ hBi =

2 0

0 2

∼

=

C

2

.

We also know hAi

∼

=

C

4

. So we know

|hA, Bi|

|hBi|

= 2.

So |hA, Bi| = 16. So this is the Sylow 2-subgroup. in fact, it is

hA, B | A

4

= B

8

= e, A

−1

BA = B

3

i

We call this the semi-dihedral group of order 16, because it is a bit like a dihedral

group.

Note that finding this subgroup was purely guesswork. There is no method

to know that A and B are the right choices.