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.
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 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 isomorphism. We just have to
check that
γ(f · v) = γ(f(α)(v))
= γ(a
0
+ a
1
α + ···+ a
n
α
n
)(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
1
= 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, 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.