Part IA — Groups
Based on lectures by J. Goedecke
Notes taken by Dexter Chua
Michaelmas 2014
These notes are not endorsed by the lecturers, and I have modified them (often
significantly) after lectures. They are nowhere near accurate representations of what
was actually lectured, and in particular, all errors are almost surely mine.
Examples of groups
Axioms for groups. Examples from geometry: symmetry groups of regular polygons,
cub e, tetrahedron. Permutations on a set; the symmetric group. Subgroups and
homomorphisms. Symmetry groups as subgroups of general permutation groups. The
M¨obius group; cross-ratios, preservation of circles, the p oint at infinity. Conjugation.
Fixed p oints of M¨obius maps and iteration. [4]
Lagrange’s theorem
Cosets. Lagrange’s theorem. Groups of small order (up to order 8). Quaternions.
Fermat-Euler theorem from the group-theoretic point of view. [5]
Group actions
Group actions; orbits and stabilizers. Orbit-stabilizer theorem. Cayley’s theorem
(every group is isomorphic to a subgroup of a permutation group). Conjugacy classes.
Cauchy’s theorem. [4]
Quotient groups
Normal subgroups, quotient groups and the isomorphism theorem. [4]
Matrix groups
The general and special linear groups; relation with the M¨obius group. The orthogonal
and special orthogonal groups. Proof (in
R
3
) that every element of the orthogonal
group is the product of reflections and every rotation in
R
3
has an axis. Basis change
as an example of conjugation. [3]
Permutations
Permutations, cycles and transpositions. The sign of a permutation. Conjugacy in
S
n
and in A
n
. Simple groups; simplicity of A
5
. [4]
Contents
0 Introduction
1 Groups and homomorphisms
1.1 Groups
1.2 Homomorphisms
1.3 Cyclic groups
1.4 Dihedral groups
1.5 Direct products of groups
2 Symmetric group I
2.1 Symmetric groups
2.2 Sign of permutations
3 Lagrange’s Theorem
3.1 Small groups
3.2 Left and right cosets
4 Quotient groups
4.1 Normal subgroups
4.2 Quotient groups
4.3 The Isomorphism Theorem
5 Group actions
5.1 Group acting on sets
5.2 Orbits and Stabilizers
5.3 Important actions
5.4 Applications
6 Symmetric groups II
6.1 Conjugacy classes in S
n
6.2 Conjugacy classes in A
n
7 Quaternions
8 Matrix groups
8.1 General and special linear groups
8.2 Actions of GL
n
(C)
8.3 Orthogonal groups
8.4 Rotations and reflections in R
2
and R
3
8.5 Unitary groups
9 More on regular polyhedra
9.1 Symmetries of the cube
9.2 Symmetries of the tetrahedron
10 M¨obius group
10.1 M¨obius maps
10.2 Fixed points of M¨obius maps
10.3 Permutation properties of M¨obius maps
10.4 Cross-ratios
11 Projective line (non-examinable)
0 Introduction
Group theory is an example of algebra. In pure mathematics, algebra (usually)
does not refer to the boring mindless manipulation of symbols. Instead, in
algebra, we have some set of objects with some operations on them. For example,
we can take the integers with addition as the operation. However, in algebra, we
allow any set and any operations, not just numbers.
Of course, such a definition is too broad to be helpful. We categorize algebraic
structures into different types. In this course, we will study a particular kind of
structures, groups. In the IB Groups, Rings and Modules course, we will study
rings and modules as well.
These different kinds of structures are defined by certain axioms. The group
axioms will say that the operation must follow certain rules, and any set and
operation that satisfies these rules will be considered to form a group. We will
then have a different set of axioms for rings, modules etc.
As mentioned above, the most familiar kinds of algebraic structures are
number systems such as integers and rational numbers. The focus of group
theory, however, is not on things that resemble “numbers”. Instead, it is the
study of symmetries.
First of all, what is a symmetry? We are all familiar with, say, the symmetries
of an (equilateral) triangle (we will always assume the triangle is equilateral). We
rotate a triangle by 120
◦
, and we get the original triangle. We say that rotating
by 120
◦
is a symmetry of a triangle. In general, a symmetry is something we do
to an object that leaves the object intact.
Of course, we don’t require that the symmetry leaves everything intact.
Otherwise, we would only be allowed to do nothing. Instead, we require certain
important things to be intact. For example, when considering the symmetries
of a triangle, we only care about how the resultant object looks, but don’t care
about where the individual vertices went.
In the case of the triangle, we have six symmetries: three rotations (rotation
by 0
◦
, 120
◦
and 240
◦
), and three reflections along the axes below:
These six together form the underlying set of the group of symmetries. A more
sophisticated example is the symmetries of
R
3
. We define these as operations on
R
3
that leave distances between points unchanged. These include translations,
rotations, reflections, and combinations of these.
So what is the operation? This operation combines two symmetries to give a
new symmetry. The natural thing to do is to do the symmetry one after another.
For example, if we combine the two 120
◦
rotations, we get a 240
◦
rotation.
Now we are studying algebra, not geometry. So to define the group, we
abstract away the triangle. Instead, we define the group to be six objects, say
{e, r, r
2
, s, rs, r
2
s}
, with rules defining how we combine two elements to get a
third. Officially, we do not mention the triangle at all when defining the group.
We can now come up with the group axioms. What rules should the set of
symmetries obey? First of all, we must have a “do nothing” symmetry. We call
this the identity element. When we compose the identity with another symmetry,
the other symmetry is unchanged.
Secondly, given a symmetry, we can do the reverse symmetry. So for any
element, there is an inverse element that, when combined with the original, gives
the identity.
Finally, given three symmetries, we can combine them, one after another. If
we denote the operation of the group as
∗
, then if we have three symmetries,
x, y, z
, we should be able to form
x ∗ y ∗ z
. If we want to define it in terms of
the binary operation
∗
, we can define it as (
x ∗ y
)
∗ z
, where we first combine
the first two symmetries, then combine the result with the third. Alternatively,
we can also define it as
x ∗
(
y ∗ z
). Intuitively, these two should give the same
result, since both are applying
x
after
y
after
z
. Hence we have the third rule
x ∗ (y ∗ z) = (x ∗ y) ∗ z.
Now a group is any set with an operation that satisfies the three rules above.
In group theory, the objective is to study the properties of groups just assuming
these three axioms. It turns out that there is a lot we can talk about.
1 Groups and homomorphisms
1.1 Groups
Definition (Binary operation). A (binary) operation is a way of combining two
elements to get a new element. Formally, it is a map ∗ : A × A → A.
Definition (Group). A group is a set
G
with a binary operation
∗
satisfying
the following axioms:
1. There is some e ∈ G such that for all a, we have
a ∗ e = e ∗ a = a. (identity)
2. For all a ∈ G, there is some a
−1
∈ G such that
a ∗ a
−1
= a
−1
∗ a = e. (inverse)
3. For all a, b, c ∈ G, we have
(a ∗ b) ∗ c = a ∗ (b ∗ c). (associativity)
Definition (Order of group). The order of the group, denoted by
|G|
, is the
number of elements in G. A group is a finite group if the order is finite.
Note that technically, the inverse axiom makes no sense, since we have not
specified what
e
is. Even if we take it to be the
e
given by the identity axiom,
the identity axiom only states there is some
e
that satisfies that property, but
there could be many! We don’t know which one
a ∗ a
−1
is supposed to be equal
to! So we should technically take that to mean there is some
a
−1
such that
a ∗ a
−1
and
a
−1
∗ a
satisfy the identity axiom. Of course, we will soon show that
identities are indeed unique, and we will happily talk about “the” identity.
Some people put a zeroth axiom called “closure”:
0. For all a, b ∈ G, we have a ∗ b ∈ G. (closure)
Technically speaking, this axiom also makes no sense — when we say
∗
is a
binary operation, by definition,
a ∗ b
must be a member of
G
. However, in
practice, we often have to check that this axiom actually holds. For example, if
we let G be the set of all matrices of the form
1 x y
0 1 z
0 0 1
under matrix multiplication, we will have to check that the product of two such
matrices is indeed a matrix of this form. Officially, we are checking that the
binary operation is a well-defined operation on G.
It is important to know that it is generally not true that
a∗b
=
b∗a
. There is
no a priori reason why this should be true. For example, if we are considering the
symmetries of a triangle, rotating and then reflecting is different from reflecting
and then rotating.
However, for some groups, this happens to be true. We call such groups
abelian groups.
Definition (Abelian group). A group is abelian if it satisfies
4. (∀a, b ∈ G) a ∗ b = b ∗ a. (commutativity)
If it is clear from context, we are lazy and leave out the operation
∗
, and
write
a ∗ b
as
ab
. We also write
a
2
=
aa
,
a
n
=
aaa · · · a
| {z }
n copies
,
a
0
=
e
,
a
−n
= (
a
−1
)
n
etc.
Example. The following are abelian groups:
(i) Z with +
(ii) Q with +
(iii) Z
n
(integers mod n) with +
n
(iv) Q
∗
with ×
(v) {−1, 1} with ×
The following are non-abelian groups:
(vi)
Symmetries of an equilateral triangle (or any
n
-gon) with composition.
(D
2n
)
(vii) 2 × 2 invertible matrices with matrix multiplication (GL
2
(R))
(viii) Symmetry groups of 3D objects
Recall that the first group axiom requires that there exists an identity element,
which we shall call
e
. Then the second requires that for each
a
, there is an inverse
a
−1
such that
a
−1
a
=
e
. This only makes sense if there is only one identity
e
, or
else which identity should a
−1
a be equal to?
We shall now show that there can only be one identity. It turns out that the
inverses are also unique. So we will talk about the identity and the inverse.
Proposition. Let (G, ∗) be a group. Then
(i) The identity is unique.
(ii) Inverses are unique.
Proof.
(i)
Suppose
e
and
e
′
are identities. Then we have
ee
′
=
e
′
, treating
e
as an
inverse, and ee
′
= e, treating e
′
as an inverse. Thus e = e
′
.
(ii)
Suppose
a
−1
and
b
both satisfy the inverse axiom for some
a ∈ G
. Then
b = be = b(aa
−1
) = (ba)a
−1
= ea
−1
= a
−1
. Thus b = a
−1
.
Proposition. Let (G, ∗) be a group and a, b ∈ G. Then
(i) (a
−1
)
−1
= a
(ii) (ab)
−1
= b
−1
a
−1
Proof.
(i) Given a
−1
, both a and (a
−1
)
−1
satisfy
xa
−1
= a
−1
x = e.
By uniqueness of inverses, (a
−1
)
−1
= a.
(ii) We have
(ab)(b
−1
a
−1
) = a(bb
−1
)a
−1
= aea
−1
= aa
−1
= e
Similarly, (
b
−1
a
−1
)
ab
=
e
. So
b
−1
a
−1
is an inverse of
ab
. By the uniqueness
of inverses, (ab)
−1
= b
−1
a
−1
.
Sometimes if we have a group
G
, we might want to discard some of the
elements. For example if
G
is the group of all symmetries of a triangle, we might
one day decide that we hate reflections because they reverse orientation. So
we only pick the rotations in
G
and form a new, smaller group. We call this a
subgroup of G.
Definition (Subgroup). A
H
is a subgroup of
G
, written
H ≤ G
, if
H ⊆ G
and
H with the restricted operation ∗ from G is also a group.
Example.
– (Z, +) ≤ (Q, +) ≤ (R, +) ≤ (C, +)
– (e, ∗) ≤ (G, ∗) (trivial subgroup)
– G ≤ G
– ({±1}, ×) ≤ (Q
∗
, ×)
According to the definition, to prove that
H
is a subgroup of
G
, we need to
make sure
H
satisfies all group axioms. However, this is often tedious. Instead,
there are some simplified criteria to decide whether H is a subgroup.
Lemma (Subgroup criteria I). Let (G, ∗) be a group and H ⊆ G. H ≤ G iff
(i) e ∈ H
(ii) (∀a, b ∈ H) ab ∈ H
(iii) (∀a ∈ H) a
−1
∈ H
Proof. The group axioms are satisfied as follows:
0. Closure: (ii)
1.
Identity: (i). Note that
H
and
G
must have the same identity. Suppose that
e
H
and
e
G
are the identities of
H
and
G
respectively. Then
e
H
e
H
=
e
H
.
Now
e
H
has an inverse in
G
. Thus we have
e
H
e
H
e
−1
H
=
e
H
e
−1
H
. So
e
H
e
G
= e
G
. Thus e
H
= e
G
.
2. Inverse: (iii)
3. Associativity: inherited from G.
Humans are lazy, and the test above is still too complicated. We thus come
up with an even simpler test:
Lemma (Subgroup criteria II). A subset H ⊆ G is a subgroup of G iff:
(I) H is non-empty
(II) (∀a, b ∈ H) ab
−1
∈ H
Proof. (I) and (II) follow trivially from (i), (ii) and (iii).
To prove that (I) and (II) imply (i), (ii) and (iii), we have
(i) H must contain at least one element a. Then aa
−1
= e ∈ H.
(iii) ea
−1
= a
−1
∈ H.
(ii) a(b
−1
)
−1
= ab ∈ H.
Proposition. The subgroups of (
Z,
+) are exactly
nZ
, for
n ∈ N
(
nZ
is the
integer multiples of n).
Proof.
Firstly, it is trivial to show that for any
n ∈ N
,
nZ
is a subgroup. Now
show that any subgroup must be in the form nZ.
Let
H ≤ Z
. We know 0
∈ H
. If there are no other elements in
H
, then
H = 0Z. Otherwise, pick the smallest positive integer n in H. Then H = nZ.
Otherwise, suppose (
∃a ∈ H
)
n ∤ a
. Let
a
=
pn
+
q
, where 0
< q < n
. Since
a − pn ∈ H
,
q ∈ H
. Yet
q < n
but
n
is the smallest member of
H
. Contradiction.
So every
a ∈ H
is divisible by
n
. Also, by closure, all multiples of
n
must be in
H. So H = nZ.
1.2 Homomorphisms
It is often helpful to study functions between different groups. First, we need to
define what a function is. These definitions should be familiar from IA Numbers
and Sets.
Definition (Function). Given two sets
X
,
Y
, a function
f
:
X → Y
sends each
x ∈ X
to a particular
f
(
x
)
∈ Y
.
X
is called the domain and
Y
is the co-domain.
Example.
–
Identity function: for any set
X
, 1
X
:
X → X
with 1
X
(
x
) =
x
is a function.
This is also written as id
X
.
–
Inclusion map:
ι
:
Z → Q
:
ι
(
n
) =
n
. Note that this differs from the
identity function as the domain and codomain are different in the inclusion
map.
– f
1
: Z → Z: f
1
(x) = x + 1.
– f
2
: Z → Z: f
2
(x) = 2x.
– f
3
: Z → Z: f
3
(x) = x
2
.
– For g : {0, 1, 2, 3, 4} → {0, 1, 2, 3, 4}, we have:
◦ g
1
(x) = x + 1 if x < 4; g
1
(4) = 4.
◦ g
2
(x) = x + 1 if x < 4; g
1
(4) = 0.
Definition (Composition of functions). The composition of two functions is a
function you get by applying one after another. In particular, if
f
:
X → Y
and
G : Y → Z, then g ◦ f : X → Z with g ◦ f(x) = g(f(x)).
Example.
f
2
◦ f
1
(
x
) = 2
x
+ 2.
f
1
◦ f
2
(
x
) = 2
x
+ 1. Note that function
composition is not commutative.
Definition (Injective functions). A function
f
is injective if it hits everything
at most once, i.e.
(∀x, y ∈ X) f(x) = f (y) ⇒ x = y.
Definition (Surjective functions). A function is surjective if it hits everything
at least once, i.e.
(∀y ∈ Y )(∃x ∈ X) f(x) = y.
Definition (Bijective functions). A function is bijective if it is both injective
and surjective. i.e. it hits everything exactly once. Note that a function has an
inverse iff it is bijective.
Example.
ι
and
f
2
are injective but not surjective.
f
3
and
g
1
are neither. 1
X
,
f
1
and g
2
are bijective.
Lemma. The composition of two bijective functions is bijective
When considering sets, functions are allowed to do all sorts of crazy things,
and can send any element to any element without any restrictions. However, we
are currently studying groups, and groups have additional structure on top of
the set of elements. Hence we are not interested in arbitrary functions. Instead,
we are interested in functions that “respect” the group structure. We call these
homomorphisms.
Definition (Group homomorphism). Let (
G, ∗
) and (
H, ×
) be groups. A
function f : G → H is a group homomorphism iff
(∀g
1
, g
2
∈ G) f(g
1
) × f(g
2
) = f(g
1
∗ g
2
),
Definition (Group isomorphism). Isomorphisms are bijective homomorphisms.
Two groups are isomorphic if there exists an isomorphism between them. We
write G
∼
=
H.
We will consider two isomorphic groups to be “the same”. For example, when
we say that there is only one group of order 2, it means that any two groups of
order 2 must be isomorphic.
Example.
– f
:
G → H
defined by
f
(
g
) =
e
, where
e
is the identity of
H
, is a
homomorphism.
–
1
G
:
G → G
and
f
2
:
Z →
2
Z
are isomorphisms.
ι
:
Z → Q
and
f
2
:
Z → Z
are homomorphisms.
– exp : (R, +) → (R
+
, ×) with exp(x) = e
x
is an isomorphism.
–
Take (
Z
4
,
+) and
H
: (
{e
ikπ/2
:
k
= 0
,
1
,
2
,
3
}, ×
). Then
f
:
Z
4
→ H
by
f(a) = e
iπa/2
is an isomorphism.
– f
:
GL
2
(
R
)
→ R
∗
with
f
(
A
) =
det
(
A
) is a homomorphism, where
GL
2
(
R
)
is the set of 2 × 2 invertible matrices.
Proposition. Suppose that f : G → H is a homomorphism. Then
(i) Homomorphisms send the identity to the identity, i.e.
f(e
G
) = e
H
(ii) Homomorphisms send inverses to inverses, i.e.
f(a
−1
) = f(a)
−1
(iii) The composite of 2 group homomorphisms is a group homomorphism.
(iv) The inverse of an isomorphism is an isomorphism.
Proof.
(i)
f(e
G
) = f(e
2
G
) = f(e
G
)
2
f(e
G
)
−1
f(e
G
) = f(e
G
)
−1
f(e
G
)
2
f(e
G
) = e
H
(ii)
e
H
= f(e
G
)
= f(aa
−1
)
= f(a)f(a
−1
)
Since inverses are unique, f(a
−1
) = f(a)
−1
.
(iii)
Let
f
:
G
1
→ G
2
and
g
:
G
2
→ G
3
. Then
g
(
f
(
ab
)) =
g
(
f
(
a
)
f
(
b
)) =
g(f(a))g(f(b)).
(iv) Let f : G → H be an isomorphism. Then
f
−1
(ab) = f
−1
n
f
f
−1
(a)
f
f
−1
(b)
o
= f
−1
n
f
f
−1
(a)f
−1
(b)
o
= f
−1
(a)f
−1
(b)
So
f
−1
is a homomorphism. Since it is bijective,
f
−1
is an isomorphism.
Definition (Image of homomorphism). If
f
:
G → H
is a homomorphism, then
the image of f is
im f = f(G) = {f (g) : g ∈ G}.
Definition (Kernel of homomorphism). The kernel of f, written as
ker f = f
−1
({e
H
}) = {g ∈ G : f(g) = e
H
}.
Proposition. Both the image and the kernel are subgroups of the respective
groups, i.e. im f ≤ H and ker f ≤ G.
Proof.
Since
e
H
∈ im f
and
e
G
∈ ker f
,
im f
and
ker f
are non-empty. Moreover,
suppose
b
1
, b
2
∈ im f
. Now
∃a
1
, a
2
∈ G
such that
f
(
a
i
) =
b
i
. Then
b
1
b
−1
2
=
f(a
1
)f(a
−1
2
) = f(a
1
a
−1
2
) ∈ im f.
Then consider
b
1
, b
2
∈ ker f
. We have
f
(
b
1
b
−1
2
) =
f
(
b
1
)
f
(
b
2
)
−1
=
e
2
=
e
. So
b
1
b
−1
2
∈ ker f.
Proposition. Given any homomorphism
f
:
G → H
and any
a ∈ G
, for all
k ∈ ker f, aka
−1
∈ ker f.
This proposition seems rather pointless. However, it is not. All subgroups
that satisfy this property are known as normal subgroups, and normal subgroups
have very important properties. We will postpone the discussion of normal
subgroups to later lectures.
Proof. f(aka
−1
) = f(a)f(k)f(a)
−1
= f(a)ef(a)
−1
= e. So aka
−1
∈ ker f.
Example. Images and kernels for previously defined functions:
(i) For the function that sends everything to e, im f = {e} and ker f = G.
(ii) For the identity function, im 1
G
= G and ker 1
G
= {e}.
(iii) For the inclusion map ι : Z → Q, we have im ι = Z and ker ι = {0}
(iv) For f
2
: Z → Z and f
2
(x) = 2x, we have im f
2
= 2Z and ker f
2
= {0}.
(v)
For
det
:
GL
2
(
R
)
→ R
∗
, we have
im det
=
R
∗
and
ker det
=
{A
:
det A
=
1} = SL
2
(R)
Proposition. For all homomorphisms f : G → H, f is
(i) surjective iff im f = H
(ii) injective iff ker f = {e}
Proof.
(i) By definition.
(ii)
We know that
f
(
e
) =
e
. So if
f
is injective, then by definition
ker f
=
{e}
. If
ker f
=
{e}
, then given
a, b
such that
f
(
a
) =
f
(
b
),
f
(
ab
−1
) =
f(a)f(b)
−1
= e. Thus ab
−1
∈ ker f = {e}. Then ab
−1
= e and a = b.
So far, the definitions of images and kernels seem to be just convenient
terminology to refer to things. However, we will later prove an important
theorem, the first isomorphism theorem, that relates these two objects and
provides deep insights (hopefully).
Before we get to that, we will first study some interesting classes of groups
and develop some necessary theory.
1.3 Cyclic groups
The simplest class of groups is cyclic groups. A cyclic group is a group of the
form
{e, a, a
2
, a
2
, · · · , a
n−1
}
, where
a
n
=
e
. For example, if we consider the
group of all rotations of a triangle, and write
r
= rotation by 120
◦
, the elements
will be {e, r, r
2
} with r
3
= e.
Officially, we define a cyclic group as follows:
Definition (Cyclic group C
n
). A group G is cyclic if
(∃a)(∀b)(∃n ∈ Z) b = a
n
,
i.e. every element is some power of a. Such an a is called a generator of G.
We write C
n
for the cyclic group of order n.
Example.
(i) Z is cyclic with generator 1 or −1. It is the infinite cyclic group.
(ii) ({+1, −1}, ×) is cyclic with generator −1.
(iii) (Z
n
, +) is cyclic with all numbers coprime with n as generators.
Notation. Given a group
G
and
a ∈ G
, we write
⟨a⟩
for the cyclic group
generated by
a
, i.e. the subgroup of all powers of
a
. It is the smallest subgroup
containing a.
Definition (Order of element). The order of an element
a
is the smallest integer
n
such that
a
n
=
e
. If
n
doesn’t exist,
a
has infinite order. Write
ord
(
a
) for the
order of a.
We have given two different meanings to the word “order”. One is the order
of a group and the other is the order of an element. Since mathematicians
are usually (but not always) sensible, the name wouldn’t be used twice if they
weren’t related. In fact, we have
Lemma. For a in g, ord(a) = |⟨a⟩|.
Proof.
If
ord
(
a
) =
∞
,
a
n
=
a
m
for all
n
=
m
. Otherwise
a
m−n
=
e
. Thus
|⟨a⟩| = ∞ = ord(a).
Otherwise, suppose
ord
(
a
) =
k
. Thus
a
k
=
e
. We now claim that
⟨a⟩
=
{e, a
1
, a
2
, · · · a
k−1
}
. Note that
⟨a⟩
does not contain higher powers of
a
as
a
k
=
e
and higher powers will loop back to existing elements. There are also no repeating
elements in the list provided since a
m
= a
n
⇒ a
m−n
= e. So done.
It is trivial to show that
Proposition. Cyclic groups are abelian.
Definition (Exponent of group). The exponent of a group
G
is the smallest
integer n such that a
n
= e for all a ∈ G.
1.4 Dihedral groups
Definition (Dihedral groups
D
2n
). Dihedral groups are the symmetries of a
regular
n
-gon. It contains
n
rotations (including the identity symmetry, i.e.
rotation by 0
◦
) and n reflections.
We write the group as
D
2n
. Note that the subscript refers to the order of
the group, not the number of sides of the polygon.
The dihedral group is not hard to define. However, we need to come up with
a presentation of D
2n
that is easy to work with.
We first look at the rotations. The set of all rotations is generated by
r
=
360
◦
n
.
This r has order n.
How about the reflections? We know that each reflection has order 2. Let
s
be our favorite reflection. Then using some geometric arguments, we can show
that any reflection can be written as a product of
r
m
and
s
for some
m
. We also
have srs = r
−1
.
Hence we can define
D
2n
as follows:
D
2n
is a group generated by
r
and
s
,
and every element can be written as a product of
r
’s and
s
’s. Whenever we see
r
n
and s
2
, we replace it by e. When we see srs, we replace it by r
−1
.
It then follows that every element can be written in the form r
m
s.
Formally, we can write D
2n
as follows:
D
2n
= ⟨r, s | r
n
= s
2
= e, srs
−1
= r
−1
⟩
= {e, r, r
2
, · · · r
n−1
, s, rs, r
2
s, · · · r
n−1
s}
This is a notation we will commonly use to represent groups. For example, a
cyclic group of order n can be written as
C
n
= ⟨a | a
n
= e⟩.
1.5 Direct products of groups
Recall that if we have to sets
X, Y
, then we can obtain the product
X × Y
=
{(x, y) : x ∈ X, y ∈ Y }. We can do the same if X and Y are groups.
Definition (Direct product of groups). Given two groups (
G, ◦
) and (
H, •
),
we can define a set
G × H
=
{
(
g, h
) :
g ∈ G, h ∈ H}
and an operation
(a
1
, a
2
) ∗ (b
1
, b
2
) = (a
1
◦ b
1
, a
2
• b
2
). This forms a group.
Why would we want to take the product of two groups? Suppose we have
two independent triangles. Then the symmetries of this system include, say
rotating the first triangle, rotating the second, or rotating both. The symmetry
group of this combined system would then be D
6
× D
6
.
Example.
C
2
× C
2
= {(0, 0), (0, 1), (1, 0), (1, 1)}
= {e, x, y, xy} with everything order 2
= ⟨x, y | x
2
= y
2
= e, xy = yx⟩
Proposition. C
n
× C
m
∼
=
C
nm
iff hcf(m, n) = 1.
Proof.
Suppose that
hcf
(
m, n
) = 1. Let
C
n
=
⟨a⟩
and
C
m
=
⟨b⟩
. Let
k
be the
order of (
a, b
). Then (
a, b
)
k
= (
a
k
, b
k
) =
e
. This is possible only if
n | k
and
m | k
, i.e.
k
is a common multiple
n
and
m
. Since the order is the minimum
value of k that satisfies the above equation, k = lcm(n, m) =
nm
hcf(n,m)
= nm.
Now consider
⟨
(
a, b
)
⟩ ≤ C
n
× C
m
. Since (
a, b
) has order
nm
,
⟨
(
a, b
)
⟩
has
nm
elements. Since
C
n
× C
m
also has
nm
elements,
⟨
(
a, b
)
⟩
must be the whole of
C
n
× C
m
. And we know that ⟨(a, b)⟩
∼
=
C
nm
. So C
n
× C
m
∼
=
C
nm
.
On the other hand, suppose
hcf
(
m, n
)
= 1. Then
k
=
lcm
(
m, n
)
=
mn
. Then
for any (
a, b
)
∈ C
n
× C
m
,we have (
a, b
)
k
= (
a
k
, b
k
) =
e
. So the order of any (
a, b
)
is at most
k < mn
. So there is no element of order
mn
. So
C
n
× C
m
is not a
cyclic group of order nm.
Given a complicated group
G
, it is sometimes helpful to write it as a product
H × K
, which could make things a bit simpler. We can do so by the following
theorem:
Proposition (Direct product theorem). Let
H
1
, H
2
≤ G
. Suppose the following
are true:
(i) H
1
∩ H
2
= {e}.
(ii) (∀a
i
∈ H
i
) a
1
a
2
= a
2
a
1
.
(iii) (∀a ∈ G)(∃a
i
∈ H
i
) a = a
1
a
2
. We also write this as G = H
1
H
2
.
Then G
∼
=
H
1
× H
2
.
Proof.
Define
f
:
H
1
×H
2
→ G
by
f
(
a
1
, a
2
) =
a
1
a
2
. Then it is a homomorphism
since
f((a
1
, a
2
) ∗ (b
1
, b
2
)) = f(a
1
b
1
, a
2
b
2
)
= a
1
b
1
a
2
b
2
= a
1
a
2
b
1
b
2
= f(a
1
, a
2
)f(b
1
, b
2
).
Surjectivity follows from (iii). We’ll show injectivity by showing that the kernel
is
{e}
. If
f
(
a
1
, a
2
) =
e
, then we know that
a
1
a
2
=
e
. Then
a
1
=
a
−1
2
. Since
a
1
∈ H
1
and
a
−1
2
∈ H
2
, we have
a
1
=
a
−1
2
∈ H
1
∩ H
2
=
{e}
. Thus
a
1
=
a
2
=
e
and ker f = {e}.
2 Symmetric group I
We will devote two full chapters to the study of symmetric groups, because
it is really important. Recall that we defined a symmetry to be an operation
that leaves some important property of the object intact. We can treat each
such operation as a bijection. For example, a symmetry of
R
2
is a bijection
f
:
R
2
→ R
2
that preserves distances. Note that we must require it to be a
bijection, instead of a mere function, since we require each symmetry to be an
inverse.
We can consider the case where we don’t care about anything at all. So a
“symmetry” would be any arbitrary bijection
X → X
, and the set of all bijections
will form a group, known as the symmetric group. Of course, we will no longer
think of these as “symmetries” anymore, but just bijections.
In some sense, the symmetric group is the most general case of a symmetry
group. In fact, we will later (in Chapter 5) show that every group can be written
as a subgroup of some symmetric group.
2.1 Symmetric groups
Definition (Permutation). A permutation of
X
is a bijection from a set
X
to
X itself. The set of all permutations on X is Sym X.
When composing permutations, we treat them as functions. So if
σ
and
ρ
are permutations, σ ◦ ρ is given by first applying ρ, then applying σ.
Theorem. Sym X with composition forms a group.
Proof. The groups axioms are satisfied as follows:
0.
If
σ
:
X → X
and
τ
:
X → X
, then
σ ◦ τ
:
X → X
. If they are both
bijections, then the composite is also bijective. So if
σ, τ ∈ Sym X
, then
σ ◦ τ ∈ Sym X.
1. The identity 1
X
: X → X is clearly a permutation, and gives the identity
of the group.
2.
Every bijective function has a bijective inverse. So if
σ ∈ Sym X
, then
σ
−1
∈ Sym X.
3. Composition of functions is associative.
Definition (Symmetric group
S
n
). If
X
is finite, say
|X|
=
n
(usually use
X
=
{
1
,
2
, · · · , n}
), we write
Sym X
=
S
n
. The is the symmetric group of degree
n.
It is important to note that the degree of the symmetric group is different
from the order of the symmetric group. For example,
S
3
has degree 3 but order
6. In general, the order of S
n
is n!.
There are two ways to write out an element of the symmetric group. The
first is the two row notation.
Notation. (Two row notation) We write 1
,
2
,
3
, · · · n
on the top line and their
images below, e.g.
1 2 3
2 3 1
∈ S
3
and
1 2 3 4 5
2 1 3 4 5
∈ S
5
In general, if σ : X → X, we write
1 2 3 · · · n
σ(1) σ(2) σ(3) · · · σ(n)
Example. For small n, we have
(i) When n = 1, S
n
=
1
1
= {e}
∼
=
C
1
.
(ii) When n = 2, S
n
=
1 2
1 2
,
1 2
2 1
∼
=
C
2
(iii) When n = 3,
S
n
=
1 2 3
1 2 3
,
1 2 3
2 3 1
,
1 2 3
3 1 2
1 2 3
2 1 3
,
1 2 3
3 2 1
,
1 2 3
1 3 2
∼
=
D
6
.
Note that
S
3
is not abelian. Thus
S
n
is not abelian for
n ≥
3 since we can
always view S
3
as a subgroup of S
n
by fixing 4, 5, 6, · · · n.
In general, we can view
D
2n
as a subgroup of
S
n
because each symmetry is
a permutation of the corners.
While the two row notation is fully general and can represent any (finite)
permutation, it is clumsy to write and wastes a lot of space. It is also very
annoying to type using L
A
T
E
X. Hence, most of the time, we actually use the
cycle notation.
Notation (Cycle notation). If a map sends 1
7→
2, 2
7→
3, 3
7→
1, then we
write it as a cycle (1 2 3). Alternatively, we can write (2 3 1) or (3 1 2), but by
convention, we usually write the smallest number first. We leave out numbers
that don’t move. So we write (1 2) instead of (1 2)(3).
For more complicated maps, we can write them as products of cycles. For
example, in S
4
, we can have things like (1 2)(3 4).
The order of each cycle is the length of the cycle, and the inverse is the cycle
written the other way round, e.g. (1 2 3)
−1
= (3 2 1) = (1 3 2).
Example.
(i)
Suppose we want to simplify (1 2 3)(1 2). Recall that composition is from
right to left. So 1 gets mapped to 3 ((1 2) maps 1 to 2, and (1 2 3) further
maps it to 3). Then 3 gets mapped to 1. 2 is mapped to 2 itself. So
(1 2 3)(1 2) = (1 3)(2)
(ii) (1 2 3 4)(1 4) = (1)(2 3 4) = (2 3 4).
Definition (
k
-cycles and transpositions). We call (
a
1
a
2
a
3
· · · a
k
) a
k
-cycle.
2-cycles are called transpositions. Two cycles are disjoint if no number appears
in both cycles.
Example. (1 2) and (3 4) are disjoint but (1 2 3) and (1 2) are not.
Lemma. Disjoint cycles commute.
Proof.
If
σ, τ ∈ S
n
are disjoint cycles. Consider any
n
. Show that:
σ
(
τ
(
a
)) =
τ
(
σ
(
a
)). If
a
is in neither of
σ
and
τ
, then
σ
(
τ
(
a
)) =
τ
(
σ
(
a
)) =
a
. Otherwise,
wlog assume that
a
is in
τ
but not in
σ
. Then
τ
(
a
)
∈ τ
and thus
τ
(
a
)
∈ σ
. Thus
σ
(
a
) =
a
and
σ
(
τ
(
a
)) =
τ
(
a
). Therefore we have
σ
(
τ
(
a
)) =
τ
(
σ
(
a
)) =
τ
(
a
).
Therefore τ and σ commute.
In general, non-disjoint cycles may not commute. For example, (1 3)(2 3) =
(1 3 2) while (2 3)(1 3) = (1 2 3).
Theorem. Any permutation in
S
n
can be written (essentially) uniquely as a
product of disjoint cycles. (Essentially unique means unique up to re-ordering of
cycles and rotation within cycles, e.g. (1 2) and (2 1))
Proof.
Let
σ ∈ S
n
. Start with (1
σ
(1)
σ
2
(1)
σ
3
(1)
· · ·
). As the set
{
1
,
2
,
3
· · · n}
is finite, for some
k
, we must have
σ
k
(1) already in the list. If
σ
k
(1) =
σ
l
(1),
with
l < k
, then
σ
k−l
(1) = 1. So all
σ
i
(1) are distinct until we get back to 1.
Thus we have the first cycle (1 σ(1) σ
2
(1) σ
3
(1) · · · σ
k−1
(1)).
Now choose the smallest number that is not yet in a cycle, say
j
. Repeat to
obtain a cycle (
j σ
(
j
)
σ
2
(
j
)
· · · σ
l−1
(
j
)). Since
σ
is a bijection, nothing in this
cycle can be in previous cycles as well.
Repeat until all
{
1
,
2
,
3
· · · n}
are exhausted. This is essentially unique
because every number
j
completely determines the whole cycle it belongs to,
and whichever number we start with, we’ll end up with the same cycle.
Definition (Cycle type). Write a permutation
σ ∈ S
n
in disjoint cycle notation.
The cycle type is the list of cycle lengths. This is unique up to re-ordering. We
often (but not always) leave out singleton cycles.
Example. (1 2) has cycle type 2 (transposition). (1 2)(3 4) has cycle type 2, 2
(double transposition). (1 2 3)(4 5) has cycle type 3, 2.
Lemma. For
σ ∈ S
n
, the order of
σ
is the least common multiple of cycle
lengths in the disjoint cycle notation. In particular, a k-cycle has order k.
Proof.
As disjoint cycles commute, we can group together each cycle when we take
powers. i.e. if σ = τ
1
τ
2
· · · τ
l
with τ
i
all disjoint cycles, then σ
m
= τ
m
1
τ
m
2
· · · τ
m
l
.
Now if cycle
τ
i
has length
k
i
, then
τ
k
i
i
=
e
, and
τ
m
i
=
e
iff
k
i
| m
. To get an
m
such that
σ
m
=
e
, we need all
k
i
to divide
m
. i.e.
m
is a common multiple of
k
i
. Since the order is the least possible
m
such that
σ
m
=
e
, the order is the
least common multiple of k
i
.
Example. Any transpositions and double transpositions have order 2.
(1 2 3)(4 5) has order 6.
2.2 Sign of permutations
To classify different permutations, we can group different permutations according
to their cycle type. While this is a very useful thing to do, it is a rather fine
division. In this section, we will assign a “sign” to each permutation, and each
permutation can either be odd or even. This high-level classification allows us
to separate permutations into two sets, which is also a useful notion.
To define the sign, we first need to write permutations as products of
transpositions.
Proposition. Every permutation is a product of transpositions.
This is not a deep or mysterious fact. All it says is that you can rearrange
things however you want just by swapping two objects at a time.
Proof.
As each permutation is a product of disjoint cycles, it suffices to prove
that each cycle is a product of transpositions. Consider a cycle (
a
1
a
2
a
3
· · · a
k
).
This is in fact equal to (
a
1
a
2
)(
a
2
a
3
)
· · ·
(
a
k−1
a
k
). Thus a
k
-cycle can be written
as a product of k − 1 transpositions.
Note that the product is not unique. For example,
(1 2 3 4 5) = (1 2)(2 3)(3 4)(4 5) = (1 2)(2 3)(1 2)(3 4)(1 2)(4 5).
However, the number of terms in the product, mod 2, is always the same.
Theorem. Writing
σ ∈ S
n
as a product of transpositions in different ways,
σ
is
either always composed of an even number of transpositions, or always an odd
number of transpositions.
The proof is rather magical.
Proof.
Write #(
σ
) for the number of cycles in disjoint cycle notation, including
singleton cycles. So #(
e
) =
n
and #((1 2)) =
n −
1. When we multiply
σ
by a
transposition τ = (c d) (wlog assume c < d),
–
If
c, d
are in the same
σ
-cycle, say, (
c a
2
· · · a
k−1
d a
k+1
· · · a
k+l
)(
c d
) =
(c a
k+1
a
k+2
· · · a
k+l
)(d a
2
a
3
· · · a
k−1
). So #(στ) = #(σ) + 1 .
– If c, d are in different σ-cycles, say
(d a
2
a
3
· · · a
k−1
)(c a
k+1
a
k+2
· · · a
k+l
)(c d)
= (c a
2
· · · a
k−1
d a
k+1
· · · a
k+l
)(c d)(c d)
= (c a
2
· · · a
k−1
d a
k+1
· · · a
k+l
) and #(στ) = #(σ) − 1.
Therefore for any transposition τ, #(στ ) ≡ #(σ) + 1 (mod 2).
Now suppose
σ
=
τ
1
· · · τ
l
=
τ
′
1
· · · τ
′
k
. Since disjoint cycle notation is unique,
#(σ) is uniquely determined by σ.
Now we can construct
σ
by starting with
e
and multiplying the transpositions
one by one. Each time we add a transposition, we increase #(
σ
) by 1 (
mod
2).
So #(
σ
)
≡
#(
e
) +
l
(
mod
2). Similarly, #(
σ
)
≡
#(
e
) +
k
(
mod
2). So
l ≡ k
(mod 2).
Definition (Sign of permutation). Viewing
σ ∈ S
n
as a product of transpositions,
σ
=
τ
1
· · · τ
l
, we call
sgn
(
σ
) = (
−
1)
l
. If
sgn
(
σ
) = 1, we call
σ
an even permutation.
If sgn(σ) = −1, we call σ an odd permutation.
While
l
itself is not well-defined, it is either always odd or always even, and
(−1)
l
is well-defined.
Theorem. For n ≥ 2, sgn : S
n
→ {±1} is a surjective group homomorphism.
Proof.
Suppose
σ
1
=
τ
1
· · · τ
l
1
and
σ
2
=
τ
′
1
· · · τ
l
2
. Then
sgn
(
σ
1
σ
2
) = (
−
1)
l
1
+l
2
=
(−1)
l
1
(−1)
l
2
= sgn(σ
1
) sgn(σ
2
). So it is a homomorphism.
It is surjective since sgn(e) = 1 and sgn((1 2)) = −1.
It is this was rather trivial to prove. The hard bit is showing that
sgn
is
well defined. If a question asks you to show that
sgn
is a well-defined group
homomorphism, you have to show that it is well-defined.
Lemma.
σ
is an even permutation iff the number of cycles of even length is
even.
Proof.
A
k
-cycle can be written as
k −
1 transpositions. Thus an even-length
cycle is odd, vice versa.
Since
sgn
is a group homomorphism, writing
σ
in disjoint cycle notation,
σ
=
σ
1
σ
2
· · · σ
l
, we get
sgn
(
σ
) =
sgn
(
σ
1
)
· · · sgn
(
σ
l
). Suppose there are
m
even-
length cycles and
n
odd-length cycles, then
sgn
(
σ
) = (
−
1)
m
1
n
. This is equal to
1 iff (−1)
m
= 1, i.e. m is even.
Rather confusingly, odd length cycles are even, and even length cycles are
odd.
Definition (Alternating group
A
n
). The alternating group
A
n
is the kernel of
sgn
, i.e. the even permutations. Since
A
n
is a kernel of a group homomorphism,
A
n
≤ S
n
.
Among the many uses of the
sgn
homomorphism, it is used in the definition
of the determinant of a matrix: if A
n×n
is a square matrix, then
det A =
X
σ∈S
n
sgn(σ)a
1σ(1)
· · · a
nσ(n)
.
Proposition. Any subgroup of
S
n
contains either no odd permutations or
exactly half.
Proof.
If
S
n
has at least one odd permutation
τ
, then there exists a bijection
between the odd and even permutations by
σ 7→ στ
(bijection since
σ 7→ στ
−1
is a well-defined inverse). So there are as many odd permutations as even
permutations.
After we prove the isomorphism theorem later, we can provide an even shorter
proof of this.
3 Lagrange’s Theorem
One can model a Rubik’s cube with a group, with each possible move correspond-
ing to a group element. Of course, Rubik’s cubes of different sizes correspond to
different groups.
Suppose I have a 4
×
4
×
4 Rubik’s cube, but I want to practice solving a
2
×
2
×
2 Rubik’s cube. It is easy. I just have to make sure every time I make a
move, I move two layers together. Then I can pretend I am solving a 2
×
2
×
2
cube. This corresponds to picking a particular subgroup of the 4
×
4
×
4 group.
Now what if I have a 3
×
3
×
3 cube? I can still practice solving a 2
×
2
×
2
one. This time, I just look at the corners and pretend that the edges and centers
do not exist. Then I am satisfied when the corners are in the right positions,
while the centers and edges can be completely scrambled. In this case, we are
not taking a subgroup. Instead, we are identifying certain moves together. In
particular, we are treating two moves as the same as long as their difference is
confined to the centers and edges.
Let
G
be the 3
×
3
×
3 cube group, and
H
be the subgroup of
G
that only
permutes the edges and centers. Then for any
a, b ∈ G
, we think
a
and
b
are “the
same” if
a
−1
b ∈ H
. Then the set of things equivalent to
a
is
aH
=
{ah
:
h ∈ H}
.
We call this a coset, and the set of cosets form a group.
An immediate question one can ask is: why not
Ha
=
{ha
:
h ∈ H}
? In
this particular case, the two happen to be the same for all possible
a
. However,
for a general subgroup
H
, they need not be. We can still define the coset
aH
=
{ah
:
h ∈ H}
, but these are less interesting. For example, the set of all
{aH}
will no longer form a group. We will look into these more in-depth in the
next chapter. In this chapter, we will first look at results for general cosets. In
particular, we will, step by step, prove the things we casually claimed above.
Definition (Cosets). Let
H ≤ G
and
a ∈ G
. Then the set
aH
=
{ah
:
h ∈ H}
is a left coset of H and Ha = {ha : h ∈ H} is a right coset of H.
Example.
(i)
Take 2
Z ≤ Z
. Then 6 + 2
Z
=
{all even numbers}
= 0 + 2
Z
. 1 + 2
Z
=
{all odd numbers} = 17 + 2Z.
(ii) Take G = S
3
, let H = ⟨(1 2)⟩ = {e, (1 2)}. The left cosets are
eH = (1 2)H = {e, (1 2)}
(1 3)H = (1 2 3)H = {(1 3), (1 2 3)}
(2 3)H = (1 3 2)H = {(2 3), (1 3 2)}
(iii)
Take
G
=
D
6
(which is isomorphic to
S
3
). Recall
D
6
=
⟨r, s | r
3
e
=
s
2
, rs
=
sr
−1
⟩
.Take
H
=
⟨s⟩
=
{e, s}
. We have left coset
rH
=
{r, rs
=
sr
−1
}
and
the right coset Hr = {r, sr}. Thus rH = Hr.
Proposition. aH = bH ⇔ b
−1
a ∈ H.
Proof.
(
⇒
) Since
a ∈ aH
,
a ∈ bH
. Then
a
=
bh
for some
h ∈ H
. So
b
−1
a
=
h ∈ H.
(
⇐
). Let
b
−1
a
=
h
0
. Then
a
=
bh
0
. Then
∀ah ∈ aH
, we have
ah
=
b
(
h
0
h
)
∈
bH. So aH ⊆ bH. Similarly, bH ⊆ aH. So aH = bH.
Definition (Partition). Let
X
be a set, and
X
1
, · · · X
n
be subsets of
X
. The
X
i
are called a partition of
X
if
S
X
i
=
X
and
X
i
∩ X
j
=
∅
for
i
=
j
. i.e. every
element is in exactly one of X
i
.
Lemma. The left cosets of a subgroup
H ≤ G
partition
G
, and every coset has
the same size.
Proof.
For each
a ∈ G
,
a ∈ aH
. Thus the union of all cosets gives all of
G
. Now
we have to show that for all
a, b ∈ G
, the cosets
aH
and
bH
are either the same
or disjoint.
Suppose that
aH
and
bH
are not disjoint. Let
ah
1
=
bh
2
∈ aH ∩ bH
. Then
b
−1
a = h
2
h
−1
1
∈ H. So aH = bH.
To show that they each coset has the same size, note that
f
:
H → aH
with
f
(
h
) =
ah
is invertible with inverse
f
−1
(
h
) =
a
−1
h
. Thus there exists a bijection
between them and they have the same size.
Definition (Index of a subgroup). The index of
H
in
G
, written
|G
:
H|
, is the
number of left cosets of H in G.
Theorem (Lagrange’s theorem). If
G
is a finite group and
H
is a subgroup of
G, then |H| divides |G|. In particular,
|H||G : H| = |G|.
Note that the converse is not true. If
k
divides
|G|
, there is not necessarily a
subgroup of order
k
, e.g.
|A
4
|
= 12 but there is no subgroup of order 6. However,
we will later see that this is true if k is a prime (cf. Cauchy’s theorem).
Proof.
Suppose that there are
|G
:
H|
left cosets in total. Since the left cosets
partition G, and each coset has size |H|, we have
|H||G : H| = |G|.
Again, the hard part of this proof is to prove that the left cosets partition
G
and have the same size. If you are asked to prove Lagrange’s theorem in exams,
that is what you actually have to prove.
Corollary. The order of an element divides the order of the group, i.e. for any
finite group G and a ∈ G, ord(a) divides |G|.
Proof.
Consider the subgroup generated by
a
, which has order
ord
(
a
). Then by
Lagrange’s theorem, ord(a) divides |G|.
Corollary. The exponent of a group divides the order of the group, i.e. for any
finite group G and a ∈ G, a
|G|
= e.
Proof.
We know that
|G|
=
k ord
(
a
) for some
k ∈ N
. Then
a
|G|
= (
a
ord(a)
)
k
=
e
k
= e.
Corollary. Groups of prime order are cyclic and are generated by every non-
identity element.
Proof.
Say
|G|
=
p
. If
a ∈ G
is not the identity, the subgroup generated by
a
must have order
p
since it has to divide
p
. Thus the subgroup generated by
a
has the same size as
G
and they must be equal. Then
G
must be cyclic since it
is equal to the subgroup generated by a.
A useful way to think about cosets is to view them as equivalence classes.
To do so, we need to first define what an equivalence class is.
Definition (Equivalence relation). An equivalence relation
∼
is a relation that
is reflexive, symmetric and transitive. i.e.
(i) (∀x) x ∼ x (reflexivity)
(ii) (∀x, y) x ∼ y ⇒ y ∼ x (symmetry)
(iii) (∀x, y, z) [(x ∼ y) ∧ (y ∼ z) ⇒ x ∼ z] (transitivity)
Example. The following relations are equivalence relations:
(i) Consider Z. The relation ≡
n
defined as a ≡
n
b ⇔ n | (a − b).
(ii)
Consider the set (formally: class) of all finite groups. Then “is isomorphic
to” is an equivalence relation.
Definition (Equivalence class). Given an equivalence relation
∼
on
A
, the
equivalence class of a is
[a]
∼
= [a] = {b ∈ A : a ∼ b}
Proposition. The equivalence classes form a partition of A.
Proof.
By reflexivity, we have
a ∈
[
a
]. Thus the equivalence classes cover the
whole set. We must now show that for all
a, b ∈ A
, either [
a
] = [
b
] or [
a
]
∩
[
b
] =
∅
.
Suppose [
a
]
∩
[
b
]
=
∅
. Then
∃c ∈
[
a
]
∩
[
b
]. So
a ∼ c, b ∼ c
. By symmetry,
c ∼ b
.
By transitivity, we have
a ∼ b
. Now for all
b
′
∈
[
b
], we have
b ∼ b
′
. Thus by
transitivity, we have
a ∼ b
′
. Thus [
b
]
⊆
[
a
]. Similarly, [
a
]
⊆
[
b
] and [
a
] = [
b
].
Lemma. Given a group
G
and a subgroup
H
, define the equivalence relation
on G with a ∼ b iff b
−1
a ∈ H. The equivalence classes are the left cosets of H.
Proof. First show that it is an equivalence relation.
(i) Reflexivity: Since aa
−1
= e ∈ H, a ∼ a.
(ii) Symmetry: a ∼ b ⇒ b
−1
a ∈ H ⇒ (b
−1
a)
−1
= a
−1
b ∈ H ⇒ b ∼ a.
(iii)
Transitivity: If
a ∼ b
and
b ∼ c
, we have
b
−1
a, c
−1
b ∈ H
. So
c
−1
bb
−1
a
=
c
−1
a ∈ H. So a ∼ c.
To show that the equivalence classes are the cosets, we have
a ∼ b ⇔ b
−1
a ∈
H ⇔ aH = bH.
Example. Consider (
Z,
+), and for fixed
n
, take the subgroup
nZ
. The cosets
are 0 +
H,
1 +
H, · · ·
(
n −
1) +
H
. We can write these as [0]
,
[1]
,
[2]
· · ·
[
n
]. To
perform arithmetic “mod
n
”, define [
a
] + [
b
] = [
a
+
b
], and [
a
][
b
] = [
ab
]. We
need to check that it is well-defined, i.e. it doesn’t depend on the choice of the
representative of [a].
If [
a
1
] = [
a
2
] and [
b
1
] = [
b
2
], then
a
1
=
a
2
+
kn
and
b
1
=
b
2
+
kn
, then
a
1
+
b
1
=
a
2
+
b
2
+
n
(
k
+
l
) and
a
1
b
1
=
a
2
b
2
+
n
(
kb
2
+
la
2
+
kln
). So [
a
1
+
b
1
] =
[a
2
+ b
2
] and [a
1
b
1
] = [a
2
b
2
].
We have seen that (
Z
n
,
+
n
) is a group. What happens with multiplication?
We can only take elements which have inverses (these are called units, cf. IB
Groups, Rings and Modules). Call the set of them
U
n
=
{
[
a
] : (
a, n
) = 1
}
. We’ll
see these are the units.
Definition (Euler totient function). (Euler totient function) ϕ(n) = |U
n
|.
Example. If p is a prime, ϕ(n) = p − 1. ϕ(4) = 2.
Proposition. U
n
is a group under multiplication mod n.
Proof. The operation is well-defined as shown above. To check the axioms:
0.
Closure: if
a, b
are coprime to
n
, then
a · b
is also coprime to
n
. So
[a], [b] ∈ U
n
⇒ [a] · [b] = [a · b] ∈ U
n
1. Identity: [1]
2.
Let [
a
]
∈ U
n
. Consider the map
U
n
→ U
n
with [
c
]
7→
[
ac
]. This is injective:
if [
ac
1
] = [
ac
2
], then
n
divides
a
(
c
1
− c
2
). Since
a
is coprime to
n
,
n
divides
c
1
− c
2
, so [
c
1
] = [
c
2
]. Since
U
n
is finite, any injection (
U
n
→ U
n
) is also a
surjection. So there exists a c such that [ac] = [a][c] = 1. So [c] = [a]
−1
.
3. Associativity (and also commutativity): inherited from Z.
Theorem (Fermat-Euler theorem). Let n ∈ N and a ∈ Z coprime to n. Then
a
ϕ(n)
≡ 1 (mod n).
In particular, (Fermat’s Little Theorem) if
n
=
p
is a prime, then for any
a
not
a multiple of p.
a
p−1
≡ 1 (mod p).
Proof.
As
a
is coprime with
n
, [
a
]
∈ U
n
. Then [
a
]
|U
n
|
= [1], i.e.
a
ϕ(n)
≡
1
(mod n).
3.1 Small groups
We will study the structures of certain small groups.
Example (Using Lagrange theorem to find subgroups). To find subgroups of
D
10
, we know that the subgroups must have size 1, 2, 5 or 10:
1: {e}
2: The groups generated by the 5 reflections of order 2
5:
The group must be cyclic since it has prime order 5. It is then generated
by an element of order 5, i.e.
r, r
2
, r
3
and
r
4
. They generate the same
group ⟨r⟩.
10: D
10
As for D
8
, subgroups must have order 1, 2, 4 or 8.
1: {e}
2: 5 elements of order 2, namely 4 reflections and r
2
.
4:
First consider the subgroup isomorphic to
C
4
, which is
⟨r⟩
. There are two
other non-cyclic group.
8: D
8
Proposition. Any group of order 4 is either isomorphic to C
4
or C
2
× C
2
.
Proof.
Let
|G|
= 4. By Lagrange theorem, possible element orders are 1 (
e
only),
2 and 4. If there is an element a ∈ G of order 4, then G = ⟨a⟩
∼
=
C
4
.
Otherwise all non-identity elements have order 2. Then
G
must be abelian
(For any
a, b
, (
ab
)
2
= 1
⇒ ab
= (
ab
)
−1
⇒ ab
=
b
−1
a
−1
⇒ ab
=
ba
). Pick
2 elements of order 2, say
b, c ∈ G
, then
⟨b⟩
=
{e, b}
and
⟨c⟩
=
{e, c}
. So
⟨b⟩ ∩ ⟨c⟩
=
{e}
. As
G
is abelian,
⟨b⟩
and
⟨c⟩
commute. We know that
bc
=
cb
has
order 2 as well, and is the only element of
G
left. So
G
∼
=
⟨b⟩ × ⟨c⟩
∼
=
C
2
× C
2
by the direct product theorem.
Proposition. A group of order 6 is either cyclic or dihedral (i.e. is isomorphic
to C
6
or D
6
). (See proof in next section)
3.2 Left and right cosets
As
|aH|
=
|H|
and similarly
|H|
=
|Ha|
, left and right cosets have the same size.
Are they necessarily the same? We’ve previously shown that they might not be
the same. In some other cases, they are.
Example.
(i)
Take
G
= (
Z,
+) and
H
= 2
Z
. We have 0 + 2
Z
= 2
Z
+ 0 = even numbers
and 1 + 2
Z
= 2
Z
+ 1 = odd numbers. Since
G
is abelian,
aH
=
Ha
for all
a, ∈ G, H ≤ G.
(ii)
Let
G
=
D
6
=
⟨r, s | r
3
=
e
=
s
2
, rs
=
sr
−1
⟩
. Let
U
=
⟨r⟩
. Since
the cosets partition
G
, so one must be
U
and the other
sU
=
{s, sr
=
r
2
s, sr
2
= rs} = Us. So for all a ∈ G, aU = Ua.
(iii)
Let
G
=
D
6
and take
H
=
⟨s⟩
. We have
H
=
{e, s}
,
rH
=
{r, rs
=
sr
−1
}
and
r
2
H
=
{r
2
, r
s
}
; while
H
=
{e, s}, Hr
=
{r, sr}
and
Hr
2
=
{r
2
, sr
2
}
.
So the left and right subgroups do not coincide.
This distinction will become useful in the next chapter.
4 Quotient groups
In the previous section, when attempting to pretend that a 3
×
3
×
3 Rubik’s
cube is a 2
×
2
×
2 one, we came up with the cosets
aH
, and claimed that these
form a group. We also said that this is not the case for arbitrary subgroup
H
,
but only for subgroups that satisfy
aH
=
Ha
. Before we prove these, we first
study these subgroups a bit.
4.1 Normal subgroups
Definition (Normal subgroup). A subgroup K of G is a normal subgroup if
(∀a ∈ G)(∀k ∈ K) aka
−1
∈ K.
We write K ◁ G. This is equivalent to:
(i) (∀a ∈ G) aK = Ka, i.e. left coset = right coset
(ii) (∀a ∈ G) aKa
−1
= K (cf. conjugacy classes)
From the example last time,
H
=
⟨s⟩ ≤ D
6
is not a normal subgroup, but
K
=
⟨r⟩ ◁ D
6
. We know that every group
G
has at least two normal subgroups
{e} and G.
Lemma.
(i) Every subgroup of index 2 is normal.
(ii) Any subgroup of an abelian group is normal.
Proof.
(i)
If
K ≤ G
has index 2, then there are only two possible cosets
K
and
G \ K
.
As
eK
=
Ke
and cosets partition
G
, the other left coset and right coset
must be G \ K. So all left cosets and right cosets are the same.
(ii) For all a ∈ G and k ∈ K, we have aka
−1
= aa
−1
k = k ∈ K.
Proposition. Every kernel is a normal subgroup.
Proof.
Given homomorphism
f
:
G → H
and some
a ∈ G
, for all
k ∈ ker f
, we
have
f
(
aka
−1
) =
f
(
a
)
f
(
k
)
f
(
a
)
−1
=
f
(
a
)
ef
(
a
)
−1
=
e
. Therefore
aka
−1
∈ ker f
by definition of the kernel.
In fact, we will see in the next section that all normal subgroups are kernels
of some homomorphism.
Example. Consider
G
=
D
8
. Let
K
=
⟨r
2
⟩
is normal. Check: Any element of
G
is either
sr
ℓ
or
r
ℓ
for some
ℓ
. Clearly
e
satisfies
aka
−1
∈ K
. Now check
r
2
:
For the case of
sr
ℓ
, we have
sr
ℓ
r
2
(
sr
ℓ
)
−1
=
sr
ℓ
r
2
r
−ℓ
s
−1
=
sr
2
s
=
ssr
−2
=
r
2
.
For the case of r
ℓ
, r
ℓ
r
2
r
−ℓ
= r
2
.
Proposition. A group of order 6 is either cyclic or dihedral (i.e.
∼
=
C
6
or D
6
).
Proof.
Let
|G|
= 6. By Lagrange theorem, possible element orders are 1
,
2
,
3 and
6. If there is an
a ∈ G
of order 6, then
G
=
⟨a⟩
∼
=
C
6
. Otherwise, we can only
have elements of orders 2 and 3 other than the identity. If
G
only has elements
of order 2, the order must be a power of 2 by Sheet 1 Q. 8, which is not the case.
So there must be an element
r
of order 3. So
⟨r⟩ ◁ G
as it has index 2. Now
G
must also have an element s of order 2 by Sheet 1 Q. 9.
Since
⟨r⟩
is normal, we know that
srs
−1
∈ ⟨r⟩
. If
srs
−1
=
e
, then
r
=
e
,
which is not true. If
srs
−1
=
r
, then
sr
=
rs
and
sr
has order 6 (lcm of the
orders of
s
and
r
), which was ruled out above. Otherwise if
srs
−1
=
r
2
=
r
−1
,
then G is dihedral by definition of the dihedral group.
4.2 Quotient groups
Proposition. Let
K ◁ G
. Then the set of (left) cosets of
K
in
G
is a group
under the operation aK ∗ bK = (ab)K.
Proof.
First show that the operation is well-defined. If
aK
=
a
′
K
and
bK
=
b
′
K
,
we want to show that
aK ∗bK
=
a
′
K ∗b
′
K
. We know that
a
′
=
ak
1
and
b
′
=
bk
2
for some
k
1
, k
2
∈ K
. Then
a
′
b
′
=
ak
1
bk
2
. We know that
b
−1
k
1
b ∈ K
. Let
b
−1
k
1
b
=
k
3
. Then
k
1
b
=
bk
3
. So
a
′
b
′
=
abk
3
k
2
∈
(
ab
)
K
. So picking a different
representative of the coset gives the same product.
1. Closure: If aK, bK are cosets, then (ab)K is also a coset
2. Identity: The identity is eK = K (clear from definition)
3. Inverse: The inverse of aK is a
−1
K (clear from definition)
4. Associativity: Follows from the associativity of G.
Definition (Quotient group). Given a group
G
and a normal subgroup
K
, the
quotient group or factor group of
G
by
K
, written as
G/K
, is the set of (left)
cosets of K in G under the operation aK ∗ bK = (ab)K.
Note that the set of left cosets also exists for non-normal subgroups (abnormal
subgroups?), but the group operation above is not well defined.
Example.
(i)
Take
G
=
Z
and
nZ
(which must be normal since
G
is abelian), the cosets
are
k
+
nZ
for 0
≤ k < n
. The quotient group is
Z
n
. So we can write
Z/
(
nZ
) =
Z
n
. In fact these are the only quotient groups of
Z
since
nZ
are
the only subgroups.
Note that if G is abelian, G/K is also abelian.
(ii)
Take
K
=
⟨r⟩ ◁ D
6
. We have two cosets
K
and
sK
. So
D
6
/K
has order 2
and is isomorphic to C
2
.
(iii)
Take
K
=
⟨r
2
⟩ ◁ D
8
. We know that
G/K
should have
8
2
= 4 elements. We
have
G/K
=
{K, rK
=
r
3
K, sK
=
sr
2
K, srK
=
sr
3
K}
. We see that all
elements (except K) has order 2, so G/K
∼
=
C
2
× C
2
.
Note that quotient groups are not subgroups of
G
. They contain different
kinds of elements. For example,
Z/nZ
∼
=
C
n
are finite, but all subgroups of
Z
infinite.
Example. (Non-example) Consider
D
6
with
H
=
⟨s⟩
.
H
is not a normal
subgroup. We have
rH ∗ r
2
H
=
r
3
H
=
H
, but
rH
=
rsH
and
r
2
H
=
srH
(by
considering the individual elements). So we have
rsH ∗ srH
=
r
2
H
=
H
, and
the operation is not well-defined.
Lemma. Given
K ◁ G
, the quotient map
q
:
G → G/K
with
g 7→ gK
is a
surjective group homomorphism.
Proof. q
(
ab
) = (
ab
)
K
=
aKbK
=
q
(
a
)
q
(
b
). So
q
is a group homomorphism.
Also for all aK ∈ G/K, q(a) = aK. So it is surjective.
Note that the kernel of the quotient map is
K
itself. So any normal subgroup
is a kernel of some homomorphism.
Proposition. The quotient of a cyclic group is cyclic.
Proof.
Let
G
=
C
n
with
H ≤ C
n
. We know that
H
is also cyclic. Say
C
n
=
⟨c⟩
and
H
=
⟨c
k
⟩
∼
=
C
ℓ
, where
kℓ
=
n
. We have
C
n
/H
=
{H, cH, c
2
H, · · · c
k−1
H}
=
⟨cH⟩
∼
=
C
k
.
4.3 The Isomorphism Theorem
Now we come to the Really Important Theorem
TM
.
Theorem (The Isomorphism Theorem). Let
f
:
G → H
be a group homomor-
phism with kernel K. Then K ◁ G and G/K
∼
=
im f.
Proof.
We have proved that
K ◁ G
before. We define a group homomorphism
θ : G/K → im f by θ(aK) = f(a).
First check that this is well-defined: If a
1
K = a
2
K, then a
−1
2
a
1
∈ K. So
f(a
2
)
−1
f(a
1
) = f(a
−1
2
a
1
) = e.
So f (a
1
) = f(a
2
) and θ(a
1
K) = θ(a
2
K).
Now we check that it is a group homomorphism:
θ(aKbK) = θ(abK) = f(ab) = f(a)f(b) = θ(aK)θ(bK).
To show that it is injective, suppose
θ
(
aK
) =
θ
(
bK
). Then
f
(
a
) =
f
(
b
). Hence
f(b)
−1
f(a) = e. Hence b
−1
a ∈ K. So aK = bK.
By definition,
θ
is surjective since
im θ
=
im f
. So
θ
gives an isomorphism
G/K
∼
=
im f ≤ H.
If
f
is injective, then the kernel is
{e}
, so
G/K
∼
=
G
and
G
is isomorphic to
a subgroup of
H
. We can think of
f
as an inclusion map. If
f
is surjective, then
im f = H. In this case, G/K
∼
=
H.
Example.
(i)
Take
f
:
GL
n
(
R
)
→ R
∗
with
A 7→ det A
,
ker f
=
SL
N
(
R
).
im f
=
R
∗
as for
all
λ ∈ R
∗
,
det
λ 0 · · · 0
0 1 · · · 0
.
.
.
.
.
.
.
.
.
.
.
.
0 0 0 1
=
λ
. So we know that
GL
n
(
R
)
/SL
n
(
R
)
∼
=
R
∗
.
(ii)
Define
θ
: (
R,
+)
→
(
C∗, ×
) with
r 7→ exp
(2
πir
). This is a group homomor-
phism since
θ
(
r
+
s
) =
exp
(2
πi
(
r
+
s
)) =
exp
(2
πir
)
exp
(2
πis
) =
θ
(
r
)
θ
(
s
).
We know that the kernel is
Z ◁ R
. Clearly the image is the unit circle
(S
1
, ×). So R/Z
∼
=
(S
1
, ×).
(iii) G
= (
Z
∗
p
, ×
) for prime
p
= 2. We have
f
:
G → G
with
a 7→ a
2
. This
is a homomorphism since (
ab
)
2
=
a
2
b
2
(
Z
∗
p
is abelian). The kernel is
{±
1
}
=
{
1
, p −
1
}
. We know that
im f
∼
=
G/ ker f
with order
p−1
2
. These
are known as quadratic residues.
Lemma. Any cyclic group is isomorphic to either
Z
or
Z/
(
nZ
) for some
n ∈ N
.
Proof.
Let
G
=
⟨c⟩
. Define
f
:
Z → G
with
m 7→ c
m
. This is a group
homomorphism since
c
m
1
+m
2
=
c
m
1
c
m
2
.
f
is surjective since
G
is by definition
all c
m
for all m. We know that ker f ◁ Z. We have three possibilities. Either
(i) ker f = {e}, so F is an isomorphism and G
∼
=
Z; or
(ii) ker f = Z, then G
∼
=
Z/Z = {e} = C
1
; or
(iii) ker f
=
nZ
(since these are the only proper subgroups of
Z
), then
G
∼
=
Z/(nZ).
Definition (Simple group). A group is simple if it has no non-trivial proper
normal subgroup, i.e. only {e} and G are normal subgroups.
Example.
C
p
for prime
p
are simple groups since it has no proper subgroups at
all, let alone normal ones. A
5
is simple, which we will prove after Chapter 6.
The finite simple groups are the building blocks of all finite groups. All finite
simple groups have been classified (The Atlas of Finite Groups). If we have
K ◁ G
with
K
=
G
or
{e}
, then we can “quotient out”
G
into
G/K
. If
G/K
is not simple, repeat. Then we can write
G
as an “inverse quotient” of simple
groups.
5 Group actions
Recall that we came up with groups to model symmetries and permutations.
Intuitively, elements of groups are supposed to “do things”. However, as we
developed group theory, we abstracted these away and just looked at how
elements combine to form new elements. Group actions recapture this idea and
make each group element correspond to some function.
5.1 Group acting on sets
Definition (Group action). Let
X
be a set and
G
be a group. An action of
G
on X is a homomorphism φ : G → Sym X.
This means that the homomorphism
φ
turns each element
g ∈ G
into a
permutation of X, in a way that respects the group structure.
Instead of writing φ(g)(x), we usually directly write g(x) or gx.
Alternatively, we can define the group action as follows:
Proposition. Let
X
be a set and
G
be a group. Then
φ
:
G → Sym X
is a
homomorphism (i.e. an action) iff
θ
:
G × X → X
defined by
θ
(
g, x
) =
φ
(
g
)(
x
)
satisfies
0. (∀g ∈ G)(x ∈ X) θ(g, x) ∈ X.
1. (∀x ∈ X) θ(e, x) = x.
2. (∀g, h ∈ G)(∀x ∈ X) θ(g, θ(h, x)) = θ(gh, x).
This criteria is almost the definition of a homomorphism. However, here we
do not explicitly require
θ
(
g, ·
) to be a bijection, but require
θ
(
e, ·
) to be the
identity function. This automatically ensures that
θ
(
g, ·
) is a bijection, since
when composed with
θ
(
g
−1
, ·
), it gives
θ
(
e, ·
), which is the identity. So
θ
(
g, ·
)
has an inverse. This is usually an easier thing to show.
Example.
(i)
Trivial action: for any group
G
acting on any set
X
, we can have
φ
(
g
) = 1
X
for all g, i.e. G does nothing.
(ii) S
n
acts on {1, · · · n} by permutation.
(iii) D
2n
acts on the vertices of a regular n-gon (or the set {1, · · · , n}).
(iv)
The rotations of a cube act on the faces/vertices/diagonals/axes of the
cube.
Note that different groups can act on the same sets, and the same group can
act on different sets.
Definition (Kernel of action). The kernel of an action
G
on
X
is the kernel of
φ, i.e. all g such that φ(g) = 1
X
.
Note that by the isomorphism theorem,
ker φ ◁ G
and
G/K
is isomorphic to
a subgroup of Sym X.
Example.
(i) D
2n
acting on {1, 2 · · · n} gives φ : D
2n
→ S
n
with kernel {e}.
(ii)
Let
G
be the rotations of a cube and let it act on the three axes
x, y, z
through the faces. We have
φ
:
G → S
3
. Then any rotation by 180
◦
doesn’t
change the axes, i.e. act as the identity. So the kernel of the action has
at least 4 elements:
e
and the three 180
◦
rotations. In fact, we’ll see later
that these 4 are exactly the kernel.
Definition (Faithful action). An action is faithful if the kernel is just {e}.
5.2 Orbits and Stabilizers
Definition (Orbit of action). Given an action
G
on
X
, the orbit of an element
x ∈ X is
orb(x) = G(x) = {y ∈ X : (∃g ∈ G) g(x) = y}.
Intuitively, it is the elements that x can possibly get mapped to.
Definition (Stabilizer of action). The stabilizer of x is
stab(x) = G
x
= {g ∈ G : g(x) = x} ⊆ G.
Intuitively, it is the elements in G that do not change x.
Lemma. stab(x) is a subgroup of G.
Proof.
We know that
e
(
x
) =
x
by definition. So
stab
(
x
) is non-empty. Suppose
g, h ∈ stab
(
x
), then
gh
−1
(
x
) =
g
(
h
−1
(
x
)) =
g
(
x
) =
x
. So
gh
−1
∈ stab
(
X
). So
stab(x) is a subgroup.
Example.
(i)
Consider
D
8
acting on the corners of the square
X
=
{
1
,
2
,
3
,
4
}
. Then
orb
(1) =
X
since 1 can go anywhere by rotations.
stab
(1) =
{e,
reflection
in the line through 1}
(ii)
Consider the rotations of a cube acting on the three axes
x, y, z
. Then
orb
(
x
) is everything, and
stab
(
x
) contains
e
, 180
◦
rotations and rotations
about the x axis.
Definition (Transitive action). An action
G
on
X
is transitive if (
∀x
)
orb
(
x
) =
X, i.e. you can reach any element from any element.
Lemma. The orbits of an action partition X.
Proof. Firstly, (∀x)(x ∈ orb(x)) as e(x) = x. So every x is in some orbit.
Then suppose
z ∈ orb
(
x
) and
z ∈ orb
(
y
), we have to show that
orb
(
x
) =
orb
(
y
). We know that
z
=
g
1
(
x
) and
z
=
g
2
(
y
) for some
g
1
, g
2
. Then
g
1
(
x
) =
g
2
(y) and y = g
−1
2
g
1
(x).
For any
w
=
g
3
(
y
)
∈ orb
(
y
), we have
w
=
g
3
g
−1
2
g
1
(
x
). So
w ∈ orb
(
x
). Thus
orb(y) ⊆ orb(x) and similarly orb(x) ⊆ orb(y). Therefore orb(x) = orb(y).
Suppose a group
G
acts on
X
. We fix an
x ∈ X
. Then by definition of
the orbit, given any
g ∈ G
, we have
g
(
x
)
∈ orb
(
x
). So each
g ∈ G
gives us
a member of
orb
(
x
). Conversely, every object in
orb
(
x
) arises this way, by
definition of
orb
(
x
). However, different elements in
G
can give us the same orbit.
In particular, if
g ∈ stab
(
x
), then
hg
and
h
give us the same object in
orb
(
x
),
since
hg
(
x
) =
h
(
g
(
x
)) =
h
(
x
). So we have a correspondence between things in
orb(x) and members of G, “up to stab(x)”.
Theorem (Orbit-stabilizer theorem). Let the group
G
act on
X
. Then there
is a bijection between
orb
(
x
) and cosets of
stab
(
x
) in
G
. In particular, if
G
is
finite, then
| orb(x)|| stab(x)| = |G|.
Proof.
We biject the cosets of
stab
(
x
) with elements in the orbit of
x
. Recall
that G : stab(x) is the set of cosets of stab(x). We can define
θ : (G : stab(x)) → orb(x)
g stab(x) 7→ g(x).
This is well-defined — if
g stab
(
x
) =
h stab
(
x
), then
h
=
gk
for some
k ∈ stab
(
x
).
So h(x) = g(k(x)) = g(x).
This map is surjective since for any
y ∈ orb
(
x
), there is some
g ∈ G
such
that
g
(
x
) =
y
, by definition. Then
θ
(
g stab
(
x
)) =
y
. It is injective since if
g(x) = h(x), then h
−1
g(x) = x. So h
−1
g ∈ stab(x). So g stab(x) = h stab(x).
Hence the number of cosets is
| orb
(
x
)
|
. Then the result follows from La-
grange’s theorem.
An important application of the orbit-stabilizer theorem is determining group
sizes. To find the order of the symmetry group of, say, a pyramid, we find
something for it to act on, pick a favorite element, and find the orbit and
stabilizer sizes.
Example.
(i)
Suppose we want to know how big
D
2n
is.
D
2n
acts on the vertices
{
1
,
2
,
3
, · · · , n}
transitively. So
| orb
(1)
|
=
n
. Also,
stab
(1) =
{e,
reflection
in the line through 1}. So |D
2n
| = | orb(1)|| stab(1)| = 2n.
Note that if the action is transitive, then all orbits have size
|X|
and thus
all stabilizers have the same size.
(ii)
Let
⟨
(1 2)
⟩
act on
{
1
,
2
,
3
}
. Then
orb
(1) =
{
1
,
2
}
and
stab
(1) =
{e}
.
orb(3) = {3} and stab(3) = ⟨(1 2)⟩.
(iii)
Consider
S
4
acting on
{
1
,
2
,
3
,
4
}
. We know that
orb
(1) =
X
and
|S
4
|
= 24.
So
| stab
(1)
|
=
24
4
= 6. That makes it easier to find
stab
(1). Clearly
S
{2,3,4}
∼
=
S
3
fix 1. So
S
{2,3,4}
≤ stab
(1). However,
|S
3
|
= 6 =
| stab
(1)
|
,
so this is all of the stabilizer.
5.3 Important actions
Given any group
G
, there are a few important actions we can define. In particular,
we will define the conjugation action, which is a very important concept on
its own. In fact, the whole of the next chapter will be devoted to studying
conjugation in the symmetric groups.
First, we will study some less important examples of actions.
Lemma (Left regular action). Any group
G
acts on itself by left multiplication.
This action is faithful and transitive.
Proof. We have
1. (∀g ∈ G)(x ∈ G) g(x) = g · x ∈ G by definition of a group.
2. (∀x ∈ G) e · x = x by definition of a group.
3. g(hx) = (gh)x by associativity.
So it is an action.
To show that it is faithful, we want to know that [(
∀x ∈ X
)
gx
=
x
]
⇒ g
=
e
.
This follows directly from the uniqueness of identity.
To show that it is transitive,
∀x, y ∈ G
, then (
yx
−1
)(
x
) =
y
. So any
x
can
be sent to any y.
Theorem (Cayley’s theorem). Every group is isomorphic to some subgroup of
some symmetric group.
Proof.
Take the left regular action of
G
on itself. This gives a group homo-
morphism
φ
:
G → Sym G
with
ker φ
=
{e}
as the action is faithful. By the
isomorphism theorem, G
∼
=
im φ ≤ Sym G.
Lemma (Left coset action). Let
H ≤ G
. Then
G
acts on the left cosets of
H
by left multiplication transitively.
Proof. First show that it is an action:
0. g(aH) = (ga)H is a coset of H.
1. e(aH) = (ea)H = aH.
2. g
1
(g
2
(aH)) = g
1
((g
2
a)H) = (g
1
g
2
a)H = (g
1
g
2
)(aH).
To show that it is transitive, given
aH, bH
, we know that (
ba
−1
)(
aH
) =
bH
.
So any aH can be mapped to bH.
In the boring case where
H
=
{e}
, then this is just the left regular action
since G/{e}
∼
=
G.
Definition (Conjugation of element). The conjugation of
a ∈ G
by
b ∈ G
is
given by
bab
−1
∈ G
. Given any
a, c
, if there exists some
b
such that
c
=
bab
−1
,
then we say a and c are conjugate.
What is conjugation? This
bab
−1
form looks familiar from Vectors and
Matrices. It is the formula used for changing basis. If
b
is the change-of-basis
matrix and
a
is a matrix, then the matrix in the new basis is given by
bab
−1
. In
this case, bab
−1
is the same matrix viewed from a different basis.
In general, two conjugate elements are “the same” in some sense. For example,
we will later show that in
S
n
, two elements are conjugate if and only if they
have the same cycle type. Conjugate elements in general have many properties
in common, such as their order.
Lemma (Conjugation action). Any group
G
acts on itself by conjugation (i.e.
g(x) = gxg
−1
).
Proof. To show that this is an action, we have
0. g(x) = gxg
−1
∈ G for all g, x ∈ G.
1. e(x) = exe
−1
= x
2. g(h(x)) = g(hxh
−1
) = ghxh
−1
g
−1
= (gh)x(gh)
−1
= (gh)(x)
Definition (Conjugacy classes and centralizers). The conjugacy classes are the
orbits of the conjugacy action.
ccl(a) = {b ∈ G : (∃g ∈ g) gag
−1
= b}.
The centralizers are the stabilizers of this action, i.e. elements that commute
with a.
C
G
(a) = {g ∈ G : gag
−1
= a} = {g ∈ G : ga = ag}.
The centralizer is defined as the elements that commute with a particular
element a. For the whole group G, we can define the center.
Definition (Center of group). The center of
G
is the elements that commute
with all other elements.
Z(G) = {g ∈ G : (∀a) gag
−1
= a} = {g ∈ G : (∀a) ga = ag}.
It is sometimes written as C(G) instead of Z(G).
In many ways, conjugation is related to normal subgroups.
Lemma. Let K ◁ G. Then G acts by conjugation on K.
Proof.
We only have to prove closure as the other properties follow from the
conjugation action. However, by definition of a normal subgroup, for every
g ∈ G, k ∈ K, we have gkg
−1
∈ K. So it is closed.
Proposition. Normal subgroups are exactly those subgroups which are unions
of conjugacy classes.
Proof.
Let
K ◁ G
. If
k ∈ K
, then by definition for every
g ∈ G
, we get
gkg
−1
∈ K
. So
ccl
(
k
)
⊆ K
. So
K
is the union of the conjugacy classes of all its
elements.
Conversely, if
K
is a union of conjugacy classes and a subgroup of
G
, then
for all k ∈ K, g ∈ G, we have gkg
−1
∈ K. So K is normal.
Lemma. Let
X
be the set of subgroups of
G
. Then
G
acts by conjugation on
X.
Proof. To show that it is an action, we have
0.
If
H ≤ G
, then we have to show that
gHg
−1
is also a subgroup. We
know that
e ∈ H
and thus
geg
−1
=
e ∈ gHg
−1
, so
gHg
−1
is non-empty.
For any two elements
gag
−1
and
gbg
−1
∈ gHg
−1
, (
gag
−1
)(
gbg
−1
)
−1
=
g(ab
−1
)g
−1
∈ gHg
−1
. So gHg
−1
is a subgroup.
1. eHe
−1
= H.
2. g
1
(g
2
Hg
−1
2
)g
−1
1
= (g
1
g
2
)H(g
1
g
2
)
−1
.
Under this action, normal subgroups have singleton orbits.
Definition (Normalizer of subgroup). The normalizer of a subgroup is the
stabilizer of the (group) conjugation action.
N
G
(H) = {g ∈ G : gHg
−1
= H}.
We clearly have
H ⊆ N
G
(
H
). It is easy to show that
N
G
(
H
) is the largest
subgroup of G in which H is a normal subgroup, hence the name.
There is a connection between actions in general and conjugation of subgroups.
Lemma. Stabilizers of the elements in the same orbit are conjugate, i.e. let
G
act on X and let g ∈ G, x ∈ X. Then stab(g(x)) = g stab(x)g
−1
.
5.4 Applications
Example. Let
G
+
be the rotations of a cube acting on the vertices. Let
X
be the set of vertices. Then
|X|
= 8. Since the action is transitive, the
orbit of element is the whole of
X
. The stabilizer of vertex 1 is the set of
rotations through 1 and the diagonally opposite vertex, of which there are 3. So
|G
+
| = | orb(1)|| stab(1)| = 8 · 3 = 24.
Example. Let
G
be a finite simple group of order greater than 2, and
H ≤ G
have index n = 1. Then |G| ≤ n!/2.
Proof.
Consider the left coset action of
G
on
H
. We get a group homomorphism
φ
:
G → S
n
since there are
n
cosets of
H
. Since
H
=
G
,
φ
is non-trivial and
ker φ
=
G
. Now
ker φ ◁ G
. Since
G
is simple,
ker φ
=
{e}
. So
G
∼
=
im φ ⊆ S
n
by the isomorphism theorem. So |G| ≤ |S
n
| = n!.
We can further refine this by considering
sgn ◦φ
:
G → {±
1
}
. The kernel
of this composite is normal in
G
. So
K
=
ker
(
sgn ◦ϕ
) =
{e}
or
G
. Since
G/K
∼
=
im
(
sgn ◦ϕ
), we know that
|G|/|K|
= 1 or 2 since
im
(
sgn ◦ϕ
) has at most
two elements. Hence for
|G| >
2, we cannot have
K
=
{e}
, or else
|G|/|K| >
2.
So we must have
K
=
G
, so
sgn
(
φ
(
g
)) = 1 for all
g
and
im φ ≤ A
n
. So
|G| ≤ n!/2
We have seen on Sheet 1 that if
|G|
is even, then
G
has an element of order
2. In fact,
Theorem (Cauchy’s Theorem). Let
G
be a finite group and prime
p
dividing
|G|
. Then
G
has an element of order
p
(in fact there must be at least
p −
1
elements of order p).
It is important to remember that this only holds for prime
p
. For example,
A
4
doesn’t have an element of order 6 even though 6
|
12 =
|A
4
|
. The converse,
however, holds for any number trivially by Lagrange’s theorem.
Proof.
Let
G
and
p
be fixed. Consider
G
p
=
G × G × · · · × G
, the set of
p
-tuples
of G. Let X ⊆ G
p
be X = {(a
1
, a
2
, · · · , a
p
) ∈ G
p
: a
1
a
2
· · · a
p
= e}.
In particular, if an element
b
has order
p
, then (
b, b, · · · , b
)
∈ X
. In fact, if
(b, b, · · · , b) ∈ X and b = e, then b has order p, since p is prime.
Now let
H
=
⟨h
:
h
p
=
e⟩
∼
=
C
p
be a cyclic group of order
p
with generator
h
(This h is not related to G in any way). Let H act on X by “rotation”:
h(a
1
, a
2
, · · · , a
p
) = (a
2
, a
3
, · · · , a
p
, a
1
)
This is an action:
0.
If
a
1
· · · a
p
=
e
, then
a
−1
1
=
a
2
· · · a
p
. So
a
2
· · · a
p
a
1
=
a
−1
1
a
1
=
e
. So
(a
2
, a
3
, · · · , a
p
, a
1
) ∈ X.
1. e acts as an identity by construction
2. The “associativity” condition also works by construction.
As orbits partition
X
, the sum of all orbit sizes must be
|X|
. We know that
|X|
=
|G|
p−1
since we can freely choose the first
p −
1 entries and the last one
must be the inverse of their product. Since
p
divides
|G|
,
p
also divides
|X|
. We
have
| orb
(
a
1
, · · · , a
p
)
|| stab
H
(
a
1
, · · · , a
p
)
|
=
|H|
=
p
. So all orbits have size 1 or
p
, and they sum to
|X|
=
p×
something. We know that there is one orbit of size
1, namely (
e, e, · · · , e
). So there must be at least
p −
1 other orbits of size 1 for
the sum to be divisible by p.
In order to have an orbit of size 1, they must look like
(a, a, · · · , a)
for some
a ∈ G, which has order p.
6 Symmetric groups II
In this chapter, we will look at conjugacy classes of
S
n
and
A
n
. It turns out this
is easy for
S
n
, since two elements are conjugate if and only if they have the same
cycle type. However, it is slightly more complicated in
A
n
. This is since while
(1 2 3) and (1 3 2) might be conjugate in
S
4
, the element needed to perform the
conjugation might be odd and not in A
n
.
6.1 Conjugacy classes in S
n
Recall σ, τ ∈ S
n
are conjugate if ∃ρ ∈ S
n
such that ρσρ
−1
= τ.
We first investigate the special case, when σ is a k-cycle.
Proposition. If (
a
1
a
2
· · · a
k
) is a
k
-cycle and
ρ ∈ S
n
, then
ρ
(
a
1
· · · a
k
)
ρ
−1
is
the k-cycle (ρ(a
1
) ρ(a
2
) · · · ρ(a
3
)).
Proof.
Consider any
ρ
(
a
1
) acted on by
ρ
(
a
1
· · · a
k
)
ρ
−1
. The three permutations
send it to
ρ
(
a
1
)
7→ a
1
7→ a
2
7→ ρ
(
a
2
) and similarly for other
a
i
s. Since
ρ
is
bijective, any
b
can be written as
ρ
(
a
) for some
a
. So the result is the
k
-cycle
(ρ(a
1
) ρ(a
2
) · · · ρ(a
3
)).
Corollary. Two elements in
S
n
are conjugate iff they have the same cycle type.
Proof.
Suppose
σ
=
σ
1
σ
2
· · · σ
ℓ
, where
σ
i
are disjoint cycles. Then
ρσρ
−1
=
ρσ
1
ρ
−1
ρσ
2
ρ
−1
· · · ρσ
ℓ
ρ
−1
. Since the conjugation of a cycle conserves its length,
ρσρ
−1
has the same cycle type.
Conversely, if σ, τ have the same cycle type, say
σ = (a
1
a
2
· · · a
k
)(a
k+1
· · · a
k+ℓ
), τ = (b
1
b
2
· · · b
k
)(b
k+1
· · · b
k+ℓ
),
if we let ρ(a
i
) = b
i
, then ρσρ
−1
= τ.
Example. Conjugacy classes of S
4
:
Cycle type Example element Size of ccl Size of centralizer Sign
(1, 1, 1, 1) e 1 24 +1
(2, 1, 1) (1 2) 6 4 −1
(2, 2) (1 2)(3 4) 3 8 +1
(3, 1) (1 2 3) 8 3 +1
(4) (1 2 3 4) 6 4 −1
We know that a normal subgroup is a union of conjugacy classes. We can now
find all normal subgroups by finding possible union of conjugacy classes whose
cardinality divides 24. Note that the normal subgroup must contain e.
(i) Order 1: {e}
(ii) Order 2: None
(iii) Order 3: None
(iv)
Order 4:
{e,
(1 2)(3 4)
,
(1 3)(2 4)
,
(1 4)(2 3)
}
∼
=
C
2
× C
2
=
V
4
is a possible
candidate. We can check the group axioms and find that it is really a
subgroup
(v) Order 6: None
(vi) Order 8: None
(vii)
Order 12:
A
4
(We know it is a normal subgroup since it is the kernel of
the signature and/or it has index 2)
(viii) Order 24: S
4
We can also obtain the quotients of
S
4
:
S
4
/{e}
∼
=
S
4
,
S
4
/V
4
∼
=
S
3
∼
=
D
6
,
S
4
/A
4
∼
=
C
2
, S
4
/S
4
= {e}.
6.2 Conjugacy classes in A
n
We have seen that
|S
n
|
= 2
|A
n
|
and that conjugacy classes in
S
n
are “nice”.
How about in A
n
?
The first thought is that we write it down:
ccl
S
n
(σ) = {τ ∈ S
n
: (∃ρ ∈ S
n
) τ = ρσρ
−1
}
ccl
A
n
(σ) = {τ ∈ A
n
: (∃ρ ∈ A
n
) τ = ρσρ
−1
}
Obviously
ccl
A
n
(
σ
)
⊆ ccl
S
n
(
σ
), but the converse need not be true since the
conjugation need to map σ to τ may be odd.
Example. Consider (1 2 3) and (1 3 2). They are conjugate in
S
3
by (2 3), but
(2 3)
∈ A
3
. (This does not automatically entail that they are not conjugate in
A
3
because there might be another even permutation that conjugate (1 2 3) and
(1 3 2). In A
5
, (2 3)(4 5) works (but not in A
3
))
We can use the orbit-stabilizer theorem:
|S
n
| = | ccl
S
n
(σ)||C
S
n
(σ)|
|A
n
| = | ccl
A
n
(σ)||C
A
n
(σ)|
We know that
A
n
is half of
S
n
and
ccl
A
n
is contained in
ccl
S
n
. So we have two
options: either
ccl
S
n
(
σ
) =
ccl
A
n
(
σ
) and
|C
S
n
(
σ
)
|
=
1
2
|C
A
n
(
σ
)
|
; or
1
2
| ccl
S
n
(
σ
)
|
=
| ccl
A
n
(σ)| and C
A
n
(σ) = C
S
n
(σ).
Definition (Splitting of conjugacy classes). When
| ccl
A
n
(
σ
)
|
=
1
2
| ccl
S
n
(
σ
)
|
, we
say that the conjugacy class of σ splits in A
n
.
So the conjugacy classes are either retained or split.
Proposition. For
σ ∈ A
n
, the conjugacy class of
σ
splits in
A
n
if and only if
no odd permutation commutes with σ.
Proof.
We have the conjugacy classes splitting if and only if the centralizer does
not. So instead we check whether the centralizer splits. Clearly
C
A
n
(
σ
) =
C
S
n
(
σ
)
∩ A
n
. So splitting of centralizer occurs if and only if an odd permutation
commutes with σ.
Example. Conjugacy classes in A
4
:
Cycle type Example | ccl
S
4
| Odd element in C
S
4
? | ccl
A
4
|
(1, 1, 1, 1) e 1 Yes (1 2) 1
(2, 2) (1 2)(3 4) 3 Yes (1 2) 3
(3, 1) (1 2 3) 8 No 4, 4
In the (3, 1) case, by the orbit stabilizer theorem,
|C
S
4
((1 2 3))
|
= 3, which is
odd and cannot split.
Example. Conjugacy classes in A
5
:
Cycle type Example | ccl
S
5
| Odd element in C
S
5
? | ccl
A
5
|
(1, 1, 1, 1, 1) e 1 Yes (1 2) 1
(2, 2, 1) (1 2)(3 4) 15 Yes (1 2) 15
(3, 1, 1) (1 2 3) 20 Yes (4 5) 20
(5) (1 2 3 4 5) 24 No 12, 12
Since the centralizer of (1 2 3 4 5) has size 5, it cannot split, so its conjugacy
class must split.
Lemma. σ = (1 2 3 4 5) ∈ S
5
has C
S
5
(σ) = ⟨σ⟩.
Proof. | ccl
S
n
(
σ
)
|
= 24 and
|S
5
|
= 120. So
|C
S
5
(
σ
)
|
= 5. Clearly
⟨σ⟩ ⊆ C
S
5
(
σ
).
Since they both have size 5, we know that C
S
5
(σ) = ⟨σ⟩
Theorem. A
5
is simple.
Proof.
We know that normal subgroups must be unions of the conjugacy classes,
must contain
e
and their order must divide 60. The possible orders are 1, 2,
3, 4, 5, 6, 10, 12, 15, 20, 30. However, the conjugacy classes 1, 15, 20, 12, 12
cannot add up to any of the possible orders apart from 1 and 60. So we only
have trivial normal subgroups.
In fact, all
A
n
for
n ≥
5 are simple, but the proof is horrible (cf. IB Groups,
Rings and Modules).
7 Quaternions
In the remaining of the course, we will look at different important groups. Here,
we will have a brief look at
Definition (Quaternions). The quaternions is the set of matrices
1 0
0 1
,
i 0
0 −i
,
0 1
−1 0
,
0 i
i 0
−1 0
0 −1
,
−i 0
0 i
,
0 −1
1 0
,
0 −i
−i 0
which is a subgroup of GL
2
(C).
Notation. We can also write the quaternions as
Q
8
= ⟨a, b : a
4
= e, b
2
= a
2
, bab
−1
= a
−1
⟩
Even better, we can write
Q
8
= {1, −1, i, −i, j, −j, k, −k}
with
(i) (−1)
2
= 1
(ii) i
2
= j
2
= k
2
= −1
(iii) (−1)i = −i etc.
(iv) ij = k, jk = i, ki = j
(v) ji = −k, kj = −i, ik = −j
We have
1 =
1 0
0 1
, i =
i 0
0 −i
, j =
0 1
−1 0
, k =
0 i
i 0
−1 =
−1 0
0 −1
, −i =
−i 0
0 i
, −j =
0 −1
1 0
, −k =
0 −i
−i 0
Lemma. If
G
has order 8, then either
G
is abelian (i.e.
∼
=
C
8
, C
4
× C
2
or
C
2
× C
2
× C
2
), or
G
is not abelian and isomorphic to
D
8
or
Q
8
(dihedral or
quaternion).
Proof. Consider the different possible cases:
– If G contains an element of order 8, then G
∼
=
C
8
.
– If all non-identity elements have order 2, then G is abelian (Sheet 1, Q8).
Let
a
=
b ∈ G \ {e}
. By the direct product theorem,
⟨a, b⟩
=
⟨a⟩ × ⟨b⟩
.
Then take
c ∈ ⟨a, b⟩
. By the direct product theorem, we obtain
⟨a, b, c⟩
=
⟨a⟩ × ⟨b⟩ × ⟨c⟩
=
C
2
× C
2
× C
2
. Since
⟨a, b, c⟩ ⊆ G
and
|⟨a, b, c⟩|
=
|G|
,
G = ⟨a, b, c⟩
∼
=
C
2
× C
2
× C
2
.
– G
has no element of order 8 but has an order 4 element
a ∈ G
. Let
H
=
⟨a⟩
. Since
H
has index 2, it is normal in
G
. So
G/H
∼
=
C
2
since
|G/H|
= 2. This means that for any
b ∈ H
,
bH
generates
G/H
. Then
(
bH
)
2
=
b
2
H
=
H
. So
b
2
∈ H
. Since
b
2
∈ ⟨a⟩
and
⟨a⟩
is a cyclic group,
b
2
commutes with a.
If b
2
= a or a
3
, then b has order 8. Contradiction. So b
2
= e or a
2
.
We also know that
H
is normal, so
bab
−1
∈ H
. Let
bab
−1
=
a
ℓ
. Since
a
and
b
2
commute, we know that
a
=
b
2
ab
−2
=
b
(
bab
−1
)
b
−1
=
ba
ℓ
b
−1
=
(bab
−1
)
ℓ
= a
ℓ
2
. So ℓ
2
≡ 1 (mod 4). So ℓ ≡ ±1 (mod 4).
◦ When l ≡ 1 (mod 4), bab
−1
= a, i.e. ba = ab. So G is abelian.
∗ If b
2
= e, then G = ⟨a, b⟩
∼
=
⟨a⟩ × ⟨b⟩
∼
=
C
4
× C
2
.
∗ If b
2
= a
2
, then (ba
−1
)
2
= e. So G = ⟨a, ba
−1
⟩
∼
=
C
4
× C
2
.
◦ If l ≡ −1 (mod 4), then bab
−1
= a
−1
.
∗
If
b
2
=
e
, then
G
=
⟨a, b
:
a
4
=
e
=
b
2
, bab
−1
=
a
−1
⟩
. So
G
∼
=
D
8
by definition.
∗ If b
2
= a
2
, then we have G
∼
=
Q
8
.
8 Matrix groups
8.1 General and special linear groups
Consider
M
n×n
(
F
), i.e. the set of
n × n
matrices over the field
F
=
R
or
C
(or
F
p
). We know that matrix multiplication is associative (since they represent
functions) but are, in general, not commutative. To make this a group, we want
the identity matrix
I
to be the identity. To ensure everything has an inverse, we
can only include invertible matrices.
(We do not necessarily need to take
I
as the identity of the group. We can,
for example, take
e
=
0 0
0 1
and obtain a group in which every matrix is of
the form
0 0
0 a
for some non-zero
a
. This forms a group, albeit a boring one
(it is simply
∼
=
R
∗
))
Definition (General linear group GL
n
(F )).
GL
n
(F ) = {A ∈ M
n×n
(F ) : A is invertible}
is the general linear group.
Alternatively, we can define
GL
n
(
F
) as matrices with non-zero determinants.
Proposition. GL
n
(F ) is a group.
Proof.
Identity is
I
, which is in
GL
n
(
F
) by definition (
I
is its self-inverse). The
composition of invertible matrices is invertible, so is closed. Inverse exist by
definition. Multiplication is associative.
Proposition. det : GL
n
(F ) → F \ {0} is a surjective group homomorphism.
Proof. det AB
=
det A det B
. If
A
is invertible, it has non-zero determinant and
det A ∈ F \ {0}.
To show it is surjective, for any
x ∈ F \ {
0
}
, if we take the identity matrix
and replace I
11
with x, then the determinant is x. So it is surjective.
Definition (Special linear group
SL
n
(
F
)). The special linear group
SL
n
(
F
) is
the kernel of the determinant, i.e.
SL
n
(F ) = {A ∈ GL
n
(F ) : det A = 1}.
So SL
n
(F ) ◁ GL
n
(F ) as it is a kernel. Note that Q
8
≤ SL
2
(C)
8.2 Actions of GL
n
(C)
Proposition.
GL
n
(
C
) acts faithfully on
C
n
by left multiplication to the vector,
with two orbits (0 and everything else).
Proof. First show that it is a group action:
1. If A ∈ GL
n
(C) and v ∈ C
n
, then Av ∈ C
n
. So it is closed.
2. Iv = v for all v ∈ C
n
.
3. A(Bv) = (AB)v.
Now prove that it is faithful: a linear map is determined by what it does
on a basis. Take the standard basis e
1
= (1
,
0
, · · · ,
0)
, · · ·
e
n
= (0
, · · · ,
1). Any
matrix which maps each e
k
to itself must be
I
(since the columns of a matrix
are the images of the basis vectors)
To show that there are 2 orbits, we know that
A
0 = 0 for all
A
. Also, as
A
is invertible,
A
v = 0
⇔
v = 0. So 0 forms a singleton orbit. Then given any
two vectors v
= w
∈ C
n
\ {
0
}
, there is a matrix
A ∈ GL
n
(
C
) such that
A
v = w
(cf. Vectors and Matrices).
Similarly, GL
n
(R) acts on R
n
.
Proposition. GL
n
(C) acts on M
n×n
(C) by conjugation. (Proof is trivial)
This action can be thought of as a “change of basis” action. Two matrices
are conjugate if they represent the same map but with respect to different bases.
The P is the base change matrix.
From Vectors and Matrices, we know that there are three different types of
orbits for GL
2
(C): A is conjugate to a matrix of one of these forms:
(i)
λ 0
0 µ
, with λ = µ, i.e. two distinct eigenvalues
(ii)
λ 0
0 λ
, i.e. a repeated eigenvalue with 2-dimensional eigenspace
(iii)
λ 1
0 λ
, i.e. a repeated eigenvalue with a 1-dimensional eigenspace
Note that we said there are three types of orbits, not three orbits. There are
infinitely many orbits, e.g. one for each of λI.
8.3 Orthogonal groups
Recall that
A
T
is defined by
A
T
ij
=
A
ji
, i.e. we reflect the matrix in the diagonal.
They have the following properties:
(i) (AB)
T
= B
T
A
T
(ii) (A
−1
)
T
= (A
T
)
−1
(iii) A
T
A = I ⇔ AA
T
= I ⇔ A
−1
= A
T
. In this case A is orthogonal
(iv) det A
T
= det A
We are now in
R
, because orthogonal matrices don’t make sense with complex
matrices.
Note that a matrix is orthogonal if the columns (or rows) form an orthonormal
basis of
R
n
:
AA
T
=
I ⇔ a
ik
a
jk
=
δ
ij
⇔
a
i
·
a
j
=
δ
ij
, where a
i
is the
i
th column
of A.
The importance of orthogonal matrices is that they are the isometries of
R
n
.
Lemma (Orthogonal matrices are isometries). For any orthogonal
A
and
x, y ∈
R
n
, we have
(i) (Ax) · (Ay) = x · y
(ii) |Ax| = |x|
Proof. Treat the dot product as a matrix multiplication. So
(Ax)
T
(Ay) = x
T
A
T
Ay = x
T
Iy = x
T
y
Then we have
|A
x
|
2
= (
A
x)
·
(
A
x) = x
·
x =
|
x
|
2
. Since both are positive, we
know that |Ax| = |x|.
It is important to note that orthogonal matrices are isometries, but not all
isometries are orthogonal. For example, translations are isometries but are not
represented by orthogonal matrices, since they are not linear maps and cannot
be represented by matrices at all! However, it is true that all linear isometries
can be represented by orthogonal matrices.
Definition (Orthogonal group O(n)). The orthogonal group is
O(n) = O
n
= O
n
(R) = {A ∈ GL
n
(R) : A
T
A = I},
i.e. the group of orthogonal matrices.
We will later show that this is the set of matrices that preserve distances in
R
n
.
Lemma. The orthogonal group is a group.
Proof.
We have to check that it is a subgroup of
GL
n
(
R
): It is non-empty,
since
I ∈
O(
n
). If
A, B ∈
O(
n
), then (
AB
−1
)(
AB
−1
)
T
=
AB
−1
(
B
−1
)
T
A
T
=
AB
−1
BA
−1
= I, so AB
−1
∈ O(n) and this is indeed a subgroup.
Proposition. det : O(n) → {±1} is a surjective group homomorphism.
Proof.
For
A ∈
O(
n
), we know that
A
T
A
=
I
. So
det A
T
A
= (
det A
)
2
= 1. So
det A = ±1. Since det(AB) = det A det B, it is a homomorphism. We have
det I = 1, det
−1 0 · · · 0
0 1 · · · 0
.
.
.
.
.
.
.
.
.
0
0 0 · · · 1
= −1,
so it is surjective.
Definition (Special orthogonal group
SO
(
n
)). The special orthogonal group is
the kernel of det : O(n) → {±1}.
SO(n) = SO
n
= SO
n
(R) = {A ∈ O(n) : det A = 1}.
By the isomorphism theorem, O(n)/SO(n)
∼
=
C
2
.
What’s wrong with matrices with determinant
−
1? Why do we want to
eliminate these? An important example of an orthogonal matrix with determinant
−
1 is a reflection. These transformations reverse orientation, and is often
unwanted.
Lemma. O(n) = SO(n) ∪
−1 0 · · · 0
0 1 · · · 0
.
.
.
.
.
.
.
.
.
0
0 0 · · · 1
SO(n)
Proof. Cosets partition the group.
8.4 Rotations and reflections in R
2
and R
3
Lemma. SO(2) consists of all rotations of R
2
around 0.
Proof.
Let
A ∈ SO
(2). So
A
T
A
=
I
and
det A
= 1. Suppose
A
=
a b
c d
.
Then
A
−1
=
d −b
−c a
.
So
A
T
=
A
−1
implies
ad − bc
= 1,
c
=
−b
,
d
=
a
.
Combining these equations we obtain
a
2
+
c
2
= 1. Set
a
=
cos θ
=
d
, and
c = sin θ = −b. Then these satisfies all three equations. So
A =
cos θ − sin θ
sin θ cos θ
.
Note that
A
maps (1
,
0) to (
cos θ, sin θ
), and maps (0
,
1) = (
− sin θ, cos θ
), which
are rotations by θ counterclockwise. So A represents a rotation by θ.
Corollary. Any matrix in O(2) is either a rotation around 0 or a reflection in a
line through 0.
Proof. If A ∈ SO(2), we’ve show that it is a rotation. Otherwise,
A =
1 0
0 −1
cos θ − sin θ
sin θ cos θ
=
cos θ − sin θ
− sin θ − cos θ
since O(2) =
SO
(2)
∪
1 0
0 −1
SO
(2). This has eigenvalues 1
, −
1. So it is a
reflection in the line of the eigenspace
E
1
. The line goes through 0 since the
eigenspace is a subspace which must include 0.
Lemma. Every matrix in SO(3) is a rotation around some axis.
Proof.
Let
A ∈ SO
(3). We know that
det A
= 1 and
A
is an isometry. The
eigenvalues
λ
must have
|λ|
= 1. They also multiply to
det A
= 1. Since we are
in
R
, complex eigenvalues come in complex conjugate pairs. If there are complex
eigenvalues
λ
and
¯
λ
, then
λ
¯
λ
=
|λ|
2
= 1. The third eigenvalue must be real and
has to be +1.
If all eigenvalues are real. Then eigenvalues are either 1 or
−
1, and must
multiply to 1. The possibilities are 1
,
1
,
1 and
−
1
, −
1
,
1, all of which contain an
eigenvalue of 1.
So pick an eigenvector for our eigenvalue 1 as the third basis vector. Then in
some orthonormal basis,
A =
a b 0
c d 0
0 0 1
Since the third column is the image of the third basis vector, and by orthogonality
the third row is 0, 0, 1. Now let
A
′
=
a b
c d
∈ GL
2
(R)
with
det A
′
= 1.
A
′
is still orthogonal, so
A
′
∈ SO
(2). Therefore
A
′
is a rotation
and
A =
cos θ − sin θ 0
sin θ cos θ 0
0 0 1
in some basis, and this is exactly the rotation through an axis.
Lemma. Every matrix in O(3) is the product of at most three reflections in
planes through 0.
Note that a rotation is a product of two reflections. This lemma effectively
states that every matrix in O(3) is a reflection, a rotation or a product of a
reflection and a rotation.
Proof.
Recall O(3) =
SO
(3)
∪
1 0 0
0 1 0
0 0 −1
SO
(3). So if
A ∈ SO
(3), we know
that
A
=
cos θ − sin θ 0
sin θ cos θ 0
0 0 1
in some basis, which is a composite of two
reflections:
A =
1 0 0
0 −1 0
0 0 1
cos θ sin θ 0
sin θ − cos θ 0
0 0 1
,
Then if
A ∈
1 0 0
0 1 0
0 0 −1
SO
(3), then it is automatically a product of three
reflections.
In the last line we’ve shown that everything in O(3)
\ SO
(3) can be written as
a product of three reflections, but it is possible that they need only 1 reflection.
However, some matrices do genuinely need 3 reflections, e.g.
−1 0 0
0 −1 0
0 0 −1
8.5 Unitary groups
The concept of orthogonal matrices only make sense if we are talking about
real matrices. If we are talking about complex, then instead we need unitary
matrices. To do so, we replace the transposition with the Hermitian conjugate.
It is defined by
A
†
= (
A
∗
)
T
with (
A
†
)
ij
=
A
∗
ji
, where the asterisk is the complex
conjugate. We still have
(i) (AB)
†
= B
†
A
†
(ii) (A
−1
)
†
= (A
†
)
−1
(iii) A
†
A = I ⇔ AA
†
= I ⇔ A
†
= A
−1
. We say A is a unitary matrix
(iv) det A
†
= (det A)
∗
Definition (Unitary group U(n)). The unitary group is
U(n) = U
n
= {A ∈ GL
n
(C) : A
†
A = I}.
Lemma.
det
: U(
n
)
→ S
1
, where
S
1
is the unit circle in the complex plane, is a
surjective group homomorphism.
Proof.
We know that 1 =
det I
=
det A
†
A
=
| det A|
2
. So
| det A|
= 1. Since
det AB = det A det B, it is a group homomorphism.
Now given
λ ∈ S
1
, we have
λ 0 · · · 0
0 1 · · · 0
.
.
.
.
.
.
.
.
.
0
0 0 0 1
∈
U(
n
). So it is surjective.
Definition (Special unitary group
SU
(
n
)). The special unitary group
SU
(
n
) =
SU
n
is the kernel of det U (n) → S
1
.
Similarly, unitary matrices preserve the complex dot product: (
A
x)
·
(
A
y) =
x · y.
9 More on regular polyhedra
In this section, we will look at the symmetry groups of the cube and the
tetrahedron.
9.1 Symmetries of the cube
Rotations
Recall that there are
|G
+
|
= 24 rotations of the group by the orbit-stabilizer
theorem.
Proposition. G
+
∼
=
S
4
, where G
+
is the group of all rotations of the cube.
Proof.
Consider
G
+
acting on the 4 diagonals of the cube. This gives a group
homomorphism
φ
:
G
+
→ S
4
. We have (1 2 3 4)
∈ im φ
by rotation around
the axis through the top and bottom face. We also (1 2)
∈ im φ
by rotation
around the axis through the mid-point of the edge connect 1 and 2. Since (1 2)
and (1 2 3 4) generate
S
4
(Sheet 2 Q. 5d),
im φ
=
S
4
, i.e.
ϕ
is surjective. Since
|S
4
| = |G
+
|, φ must be an isomorphism.
All symmetries
Consider the reflection in the mid-point of the cube
τ
, sending every point to
its opposite. We can view this as
−I
in
R
3
. So it commutes with all other
symmetries of the cube.
Proposition.
G
∼
=
S
4
× C
2
, where
G
is the group of all symmetries of the cube.
Proof.
Let
τ
be “reflection in mid-point” as shown above. This commutes with
everything. (Actually it is enough to check that it commutes with rotations
only)
We have to show that
G
=
G
+
⟨τ⟩
. This can be deduced using sizes: since
G
+
and
⟨τ⟩
intersect at
e
only, (i) and (ii) of the Direct Product Theorem gives
an injective group homomorphism
G
+
× ⟨τ⟩ → G
. Since both sides have the
same size, the homomorphism must be surjective as well. So
G
∼
=
G
+
× ⟨τ⟩
∼
=
S
4
× C
2
.
In fact, we have also proved that the group of symmetries of an octahedron
is
S
4
× C
2
since the octahedron is the dual of the cube. (if you join the centers
of each face of the cube, you get an octahedron)
9.2 Symmetries of the tetrahedron
Rotations
Let 1
,
2
,
3
,
4 be the vertices (in any order).
G
+
is just the rotations. Let it act
on the vertices. Then orb(1) = {1, 2, 3, 4} and stab(1) = { rotations in the axis
through 1 and the center of the opposite face } = {e,
2π
3
,
4π
3
}
So |G
+
| = 4 · 3 = 12 by the orbit-stabilizer theorem.
The action gives a group homomorphism
φ
:
G
+
→ S
4
. Clearly
ker φ
=
{e}
.
So
G
+
≤ S
4
and
G
+
has size 12. We “guess” it is
A
4
(actually it must be
A
4
since that is the only subgroup of
S
4
of order 12, but it’s nice to see why that’s
the case).
If we rotate in an axis through 1, we get (2 3 4)
,
(2 4 3). Similarly, rotating
through other axes through vertices gives all 3-cycles.
If we rotate through an axis that passes through two opposite edges, e.g.
through 1-2 edge and 3-4 edge, then we have (1 2)(3 4) and similarly we obtain
all double transpositions. So
G
+
∼
=
A
4
. This shows that there is no rotation
that fixes two vertices and swaps the other two.
All symmetries
Now consider the plane that goes through 1, 2 and the mid-point of 3 and 4.
Reflection through this plane swaps 3 and 4, but doesn’t change 1
,
2. So now
stab
(1) =
⟨
(2 3 4)
,
(3
,
4)
⟩
∼
=
D
6
(alternatively, if we want to fix 1, we just move
2, 3, 4 around which is the symmetries of the triangular base)
So
|G|
= 4
·
6 = 24 and
G
∼
=
S
4
(which makes sense since we can move
any of its vertices around in any way and still be a tetrahedron, so we have all
permutations of vertices as the symmetry group)
10 M¨obius group
10.1 M¨obius maps
We want to study maps
f
:
C → C
in the form
f
(
z
) =
az+b
cz+d
with
a, b, c, d ∈ C
and ad − bc = 0.
We impose
ad − bc
= 0 or else the map will be constant: for any
z, w ∈ C
,
f
(
z
)
− f
(
w
) =
(az+b)(cw+d)−(aw+b)(cz+d)
(cw+d)(cz+d)
=
(ad−bc)(z−w)
(cw+d)(cz+d)
. If
ad − bc
= 0, then
f
is constant and boring (more importantly, it will not be invertible).
If
c
= 0, then
f
(
−
d
c
) involves division by 0. So we add
∞
to
C
to form
the extended complex plane (Riemann sphere)
C ∪ {∞}
=
C
∞
(cf. Vectors and
Matrices). Then we define
f
(
−
d
c
) =
∞
. We call
C
∞
a one-point compactification
of
C
(because it adds one point to
C
to make it compact, cf. Metric and Topology).
Definition (M¨obius map). A M¨obius map is a map from
C
∞
→ C
∞
of the form
f(z) =
az + b
cz + d
,
where
a, b, c, d ∈ C
and
ad − bc
= 0, with
f
(
−
d
c
) =
∞
and
f
(
∞
) =
a
c
when
c
= 0.
(if c = 0, then f (∞) = ∞)
Lemma. The M¨obius maps are bijections C
∞
→ C
∞
.
Proof.
The inverse of
f
(
z
) =
az+b
cz+d
is
g
(
z
) =
dz−b
−cz+a
, which we can check by
composition both ways.
Proposition. The M¨obius maps form a group
M
under function composition.
(The M¨obius group)
Proof. The group axioms are shown as follows:
0.
If
f
1
(
z
) =
a
1
z+b
1
c
1
z+d
1
and
f
2
(
z
) =
a
2
z+b
2
c
2
z+d
2
, then
f
2
◦f
1
(
z
) =
a
2
a
1
z+b
1
c
1
z+d
1
+ b
2
c
2
a
1
z+b
1
c
1
z+d
1
+ d
2
=
(a
1
a
2
+ b
2
c
1
)z + (a
2
b
1
+ b
2
d
1
)
(c
2
a
1
+ d
2
c
1
)z + (c
2
b
1
+ d
1
d
2
)
. Now we have to check that
ad − bc
= 0:
we have (
a
1
a
2
+
b
2
c
1
)(
c
2
b
1
+
d
1
d
2
)
−
(
a
2
b
1
+
b
2
d
1
)(
c
2
a
1
+
d
2
c
1
) = (
a
1
d
1
−
b
1
c
1
)(a
2
d
2
− b
2
c
2
) = 0.
(This works for
z
=
∞, −
d
1
c
1
. We have to manually check the special cases,
which is simply yet more tedious algebra)
1. The identity function is 1(z) =
1z+0
0+1
which satisfies ad − bc = 0.
2.
We have shown above that
f
−1
(
z
) =
dz−b
−cz+a
with
da − bc
= 0, which are
also M¨obius maps
3. Composition of functions is always associative
M
is not abelian. e.g.
f
1
(
z
) = 2
z
and
f
2
(
z
) =
z
+ 1 are not commutative:
f
1
◦ f
2
(z) = 2z + 2 and f
2
◦ f
1
(z) = 2z + 1.
Note that the point at “infinity” is not special.
∞
is no different to any other
point of the Riemann sphere. However, from the way we write down the M¨obius
map, we have to check infinity specially. In this particular case, we can get quite
far with conventions such as
1
∞
= 0,
1
0
= ∞ and
a·∞
c·∞
=
a
c
.
Clearly
az+b
cz+d
=
λaz+λb
λcz+λd
for any
λ
= 0. So we do not have a unique represen-
tation of a map in terms of
a, b, c, d
. But
a, b, c, d
does uniquely determine a
M¨obius map.
Proposition. The map
θ
:
GL
2
(
C
)
→ M
sending
a b
c d
7→
az + b
cz + d
is a
surjective group homomorphism.
Proof.
Firstly, since the determinant
ad−bc
of any matrix in
GL
2
(
C
) is non-zero,
it does map to a M¨obius map. This also shows that θ is surjective.
We have previously calculated that
θ(A
2
) ◦ θ(A
1
) =
(a
1
a
2
+ b
2
c
1
)z + (a
2
b
1
+ b
2
d
1
)
(c
2
a
1
+ d
2
c
1
)z + (c
2
b
1
+ d
1
d
2
)
= θ(A
2
A
1
)
So it is a homomorphism.
The kernel of θ is
ker(θ) =
A ∈ GL
2
(C) : (∀z) z =
az + b
cz + d
We can try different values of
z
:
z
=
∞ ⇒ c
= 0;
z
= 0
⇒ b
= 0;
z
= 1
⇒ d
=
a
.
So
ker θ = Z = {λI : λ ∈ C, λ = 0},
where I is the identity matrix and Z is the centre of GL
2
(C).
By the isomorphism theorem, we have
M
∼
=
GL
2
(C)/Z
Definition (Projective general linear group
PGL
2
(
C
)). (Non-examinable) The
projective general linear group is
PGL
2
(C) = GL
2
(C)/Z.
Since
f
A
=
f
B
iff
B
=
λA
for some
λ
= 0 (where
A, B
are the corresponding
matrices of the maps), if we restrict
θ
to
SL
2
(
C
), we have
θ|
SL
2
(C)
:
SL
2
(
C
)
→ M
is also surjective. The kernel is now just {±I}. So
M
∼
=
SL
2
(C)/{±I} = PSL
2
(C)
Clearly PSL
2
(C)
∼
=
PGL
2
(C) since both are isomorphic to the M¨obius group.
Proposition. Every M¨obius map is a composite of maps of the following form:
(i) Dilation/rotation: f(z) = az, a = 0
(ii) Translation: f(z) = z + b
(iii) Inversion: f(z) =
1
z
Proof. Let
az+b
cz+d
∈ M.
If c = 0, i.e. g(∞) = ∞, then g(z) =
a
d
z +
b
d
, i.e.
z 7→
a
d
z 7→
a
d
z +
b
d
.
If
c
= 0, let
g
(
∞
) =
z
0
, Let
h
(
z
) =
1
z−z
0
. Then
hg
(
∞
) =
∞
is of the above form.
We have
h
−1
(
w
) =
1
w
+
z
0
being of type (iii) followed by (ii). So
g
=
h
−1
(
hg
) is
a composition of maps of the three forms listed above.
Alternatively, with sufficient magic, we have
z 7→ z +
d
c
7→
1
z +
d
c
7→ −
ad + bc
c
2
(z +
d
c
)
7→
a
c
−
ad + bc
c
2
(z +
d
c
)
=
az + b
cz + d
.
Note that the non-calculation method above can be transformed into another
(different) composition with the same end result. So the way we compose a
M¨obius map from the “elementary” maps are not unique.
10.2 Fixed points of M¨obius maps
Definition (Fixed point). A fixed point of f is a z such that f(z) = z.
We know that any M¨obius map with
c
= 0 fixes
∞
. We also know that
z → z
+
b
for any
b
= 0 fixes
∞
only, where as
z 7→ az
for
a
= 0
,
1 fixes 0 and
∞
. It turns out that you cannot have more than two fixed points, unless you
are the identity.
Proposition. Any M¨obius map with at least 3 fixed points must be the identity.
Proof.
Consider
f
(
z
) =
az+b
cz+d
. This has fixed points at those
z
which satisfy
az+b
cz+d
=
z ⇔ cz
2
+ (
d − a
)
z − b
= 0. A quadratic has at most two roots, unless
c = b = 0 and d = a, in which the equation just says 0 = 0.
However, if c = b = 0 and d = a, then f is just the identity.
Proposition. Any M¨obius map is conjugate to
f
(
z
) =
νz
for some
ν
= 0 or to
f(z) = z + 1.
Proof.
We have the surjective group homomorphism
θ
:
GL
2
(
C
)
→ M
. The
conjugacy classes of GL
2
(C) are of types
λ 0
0 µ
7→ g(z) =
λz + 0
0z + µ
=
λ
µ
z
λ 0
0 λ
7→ g(z) =
λz + 0
0z + λ
= 1z
λ 1
0 λ
7→ g(z) =
λz + 1
λ
= z +
1
λ
But the last one is not in the form
z
+ 1. We know that the last
g
(
z
) can
also be represented by
1
1
λ
0 1
, which is conjugate to
1 1
0 1
(since that’s its
Jordan-normal form). So z +
1
λ
is also conjugate to z + 1.
Now we see easily that (for
ν
= 0
,
1),
νz
has 0 and
∞
as fixed points,
z
+ 1
only has ∞. Does this transfer to their conjugates?
Proposition. Every non-identity has exactly 1 or 2 fixed points.
Proof.
Given
f ∈ M
and
f
=
id
. So
∃h ∈ M
such that
hfh
−1
(
z
) =
νz
. Now
f
(
w
) =
w ⇔ hf
(
w
) =
h
(
w
)
⇔ hfh
−1
(
h
(
w
)) =
h
(
w
). So
h
(
w
) is a fixed point
of
hfh
−1
. Since
h
is a bijection,
f
and
hfh
−1
have the same number of fixed
points.
So
f
has exactly 2 fixed points if
f
is conjugate to
νz
, and exactly 1 fixed
point if f is conjugate to z + 1.
Intuitively, we can show that conjugation preserves fixed points because if we
conjugate by
h
, we first move the Riemann sphere around by
h
, apply
f
(that
fixes the fixed points) then restore the Riemann sphere to its original orientation.
So we have simply moved the fixed point around by h.
10.3 Permutation properties of M¨obius maps
We have seen that the M¨obius map with three fixed points is the identity. As a
corollary, we obtain the following.
Proposition. Given
f, g ∈ M
. If
∃z
1
, z
2
, z
3
∈ C
∞
such that
f
(
z
i
) =
g
(
z
i
), then
f = g. i.e. every M¨obius map is uniquely determined by three points.
Proof.
As M¨obius maps are invertible, write
f
(
z
i
) =
g
(
z
i
) as
g
−1
f
(
z
i
) =
z
i
. So
g
−1
f has three fixed points. So g
−1
f must be the identity. So f = g.
Definition (Three-transitive action). An action of
G
on
X
is called three-
transitive if the induced action on
{
(
x
1
, x
2
, x
3
)
∈ X
3
:
x
i
pairwise disjoint}
,
given by g(x
1
, x
2
, x
3
) = (g(x
1
), g(x
2
), g(x
3
)), is transitive.
This means that for any two triples
x
1
, x
2
, x
3
and
y
1
, y
2
, y
3
of distinct elements
of X, there exists g ∈ G such that g(x
i
) = y
i
.
If this g is always unique, then the action is called sharply three transitive
This is a really weird definition. The reason we raise it here is that the
M¨obius map satisfies this property.
Proposition. The M¨obius group M acts sharply three-transitively on C
∞
.
Proof.
We want to show that we can send any three points to any other three
points. However, it is easier to show that we can send any three points to 0
,
1
, ∞
.
Suppose we want to send
z
1
→ ∞, z
2
7→
0
, z
3
7→
1. Then the following works:
f(z) =
(z − z
2
)(z
3
− z
1
)
(z − z
1
)(z
3
− z
2
)
If any term
z
i
is
∞
, we simply remove the terms with
z
i
, e.g. if
z
1
=
∞
, we have
f(z) =
z−z
2
z
3
−z
2
.
So given also
w
1
, w
2
, w
3
distinct in
C
∞
and
g ∈ M
sending
w
1
7→ ∞, w
2
7→
0, w
3
7→ 1, then we have g
−1
f(z
i
) = w
i
.
The uniqueness of the map follows from the fact that a M¨obius map is
uniquely determined by 3 points.
3 points not only define a M¨obius map uniquely. They also uniquely define
a line or circle. Note that on the Riemann sphere, we can think of a line as a
circle through infinity, and it would be technically correct to refer to both of
them as “circles”. However, we would rather be clearer and say “line/circle”.
We will see how M¨obius maps relate to lines and circles. We will first recap
some knowledge about lines and circles in the complex plane.
Lemma. The general equation of a circle or straight line in C is
Az¯z +
¯
Bz + B¯z + C = 0,
where A, C ∈ R and |B|
2
> AC.
A
= 0 gives a straight line. If
A
= 0
, B
= 0, we have a circle centered at the
origin. If C = 0, the circle passes through 0.
Proof.
This comes from noting that
|z − B|
=
r
for
r ∈ R >
0 is a circle;
|z − a|
=
|z − b|
with
a
=
b
is a line. The detailed proof can be found in Vectors
and Matrices.
Proposition. M¨obius maps send circles/straight lines to circles/straight lines.
Note that it can send circles to straight lines and vice versa.
Alternatively, M¨obius maps send circles on the Riemann sphere to circles on
the Riemann sphere.
Proof.
We can either calculate it directly using
w
=
az+b
cz+d
⇔ z
=
dw−b
−cw+a
and
substituting
z
into the circle equation, which gives
A
′
w ¯w
+
¯
B
′
w
+
B
′
¯w
+
C
′
= 0
with A
′
, C
′
∈ R.
Alternatively, we know that each M¨obius map is a composition of translation,
dilation/rotation and inversion. We can check for each of the three types. Clearly
dilation/rotation and translation maps a circle/line to a circle/line. So we simply
do inversion: if w = z
−1
Az¯z +
¯
Bz + B¯z + C = 0
⇔ Cw ¯w + Bw +
¯
B ¯w + A = 0
Example. Consider
f
(
z
) =
z−i
z+i
. Where does the real line go? The real line
is simply a circle through 0
,
1
, ∞
.
f
maps this circle to the circle containing
f(∞) = 1, f(0) = −1 and f(1) = −i, which is the unit circle.
Where does the upper half plane go? We know that the M¨obius map is
smooth. So the upper-half plane either maps to the inside of the circle or the
outside of the circle. We try the point
i
, which maps to 0. So the upper half
plane is mapped to the inside of the circle.
10.4 Cross-ratios
Finally, we’ll look at an important concept known as cross-ratios. Roughly
speaking, this is a quantity that is preserved by M¨obius transforms.
Definition (Cross-ratios). Given four distinct points
z
1
, z
2
, z
3
, z
4
∈ C
∞
,
their
cross-ratio is [
z
1
, z
2
, z
3
, z
4
] =
g
(
z
4
), with
g
being the unique M¨obius map that
maps z
1
7→ ∞, z
2
7→ 0, z
3
7→ 1. So [∞, 0, 1, λ] = λ for any λ = ∞, 0, 1. We have
[z
1
, z
2
, z
3
, z
4
] =
z
4
− z
2
z
4
− z
1
·
z
3
− z
1
z
3
− z
2
(with special cases as above).
We know that this exists and is uniquely defined because
M
acts sharply
three-transitively on C
∞
.
Note that different authors use different permutations of 1
,
2
,
3
,
4, but they
all lead to the same result as long as you are consistent.
Lemma. For z
1
, z
2
, z
3
, z
4
∈ C
∞
all distinct, then
[z
1
, z
2
, z
3
, z
4
] = [z
2
, z
1
, z
4
, z
3
] = [z
3
, z
4
, z
1
, z
2
] = [z
4
, z
3
, z
2
, z
1
]
i.e. if we perform a double transposition on the entries, the cross-ratio is retained.
Proof. By inspection of the formula.
Proposition. If f ∈ M, then [z
1
, z
2
, z
3
, z
4
] = [f(z
1
), f(z
2
), f(z
3
), f(z
4
)].
Proof.
Use our original definition of the cross ratio (instead of the formula). Let
g be the unique M¨obius map such that [z
1
, z
2
, z
3
, z
4
] = g(z
4
) = λ, i.e.
z
1
g
7−→ ∞
z
2
7→ 0
z
3
7→ 1
z
4
7→ λ
We know that gf
−1
sends
f(z
1
)
f
−1
7−−→ z
1
g
7−→ ∞
f(z
2
)
f
−1
7−−→ z
2
g
7−→ 0
f(z
3
)
f
−1
7−−→ z
3
g
7−→ 1
f(z
4
)
f
−1
7−−→ z
4
g
7−→ λ
So [f (z
1
), f(z
2
), f(z
3
), f(z
4
)] = gf
−1
f(z
4
) = g(z
4
) = λ.
In fact, we can see from this proof that: given
z
1
, z
2
, z
3
, z
4
all distinct and
w
1
, w
2
, w
3
, w
4
distinct in
C
∞
, then
∃f ∈ M
with
f
(
z
i
) =
w
i
iff [
z
1
, z
2
, z
3
, z
4
] =
[w
1
, w
2
, w
3
, w
4
].
Corollary. z
1
, z
2
, z
3
, z
4
lie on some circle/straight line iff [z
1
, z
2
, z
3
, z
4
] ∈ R.
Proof.
Let
C
be the circle/line through
z
1
, z
2
, z
3
. Let
g
be the unique M¨obius
map with
g
(
z
1
) =
∞
,
g
(
z
2
) = 0,
g
(
z
3
) = 1. Then
g
(
z
4
) = [
z
1
, z
2
, z
3
, z
4
] by
definition.
Since we know that M¨obius maps preserve circle/lines,
z
4
∈ C ⇔ g
(
z
4
) is on
the line through ∞, 0, 1, i.e. g(z
4
) ∈ R.
11 Projective line (non-examinable)
We have seen in matrix groups that
GL
2
(
C
) acts on
C
2
, the column vectors.
Instead, we can also have GL
2
(C) acting on the set of 1-dimensional subspaces
(i.e. lines) of C
2
.
For any v
∈ C
2
, write the line generated by v as
⟨
v
⟩
. Then clearly
⟨
v
⟩
=
{λ
v :
λ ∈ C}
. Now for any
A ∈ GL
2
(
C
), define the action as
A⟨
v
⟩
=
⟨A
v
⟩
. Check
that this is well-defined: for any
⟨
v
⟩
=
⟨
w
⟩
, we want to show that
⟨A
v
⟩
=
⟨A
w
⟩
.
This is true because
⟨
v
⟩
=
⟨
w
⟩
if and only if w =
λ
v for some
λ ∈ C \ {
0
}
, and
then ⟨Aw⟩ = ⟨Aλv⟩ = ⟨λ(Av)⟩ = ⟨Av⟩.
What is the kernel of this action? By definition the kernel has to fix all lines.
In particular, it has to fix our magic lines generated by
1
0
,
0
1
and
1
1
. Since
we want
A⟨
1
0
⟩
=
⟨
1
0
⟩
, so we must have
A
1
0
=
λ
0
for some
λ
. Similarly,
A
0
1
=
0
µ
. So we can write
A
=
λ 0
0 µ
. However, also need
A⟨
1
1
⟩
=
⟨
1
1
⟩
.
Since
A
is a linear function, we know that
A
1
1
=
A
1
0
+
A
0
1
=
λ
µ
. For the final
vector to be parallel to
1
1
, we must have
λ
=
µ
. So
A
=
λI
for some
I
. Clearly
any matrix of this form fixes any line. So the kernel Z = {λI : λ ∈ C \ {0}}.
Note that every line is uniquely determined by its slope. For any v =
(
v
1
, v
2
)
,
w = (
w
1
, w
2
), we have
⟨
v
⟩
=
⟨
w
⟩
iff
z
1
/z
2
=
w
1
/w
2
. So we have a
one-to-one correspondence from our lines to C
∞
, that maps ⟨
z
1
z
2
⟩ ↔ z
1
/z
2
.
Finally, for each A ∈ GL
2
(C), given any line ⟨
z
1
⟩, we have
a b
c d
z
1
=
az + b
cz + d
↔
az + b
cz + d
So
GL
2
(
C
) acting on the lines is just “the same” as the M¨obius groups acting
on points.