1Groups and homomorphisms
IA Groups
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.