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MATH40003 Linear Algebra and Groups
1 Groups and Subgroups
Lecture
10
1.1 Binary operations; groups; basic facts
Definition 1.1. Suppose S is a set. A binary operation ∗ on S assigns to each ordered
pair (a, b) of elements of S an element a∗b of S . More formally, ∗ is a function S×S → S .
(Where S × S is the Cartesian product set {(a, b) : a, b ∈ S} .)
This is a very general notion. Here are some examples involving M2(R): not all of them will
lead to groups!
Examples: Let S =M2(R). The following are binary operations on S .
(1) a ∗ b = ab , the usual matrix multiplication.
(2) ∗ is matrix addition.
(3) a ∗ b = a for all a, b ∈ A (‘projection onto the first coordinate’)
(4) a ∗ b = ab− ba (called the Lie product).
(5) Let S1 = {a ∈ M2(R) : a is invertible} . For a, b ∈ S1 let a ∗ b be the usual matrix
product ab . As the product of two invertible matrices is invertible, this is a binary operation
on S1 .
(6) If in (5) we had taken a ∗ b = a + b , this would not be a binary operation on S1 as the
sum of two invertible matrices need not be invertible.
You will have seen this before:
Definition 1.2. A binary operation ∗ on a set S is associative if
for all a, b, c ∈ S we have a ∗ (b ∗ c) = (a ∗ b) ∗ c.
Exercise: Which of the binary operations above are associative?
Associativity means that we can unambiguously write an expression such as
((a1 ∗ a2) ∗ (a3 ∗ a4)) ∗ a5
as a1 ∗ a2 ∗ a3 ∗ a4 ∗ a5 . The bracketing is redundant. From now on, we will usually do this
when dealing with associative operations.
Puzzle: How many ways are there of bracketing a1 ∗ a2 ∗ . . . ∗ an? Try it for n = 4, 5. Can
you find a general expression?
Now, here is the main definition.
Definition 1.3. A group (G, ∗) consists of a set G with a binary operation ∗ on G satisfying
the following:
2
G1 (Associativity) For all g, h, k ∈ G we have
(g ∗ h) ∗ k = g ∗ (h ∗ k).
G2 (Identity axiom) There exists an element e ∈ G such that for all g ∈ G we have
e ∗ g = g ∗ e = g.
We will show that G1 and G2 imply that there is a unique such e ∈ G , which we will denote
by e or eG and call it the identity element of the group.
G3 (Existence of inverses) With e as in G2, for all g ∈ G there is an element h ∈ G such
that
g ∗ h = h ∗ g = e.
We will show that the h here is uniquely determined by g : we call it the inverse of g and
usually denote it by h−1.
Remarks: Here, G1,G2,G3 are called the group axioms. In some books you will see that
the following is also included as an axiom:
G0 (Closure) If g, k ∈ G , then g ∗ k ∈ G .
It’s not necessary to include this as we are assuming that ∗ is a binary operation on G and
so G0 holds anyway.
Remarks 1.4. Here are some remarks on simplifications to notation and terminology which
we shall use throughout.
(1) It is more common to use . instead of ∗ for the group operation: so from now on, you
will see g.h rather than g ∗ h . In fact, just as in ordinary arithmetic, we often omit the .
and just write gh instead of g.h . We call the operation the product in the group.
(2) A group (G, ∗) is called abelian or commutative if for all g, h ∈ G we have g ∗ h = h ∗ g .
In such cases, we often write the operation as +. When we do this we write the identity
element as 0 and the inverse of g as −g .
Remarks 1.5. We now justify the claims made in Definition 1.3. Suppose (G, .) is a group.
(1) We have to show the following, using axioms G1 and G2.
If e, e′ ∈ G and for all g ∈ G
e.g = g.e = g and e′.g = g.e′ = g
then e = e′ .
3
Proof: Using these equations, we have e = e.e′ = e′ (exercise: say which of the equations
were used here!) 2
(2) We have to show the following using the group axioms.
Suppose g, g′, g′′ ∈ G and
gg′
(1)
= e
(2)
= g′g and gg′′
(3)
= e
(4)
= g′′g.
Then g′ = g′′ .
Proof: We have
(g′g)g′′
(2)
= eg′′ = g′′ and g′(gg′′)
(3)
= g′e = g′.
So by Associativity g′ = g′′ .
[Exercise: Why is the following not a proof? ‘From the equations, we have g′ = g−1 and
g′′ = g−1 , so g′ = g′′ .’]
We will use the following all the time, usually without mentioning that it’s what we are
using. It’s about manipulating equations in a group.
Lemma 1.6. (Equations in groups) Suppose (G, .) is a group and g, h ∈ G.
(1) For x ∈ G we have gx = h⇔ x = g−1h
(2) For y ∈ G we have yg = h⇔ y = hG−1 .
Proof: (1) ⇒: Multiply on the left by g−1 :
gx = h⇒ g−1(gx) = g−1h G1⇒ (g−1g)x = g−1h⇒ ex = g−1h⇒ x = g−1h.
⇐ : Check that all of the above arrows can be reversed.
(2) Similar, but multiply on the right by g−1 . 2
The following will look very familiar from matrix multiplication.
Lemma 1.7. (Inverse of a product) Suppose (G, .) is a group.
(1) If g, h ∈ G, then (gh)−1 = h−1g−1 .
(2) If g1, . . . , gn ∈ G, then
(g1g2 . . . gn)
−1 = g−1n . . . g
−1
2 g
−1
1 .
Proof: (1) Note that
(h−1g−1)(gh) = h−1(g−1)g)h = h−1eh = h−1h = e.
Similarly gh(h−1g−1) = e .
(2) Prove this by induction on n . The base case n = 2 is (1). 2
[Before the next lecture, review the material in the Introductory Module about Bijections
(Sections 2.3 and 2.4 there).]
4
1.2 Examples of groups
Examples 1.8. Examples from fields Lecture
11
(1) R with the operation + is a group. The identity element is 0 and the inverse of a ∈ R
is −a . We refer to this as the additive group of the field R .
(2) R× = R \ {0} with the operation . is a group. The identity element is 1 and the inverse
of a ∈ R× is a−1 . We refer to this as the multiplicative group of the field R .
(3) Examples (1), (2) work with any field, not just R .
(4) If F is any field and n ∈ N , then (F n,+) is a group.
(5) If V is a vector space (over a field F ) then (V,+) is a group.
(6) Let n ∈ N and suppose F is a field. The general linear group GL(n, F ) (or GLn(F )) is
the set G of n × n invertible matrices (over F ) and operation matrix multiplication. This
is a group:
• Binary operation: If A,B ∈ G , you can check that AB ∈ G (because the inverse of
AB is B−1A−1 ).
• G1, Associativity: If A,B,C ∈ G , then A(BC) = (AB)C : this is a general property
of matrix multiplication.
• G2, Existence of identity element: The identity matrix In ∈ G has the required
property.
• G3, Existence of inverses: this follows from the definition of G .
Note: GL(1, F ) is the multiplicative group F× .
Definition 1.9. The Symmetric Groups.
Suppose X is any (non-empty) set. (For example, X = {1, 2, . . . , n} .) A permutation of X
is a bijection α : X → X .
For example, if X = {1, 2, 3, 4} let α be the map denoted by:(
1 2 3 4
2 4 1 3
)
with α(1) = 2, α(2) = 4, etc.
More generally we use the following ‘two-row’ notation for permutations on the set X =
[n] = {1, 2, . . . , n} . We denote a permutation α : [n]→ [n] by:(
1 2 . . . n
α(1) α(2) . . . α(n)
)
.
5
Note that the second row here is a ‘rearrangement’ of 1, 2, 3, . . . , n .
Exercise: If α, β : X → X are permutations, then so is the composition α ◦ β : X → X
(α ◦ β)(x) = α(β(x)), for x ∈ X.
- See the Introductory Module if you are unsure about this.
Example: If
α =
(
1 2 3 4
2 4 1 3
)
and β =
(
1 2 3 4
2 1 4 3
)
then
α ◦ β =
(
1 2 3 4
4 2 3 1
)
and β ◦ α =
(
1 2 3 4
1 3 2 4
)
.
Let Sym(X) denote the set of all permutations of X . If n ∈ N and X = [n] = {1, 2, . . . , n} ,
we will also denote this by Sym(n) or Sn .
Theorem 1.10. Suppose X is any (non-empty) set. Then Sym(X) with the operation ◦ of
composition is a group. It is called the symmetric group on X .
Proof: By the above exercise, we know that ◦ is a binary operation on Sym(X). We proceed
to check the group axioms.
G1 (Associativity): Composition of functions is associative: if α, β, γ ∈ Sym(X), then
α ◦ (β ◦ γ) and (α ◦ β) ◦ γ are the same function X → X (for x ∈ X they both send x to
α(β(γ(x)))).
G2 (identity element): The identtity function ι : X → X , with ι(x) = x for all x ∈ X , is in
Sym(X) and has the required properties.
G3 (Existence of inverses): If α ∈ Sym(X), then it is a bijection and so has an inverse α−1
which is also a bijection: α−1(y) = x⇔ α(x) = y . 2
Remarks: If α, β ∈ Sym(X) we will often write αβ instead of α ◦ β . Remember that this
is composition of functions and so this means ‘apply β first, then α .’
Warning: Some books or lecture notes will use the notation the other way round.
Example/ Exercise: Consider the following elements of S6 :
α =
(
1 2 3 4 5 6
2 3 4 5 6 1
)
and β =
(
1 2 3 4 5 6
1 6 5 4 3 2
)
.
Compute αβ, βα, α−1, β−1, αβα−1.
Label the vertices of a regular hexagon 1, 2, 3, 4, 5, 6 (clockwise) and think about how the
permutations α, β and these other permutations correspond to symmetries of the regular
hexagon.
Definition 1.11. We say that a group (G, .) is a finite group if the set G is a finite set. In
this case, the order of (G, .) is |G| , the number of elements in the group.
6
Theorem 1.12. If n ∈ N, then |Sn| = n!
Proof: We have to count the number of permutations
α =
(
1 2 . . . n
a1 a2 . . . an
)
where a1, . . . , an are 1, . . . , n in some order.
There are
n choices for a1 ;
n− 1 choices for a2 ;
n− 2 choices for a3 ;
. . .
So there are a total of n(n− 1)(n− 2) . . . 2.1 = n! possibilities for α . 2
1.3 Powers and Subgroups
Lecture
12Definition 1.13. Suppose (G, ·) is a group. For g ∈ G , we let
g0 = e, g1 = g, g2 = g · g, g3 = g · g · g, . . . .
More precisely, for n ∈ N we define inductively
g0 = e, g1 = g and gn+1 = gn · g.
We also define g−n = (g−1)n .
Using this as our definition, we can then prove the following.
Lemma 1.14. With this notation, if m,n ∈ Z, then
(o) gm+1 = gmg and gmg = ggm ;
(i) (gm)−1 = g−m ;
(ii) gmgn = gm+n = gngm ;
(iii) gmn = (gm)n .
Example: Let
g =
(
1 2 3 4
2 3 4 1
)
∈ S4.
Then
g2 =
(
1 2 3 4
3 4 1 2
)
, g3 =
(
1 2 3 4
4 1 2 3
)
= g−1, g4 = ι, g5 = g, . . . , g19 = g3
7
the last of these because 19 = 4.4 + 3 and so
g19 = (g4)4g3 = ι4g3 = g3
using the lemma.
Similarly if n ≡ k (mod 4), gn = gk .
Proof of Lemma: It’s not trivial to get all of the details correct, though it is quite tedious.
The proof should remind you of some things you discussed around Peano’s axioms in the
Introductory module.
(o) For the first part, if m ≥ 0, this is by definition. So assume m = −k where k ∈ N
and prove by induction on k that g−k+1 = g−kg . The base case is k = 1, which follows
from g0 = e = g−1g . For the inductive step suppose we know g−k+1 = g−kg . Then
g−(k+1)g = (g−1)k+1g = ((g−1)kg−1)g = g−k , as required (the first two equalities comes from
the definitions). The second part is an exercise.
(i) Prove for m ≥ 0 that (gm)(g−m) = e , by induction on m . Then deduce for m < 0.
(ii) We first prove this for n ≥ 0. We do this by induction on n , regarding m as fixed.
Base case n = 0: We have gm+0 = gm and gmg0 = gme = gm , so gm+0 = gmg0 .
Inductive step: Suppose we know that gm+n = gmgn . Then
gm+(n+1) = g(m+n)+1
(o)
= gm+ng = (gmgn)g = gm(gng) = gmgn+1
as required.
Exercise: Complete the proof of (ii) by considering the case n < 0.
(iii) Similar, using (ii). 2
Remark: on additive notation. If our group is (G,+) and n ∈ N , we write g + . . .+ n (n
times) as ng (not gn ).
Definition 1.15. Suppose (G, .) is a group and H ⊆ G . We say that H is a subgroup of
G (or of (G, .)) if H with the binary operation from G is a group, that is:
for all h1, h2 ∈ H we have h1.h2 ∈ H
and (H, .) satisfies axioms G1,G2,G3. We will write H ≤ G to indicate that H is a subgroup
of G .
Examples: (1) G is a subgroup of G ;
(2) {e} is a subgroup of G .
In practice, what we use to decide whether a given subset of a group is a subgroup is the
direction ⇐ of the following.
8
Theorem 1.16. (Test for a subgroup) Suppose (G, .) is a group and H ⊆ G. Then H is a
subgroup of G if and only if
(1) H ̸= ∅;
(2) for all h1, h2 ∈ H we have h1h2 ∈ H (closure under the group operation);
(3) for all h ∈ H we have h−1 ∈ H (closure under inverses).
Example: If g ∈ G , let H = {gm : m ∈ Z} . This is a subgroup of G , by 1.16 and 1.14.
Proof of 1.16: ⇐: Suppose (1), (2), (3) hold for H . By (2), the multiplication . in G gives
a binary operation on H . So we need to check that (H, .) satisfies the group axioms.
G1 follows from associativity in G .
For G2, it is enough to show that eG ∈ H . By (1), there is some h ∈ H . By (3), h−1 ∈ H .
Then by (2), hh−1 ∈ H . So eG ∈ H .
Then G3 follows from (3).
⇒: If H is a subgroup of G , then (2) holds by definition.
By G2, we know that H ̸= ∅ , so (1) holds.
For (3) we first show that eG ∈ H . Let h ∈ H . As (H, .) is a group, there is some x ∈ H
with hx = h . But the only solution to this equation in G is x = eG . So eG ∈ H . Similarly,
the only solution to hh′ = eG in G is h′ = h−1 . As H is a group, there must be a solution
in H , so h−1 ∈ H , as required for (3). 2
Remark: The direction ⇒ shows that if H is a subgroup of G then eG ∈ H and inverses
are the same in H as they are in G .
Before we give some examples of using the test, we formalise the example given before the
above proof by giving some terminology.
Definition 1.17. (1) Suppose (G, .) is a group and g ∈ G . The cyclic subgroup generated
by g is ⟨g⟩ = {gm : m ∈ Z}.
(2) We say that a group G is cyclic if there is g ∈ G with ⟨g⟩ = G . In this case, g is called
a generator of G .
Examples 1.18. (1) Let G = GL(n,R) (the group of n × n invertible matrices over R).
We give some subgroups.
(1) Let H = {g ∈ G : det(g) = 1} . We use the subgroup test to check that this is a subgroup
of G :
H ̸= ∅ as In ∈ H .
H is closed under . as det(g1g2) = det(g1)det(g2) so if g1, g2 ∈ H , then g1g2 ∈ H .
H is closed under inverses as det(g−1) = 1/det(g) for g ∈ G , so if h ∈ H , then h−1 ∈ H .
9
(2) You can use the subgroup test to show that
K = {
(
cos θ − sin θ
sin θ cos θ
)
: θ ∈ R} ≤ GL2(R).
Note K consists of linear maps R2 → R2 given by rotations about 0. The subgroup
K is abelian, but not cyclic (for example, K is uncountable, whereas any cyclic group is
countable).
(3) We have:
2Z ≤ Z ≤ Q ≤ R ≤ (C,+),
where 2Z consists of the even integers. Note that Z = ⟨1⟩ and 2Z = ⟨2⟩ are cyclic, but
none of the other groups is cyclic.
(4) Consider the unit circle in the complex plane:
U = {eiθ : θ ∈ R} = {z ∈ C : |z| = 1} = {z ∈ C : zz¯ = 1}.
This is a subgroup of (C×, .), the multiplicative group of the field of complex numbers. It is
abelian, but not cyclic.
[Puzzle: What is the relationship to the group K in (2)?]
(5) Let n ∈ N and ζ = e2πi/n ∈ C× . So ζn = 1. Then ⟨ζ⟩ ≤ U and
⟨ζ⟩ = {1, ζ, ζ2, . . . , ζn−1}
is a cyclic group of order n .
[Note that if you draw the elements of ⟨ζ⟩ on an Argand diagram, they form the vertices of
a regular n-gon on the unit circle.]
(6) Suppose F is any field and n ∈ N . Consider the group (F n,+). Any subspace of F n is
a subgroup of this, but the converse is not necessarily true. For example Q2 is a subgroup
of R2 , but it is not a subspace.
