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University of Cambridge Lecture Notes Michaelmas 2011
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Michaelmas 2011
GroupsUniversity of Cambridge Part IA Mathematical Tripos
Lecturer: Prof. J. Saxl
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στ
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Notes by L. Shen
Last Updated: 22/1/2013
Part IA: Groups
Contents
1 Groups and Permutations 3
1.1 Group Axiom 3
1.2 Permutations, cycles and transpositions 5
1.3 Sign of a permutation 7
1.4 Subgroups and homomorphisms 10
1.5 Modular Arithmetic 12
1.6 Cyclic and Dihedral groups 13
2 Lagrange’s Theorem and Small Groups 15
2.1 Lagrange’s Theorem 15
2.2 Euler-Fermat Theorem 18
2.3 Direct Product of Groups 19
2.4 Groups of Small Order 20
3 Group Actions 23
3.1 Orbits and transitivity 24
3.2 Groups of symmetries of a regular n-gon 25
3.3 Actions from groups and Cayley’s theorem 27
3.4 Conjugacy classes in Sn and An 30
3.5 Cauchy’s theorem 31
4 Homomorphisms, Normal subgroups and Quotients 33
4.1 Homomorphisms and Normal subgroups 33
4.2 Quotient groups 34
4.3 Simplicity of A5 37
5 Matrix and Orthogonal Groups 39
5.1 Matrix Groups 39
5.2 Orthogonal Groups 41
6 Mobius transformations 45
6.1 Mobius group 45
6.2 Cross ratios 51
– 1 –
Part IA: Groups
Schedule
Groups and Permutations
Axioms for groups. Permutations, cycles and transpositions. The sign of a permuta-
tion. Permutations of a set; the symmetric group. Subgroups and homomorphisms.
Symmetry groups as subgroups of general permutation groups. [3]
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. Examples from geo-
metry: symmetry groups of regular polygons, cube, tetrahedron. Cayley’s theorem
(every group is isomorphic to a subgroup of a permutation group). Conjugacy classes.
Cauchy’s theorem. Conjugacy in Sn and in An. Simple groups; simplicity of A5. [5]
Quotient groups
Normal subgroups, quotient groups and the isomorphism theorem. [4]
Matrix groups
The orthogonal and special orthogonal groups. Proof (in R3) that every element of the
orthogonal group is the product of reflections and every rotation in R3 has an axis.
Basis change as an example of conjugation. [3]
Mobius transformations
The Mobius group; cross-rations, preservation of circles, the point at infinity. Con-
jugation. Fixed points of Mobius maps and iteration. The general and special linear
groups; relation with the Mobius group. [4]
Disclaimer
These unofficial notes are based on Groups given by Prof. Saxl during Michaelmas
2011. Typos and mistakes are my fault entirely, comments to [email protected].
Li Shen, Pembroke College
– 2 –
Part IA: Groups
1 | Groups and Permutations
1.1 Group Axiom
Definition. We say that (G, ∗) is a group if G is a set and ∗ is a binary operation such
that the following properties are satisfied:
(a) Closure: if a, b ∈ G, then a ∗ b ∈ G;(b) Associativity: if a, b, c ∈ G, then (a ∗ b) ∗ c = a ∗ (b ∗ c);(c) Identity: ∃!e : a ∗ e = a = e ∗ a for any a ∈ G(d) Inverse: if a ∈ G, ∃!a−1 : a ∗ a−1 = e = a−1 ∗ a.
Examples of Groups. (Z,+), (Q,+), (R,+), (Q \ {0} ,×), ({±1} ,×), ({e} , ∗) and
the set {(a b
c d
)
| ad− bc = 1, a, b, c, d ∈ F
}
with respect to matrix multiplications.
Lemma 1.1 (Properties of Group elements).
(a) Uniqueness of inverse and identity.
(b) Inverse property. For a, b ∈ G, (ab)−1 = b−1a−1 and (a−1)−1
= a.
(c) Cancellation property. For a, b, x ∈ G then ax = bx⇒ a = b.
Proof. (a) Assume ∃e′ : ae′ = a∀a ∈ G, thus e′e = e = e′. Similarly, assume ∃b′ : b′a =
e∀a ∈ G, then b′e = b′(ab) = eb = b.
(b) Consider (ab) b−1a−1 = e and a−1a = e = aa−1.
(c) Consider (ax)x−1 = (bx)x−1.
Definition. Let A,B be sets. A function f : A → B is a rule which assigns to each
element a ∈ A, a unique element f(a) of B.
A function from A to B is bijective if, for each b ∈ B there exists a unique image a ∈ Awith f(a) = b.
If g : A → B and f : B → C, the composite function f ◦ g : A → C is defined by
f ◦ g(a) = f(g(a)) for any a ∈ A.
Note. If f : A→ B is bijective, then |A| = |B|. (|G| represents the size of G i.e. the
number of elements in the set G)
– 3 –
Part IA: Groups
Lemma 1.2. If g : A → B is a bijection and also f : B → C is a bijection, then
f ◦ g : A→ C is a bijection.
Proof. f ◦ g : A→ C is well-defined by definition. Consider that ∃b ∈ B : f(b) = c and
∃a ∈ A : g(a) = b =⇒ ∃c ∈ C : f ◦ g(a) = c.
Now a is unique since
f ◦ g(a1) = f ◦ g(a2) =⇒ g(a1) = g(a2)
and thus a1 = a2 due to injectivity of both f and g.
Remarks. g : A → B is surjective if ∀b ∈ B∃a ∈ A : g(a) = b. The function g is
said to be injective if whenever a1 6= a2, we have g(a1) 6= g(a2) and bijective if it is
both injective and surjective. We have essentially proved in (1.2) that the composite
of injections (surjections) is an injection (surjection).
Definition. A bijection from X onto X is a permutation of X .
Theorem 1.3. Let Sym(X) be the set of all permutations of the setX , then (Sym(X), ◦)is a group.
Proof. Closure: follows from (1.2);
Associativity: if f, g, h ∈ Sym(X), then
(f ◦ g) ◦ h(a) = (f ◦ g) (h(a)) .= f(g(h(a)))
= f ((g ◦ h) (a))= f ◦ (g ◦ h)(a).
Identity: consider the function i : X → X , x 7→ x∀x ∈ X , then f ◦ i = f = i◦f∀x ∈ X .
Inverse: let f ∈ Sym(X), consider the function g : X → X , b 7→ b−1 ∀b ∈ X where b−1
is the unique f pre-image of b. g is well-defined and ∈ Sym(X), and thus
f ◦ g(x) = x = g ◦ f(x) =⇒ f ◦ g = e = g ◦ f
and so g = f−1.
Notation. If X is a finite set of n elements, we often write Sn for the group Sym(X)
where n is the degree of Sym(X). For example, the order of Sn is n!.
– 4 –
Part IA: Groups
For permutations f ∈ Sn,
f =
(1, 2, . . . , n
f(1), f(2), . . . , f(n)
)
.
Example. n = 3, X = {1, 2, 3}
e =
(1 2 3
1 2 3
)
, f =
(1 2 3
1 3 2
)
, g =
(1 2 3
2 3 1
)
,
g2 =
(1 2 3
3 1 2
)
, fg =
(1 2 3
3 2 1
)
, gf =
(1 2 3
2 1 3
)
,
where f = f−1 and g2 = g−1 and thus |S3| = 6.
Notation. g ∈ G, n ∈ Z, if n ∈ N, then
gn =
g ∗ g ∗ · · · ∗ g︸ ︷︷ ︸
n
if n > 0,
e if n = 0,
g−1 ∗ · · · ∗ g−1
︸ ︷︷ ︸
|n|
= (g−1)|n| = (g|n|)−1 if n < 0.
Exercise. G is a group, g ∈ G, i, j ∈ Z, then gigj = gi+j and (gi)j = gij.
Definition. Let g ∈ G. If ∃n ∈ Z n > 0, with gn = e, the least such integer is the
order of G, o(g). If there is no such n, we say the order of g is infinite.
1.2 Permutations, cycles and transpositions
Definition. If σ ∈ Sn is a k-cycle for some k ∈ N, then we write σ = (a1a2 . . . ak),
where
σ(ai) =
ai+1 if 1 6 i < k
a1 if i = k
ai if i ∈ X\ {a1, . . . , ak}where a1, a2, . . . , an are distinct points of X .
Remarks. The order of any k-cycle is k and we can see
(a1a2 . . . ak)−1 = (akak−1 . . . a1)
– 5 –
Part IA: Groups
and that
(a1a2 . . . ak) = (a2a3 . . . aka1) .
Definition. The two cycles σ = (a1 . . . ak) and τ = (b1 · · · bl) are disjoint if no ai equalsto a bj .
Lemma 1.4. Disjoint cycles σ, τ in Sn commute; i.e. στ = τσ.
Remark. If σ, τ are not disjoint, they do not commute in general.
Proof. Let x ∈ X, then we have 3 cases:
(a) If x ∈ {a1, . . . , ak}, then στ(x) = σ(x) and τσ(x) = σ(x).
(b) If x ∈ {b1, . . . , bl}, τσ(x) = τ(x). στ(x) = τ(x).
(c) If x ∈ X \ {a1, . . . , ak, b1, . . . , bl}, then x is fixed by both σ and τ, hence by both
στ and τσ.
Notation. Cycles of length 1 are often deleted, e.g.(1 2 3 4 5 6 7 8
2 4 7 1 5 8 6 3
)
= (124)(3768)
Theorem 1.5. Any permutation of the finite set X can be written as a unique product
of disjoint cycles.
Proof. Let π ∈ Sn and a ∈ X , consider a, π(a), π2(a), . . . . Since X is finite, ∃ smallest
positive integer l such that
πℓ(a) ∈{a, π2(a), . . . , πl−1(a)
}.
Then πl(a) = a : if πl(a) = πj(a) with 0 6 l < j, then πl−j(a) = a, so j = 0 by the
minimality of l.
If ∃b ∈ X not obtained yet, consider b, π(b), . . . , πk−1(b) where k is the smallest positive
integer such that πk(b) ∈{b, π (b) , . . . , πk−1 (b)
}, then πk(b) = b by a similar argument.
Writing {ali=0} = πi(a) and {bkj=0} = πj(b), noting that the bjs are disjoint from ai,
and since X is finite, this stops with all points of X listed.
Lemma 1.6. The order of π ∈ Sn is the lcm of the cycle lengths in the disjoint cycle
expression of π.
Proof. The disjoint cycles in π commute, so πm = e ⇐⇒ σm = e for each of the cycles
σ ∈ π. Also σm = e ⇐⇒ the length of σ divides m.
Definition. A 2-cycle is called a transposition.
– 6 –
Part IA: Groups
Lemma 1.7. Every permutation in Sn can be written as a product of transpositions.
Proof. Assuming (1.6), then
(a1a2 . . . ak) = (a1a2) (a2a3) · · · (ak−1ak)︸ ︷︷ ︸
k−1 transpositions
Remark. A k-cycle can be written as a product of k−1 transpositions, but not uniquely.Since e = (12)(12) = (13)(12)(12)(13).
1.3 Sign of a permutation
Definition. The sign of the permutation π is
sgn(π) = (−1)k
where k is the number of transpositions with product π.
Lemma 1.8. sgn:Sn → ±1 is a well defined function, i.e. if π = τ1τ2 · · · τk = τ ′1τ′2 · · · τ ′l
then (−1)k = (−1)l, with τi, τ ′j transpositions.
Proof. Let τ be any transposition, σ any permutation and c(σ) the number of cycles in
the disjoint cycle expression for σ (including the 1-cycle). Then claim c(στ) = c(σ)±1.
(a) If r, s are in different σ cycles, these become fused into one στ cycle, i.e. (left)
στ = (r s+ 1 · · · s r + 1 · · · r − 1)
(b) If r, s are in the same σ cycle, it splits into στ cycles, i.e. (right)
στ = (r s+ 1 · · · r − 1) (s r + 1 · · · s− 1)
r + 1
rr − 1
s− 1
s
s+ 1
στ
σ τ
b
b
b
bb
b
s+ 1 s
r + 1rr − 1b
bb
bb
– 7 –
Part IA: Groups
Hence (−1)c(στ) = − (−1)c(σ) and thus if we let σ = ıτ1τ2 · · · τk−1 and τ = τk, then
(−1)c(ıτ1τ2···τk) = − (−1)c(ıτ1τ2···τk−1)
= (−1)c(ı) (−1)k
(−1)c(ıτ ′1···τ ′k) = (−1)c(ı) (−1)l
so (−1)k = (−1)l.
Definition. π ∈ Sn is even if sgn(π) = +1 and odd if sgn(π) = −1.
Example. In S3, the elements e, (123) and (132) are even and the elements (12), (13)
and (23) are odd.
Lemma 1.9. sgn(π1 ◦ π2) = sgn(π1)sgn(π2).
Proof. If π1 = τ1 · · · τk, π2 = τ ′1 · · · τ ′ℓ with τi, τ ′j transpositions, then
π1π2 = τ1 · · · τkτ ′1 · · · τ ′l ,
so sgn(π1 ◦ π2) = (−1)k+l = (−1)k · (−1)l = sgn(π1)sgn(π2).
Corollary 1.10. The set An of all even permutations in Sn is a group under composition,
called the alternating group with |An| = n!/2.
Proof. Since
|An| = |{even permutations in Sn}| = |{odd permutations in Sn}| ,
let τ be an odd permutation and consider the mapping π → πτ , which takes {even in Sn} 7→{odd in Sn}, is a bijection between the 2 sets, so they have the same size n!/2.
Remarks. Note that a k-cycle is odd ⇐⇒ k is an even number (k− 1 transpositions).
Hence a permutation is odd ⇐⇒ the number of cycles of even length in its disjoint
cycle expression is odd.
Example. |A4| = 4!/2 = 12, the elements are
e, (12)(34) (123) (132)
(13)(24) (124) (142)
(14)(23) (134) (143)
(234) (243)
.
– 8 –
Part IA: Groups
*An alternative treatment of sgn
Definition. Given σ ∈ Sn, let x1, . . . , xn be distinct positive integers. Define
ǫ(σ) =∏
16i<j6n
xσ(j) − xσ(i)xj − xi
.
Example. (123), ǫ((123)) = 3−22−1· 1−23−1· 1−33−2
= +1
Lemma 1.11. ǫ(σ) = (−1)N(σ), where N(σ) = |{i < j : σ(i) > σ(j)}|.Proof. If r < s, xs − xr occurs once in denominator, and exactly one of xs − xr and
xr − xs occur in the numerator:
if
{
σ−1(r) < σ−1(s) ⇒ xs − xrσ−2(s) < σ−1(r) ⇒ xr − xs
so ǫ(σ) = (−1)N(σ) where N(σ) = |{1 6 i < j 6 n : σ(i) > σ(j)}|.Lemma 1.12. ǫ is multiplicative, i.e. for any σ, π ∈ Sn, ǫ(σπ) = ǫ(σ)ǫ(π)
Proof.
ǫ(σπ) =∏
16i<j6n
σπ(j)− σπ(i)j − i
=∏
16i<j6n
σ(j)− σ(i)j − i
︸ ︷︷ ︸
ǫ(σ)
∏
16i<j6n
σπ(j)− σπ(i)σ(j)− σ(i)
︸ ︷︷ ︸
ǫ(π)
= ǫ(σ)ǫ(π).
the latter seen by taking xi = σ(i), hence xπ(i) = σ(π(i)).
Lemma 1.13. ǫ(τ) = −1 for any transposition τ .
Proof. Let τ = (12),
ǫ(τ) =1− 2
2− 1
3− 2
3− 1
4− 2
4− 1· · · n− 2
n− 1
3− 1
3− 2· · · n− 1
n− 2· · · 4− 3
4− 3= −1
(1l) = (2l)(12)(2l), thus ǫ(1l) = ǫ(12) and therefore (kl) = (1k)(1l)(1k) → ǫ (kl) =
ǫ(1l) = ǫ(12) = −1.Corollary 1.14. ǫ(σ) = sgn(σ) for any σ ∈ Sn.
Proof. This is true if σ is a transposition, since both ǫ and sgn are multiplicative, and
each permutation is a product of transposition.
– 9 –
Part IA: Groups
1.4 Subgroups and homomorphisms
Definition. The subset H of G (H ⊆ G) is a subgroup of G (H 6 G) if h1, h2 ∈ H ⇒h1h2 ∈ H and that e, h−1 ∈ H .
Remark. If G is finite and non-empty, then the first condition is sufficient.
Example. (a) (Z,+) < (Q,+) < (R,+);
(b) ({±1} ,×) < (Q \ {0} ,×) but ({±1} ,×) ≮ (Q,+) due to different operations;
(c) An < Sn;
(d) {e} 6 G;
(e) Let g ∈ G, we write 〈g〉 = {gj : j ∈ Z} 6 G as the cyclic subgroup of G generated
by g. This is the smallest subgroup of G which contains g.
More generally, let I be an indexing set. Write 〈gj|j ∈ I〉 for the smallest subgroup of
G containing each gj with j ∈ I. Note that
〈gj : j ∈ I〉 =⋂
H⊆G
H
such that all gj ∈ H .
Definition. We say that the group is generated by its elements g1, . . . , gn if there is no
proper subgroup of G containing all the gi; so 〈g1, . . . , gn〉 = G.
Example. Sn is generated by the set of its transpositions: Sn = 〈(r s) : 1 6 r < s 6 n〉.(For any element of Sn can be written as a product of transpositions.)
Definition. Let (G1, ∗1) and (G2, ∗2) be groups. The mapping
θ : G1 → G2
is a homomorphism if θ (a ∗1 b) = θ(a) ∗2 θ(b) for any a, b ∈ G1.
Example. sgn: (Sn, ◦)→ ({±1} ,×) is a homomorphism.
Definition. A mapping θ is an isomorphism from G1 to G2 if it is a homomorphism
and is bijective. We write (G1, ∗1) ≃ (G2, ∗2) or just G1 ≃ G2 if there is an isomorphism
from G1 to G2.
– 10 –
Part IA: Groups
Example. Let (G1, ∗1) and (G2, ∗2) be groups, where
G1 =
{(1 a
0 1
)
: a ∈ R
}
,
∗1 is the multiplication of matrices, G2 = R and ∗2 is addition. We have the mapping
θ : G1 → G2 defined by θ
[(1 a
0 1
)]
= a.
θ is a homomorphism since
θ
[(1 a
0 1
)(1 b
0 1
)]
= θ
(1 a + b
0 1
)
= θ
(1 a
0 1
)
+ θ
(1 b
0 1
)
= a+ b.
θ is also a bijection, thus it is also an isomorphism.
�
Lemma 1.15. Isomorphism is an equivalence relation on the set of groups.
(a) Reflexive. If G is a group, G ≃ G.
(b) Symmetric. If G1, G2 are groups with G1 ≃ G2 then G2 ≃ G1.
(c) Transitive. If G1, G2, G3 are groups with G1 ≃ G2, G2 ≃ G3 then G1 ≃ G3.
Proof. (a) e : G→ G, which takes g 7→ g for g ∈ G, is an isomorphism.
(b) If θ : G1 → G2 is an isomorphism, then θ has an inverse θ−1 : G2 → G1. Then θ−1
is a bijection, and it is a homomorphism.
Let a2, b2 ∈ G2, let a1, b1 be the pre-images of θ, then
θ−1(a2 ∗2 b2) = θ−1 [θ (a1) ∗2 θ (b1)]= θ−1 [θ (a1 ∗1 b1)]= a1 ∗1 b1= θ−1 (a2) ∗1 θ−1 (b2) .
(c) If θ : (G1, ∗1) → (G2, ∗2) and φ : (G2, ∗2) → (G3, ∗3) are homomorphisms, then
φθ : G1 → G2 is a homomorphism. (And, of course, if both θ, φ are bijective, so is φθ)
– 11 –
Part IA: Groups
Let a, b ∈ G1, then
φθ (a ∗1 b) = φ [θ (a ∗1 b)]= φ [θ (a) ∗2 θ (b)]= φθ(a) ∗3 φθ(b)
1.5 Modular Arithmetic
Definition. Let n ∈ N, X = Z, for a, b ∈ Z, define the equivalence relation on Z:
a ∼ b (mod n) ⇐⇒ if n divides a− b.
Proof. reflexive: a ∼ a ∀a n|0;symmetric: if a ∼ b, then b ∼ a;
transitive: if a ∼ b ∼ c, then a ∼ c.
We write [a] = {b ∈ Z : a ≡ b (mod n)}, hence [0], [1], [2], . . . , [n−1] are the equivalenceclasses for modulo n.
Addition modulo n is written as [i] + [j] = [i+ j].
Remark. Addition (mod n) is well defined since [i1] = [i2], [j1] = [j2] ⇒ [i1 + j1] =
[i2+j2]. Noting that if i2− i1 = rn and j2−j1 = sn, then (i2+j2)−(i1+j1) = (r+s)n.
Lemma 1.16. Integers modulo n form a group under addition modulo n, usually
denoted by Zn (or Z/nZ or Z/ 〈n〉). This is a cyclic group, generated by [1].
Proof. Let Zn = {[0], [1], [2], . . . , [n− 1]}, the addition + (mod n) gives a binary oper-
ation, which is associative (since the addition on Z is) with identity [0], and the inverse
of [i] is [n− i].And the powers of [1] in Zn are precisely [0], [1], [2], . . . , [n − 1], so all elements of Zn
are ’powers’ of [1], so Zn is cyclic.
– 12 –
Part IA: Groups
1.6 Cyclic and Dihedral groups
Definition. The cyclic group, Cn, is defined abstractly as
Cn = 〈g : gn = e〉 .where g is the generator, gn = e is the relation.
Lemma 1.17. Any 2 cyclic groups of the same order are isomorphic. Any infinite
cyclic group is isomorphic to (Z,+).
Proof. If G = 〈g〉 for some g ∈ G, with |G| = n, n ∈ N, then G = {e, g, g2, . . . , gn−1}with o(g) = n.
The mapping θ : G→ Zn which takes gj 7→ [j] is an isomorphism.
It is well defined: gj = gk, then gj−k = e, from which it follows that o(g) divides j − k,so j ≡ k (mod n), so [j] = [k]. It is injective (reverse the argument) and obviously
surjective. It is an homomorphism, so also an isomorphism.
Lemma 1.18. For g ∈ G, then gm = e ⇐⇒ n, the order of g, o(g) divides m.
Proof. If m = nq, then gm = (gn)q = eq = e, and if gm = e, write m = nq + r with
q, r ∈ N and 0 6 r < n then gr = gm−nq = gm (gn)−q = e, so r = 0 by minimality of n,
the order of g.
Theorem 1.19. Any subgroup of a cyclic group is cyclic.
Proof. See Examples Sheet 1.
Definition. The Dihedral group, D2n is the subgroup of Sn consisting of symmetries of
a regular n-gon.
Take the n points for Sn to lie uniformly distributed along the unit circle, D2n consists
of permutations of the points not destroying the symmetry of the n-gon.
12
3b
b b
12
34
b b
bb
– 13 –
Part IA: Groups
For n = 3, we have the dihedral group: D6 = {e, (123), (132), (23), (13), (12)} where(123), (132) are rotations and the transpositions are reflections, thus D6 ⊆ S3 ⇒ D6 =
S3.
Similarly for n = 4, we have D8 = {e, σ, σ2, σ3, τ, τσ, τσ2, τσ3} where σ = (1234) and
τ = (13).
Remarks. In general, we have n rotations, forming a cyclic subgroup of order n and n
reflections.
If τ is a reflection, e.g. one fixing the point 1, all the reflections τ, τσ, . . . , τσn−1 (in
fact, this is the full group of symmetries). Note that στ = τσ−1, τστ = σ−1.
1
στ
In fact, we can also define the dihedral groups abstractly
D2n =⟨s, t : sn = e, t2 = e, tst = s−1
⟩.
This group has 2n elements, namely e, s, . . . , sn−1, ts, . . . , tsn−1.
Remark. Any two dihedral groups of the same order are isomorphic.
– 14 –
Part IA: Groups
2 | Lagrange’s Theorem and Small Groups
Lemma 2.1 (Subgroup condition). Let G be a group, then if
a, b ∈ H ⇒ a−1b ∈ H
for ∅ 6= H ⊆ G, then H 6 G.
Proof. Let h, k ∈ H . Then letting a = b = h, we obtain h−1h ∈ H , and this is the
identity, so e ∈ H . Next, letting a = h, b = e, we obtain h−1e = h−1 ∈ H . Finally,
letting a = h−1, b = k, we obtain (h−1)−1k = hk ∈ H . Thus all three conditions are
satisfied and H is a subgroup.
Example. The subgroups of S3 are 1 × S3 (order 6), 1 × A3 (order 3), 3 × C2 (order
2) and the identity set {e}.
Definition. If H 6 G and g ∈ G, the left coset gH = {gh|h ∈ H}.
Example. An 6 Sn has precisely two different cosets, H and τH, where τ is any odd
permutation, such as (12).
Example. G = S3 and H = 〈(12)〉. Now |H| = 2 and |G| = 6, so we are expecting 3
left cosets. H = {e, (12)}, (123)H = {(123) , (13)} and (132)H = {(132) , (23)}.
2.1 Lagrange’s Theorem
Theorem 2.2 (Lagrange). If H is a subgroup of the finite group G, then the order of
H divides |G|.
g
e
H
gH
b
b
We need two lemmas first.
– 15 –
Part IA: Groups
Lemma 2.3. If H 6 G, then all the cosets have the same size, i.e. |H|.
Proof. Let g ∈ G, if gH is a coset, then consider the map
θ : H → gH defined by θ(h) = gh.
Injectivity. If gh1 = gh2 then h1 = h2 so this map is injective.
Surjectivity. If y ∈ gH then y = gh for some h ∈ H , so y is the image of h; therefore
the map is surjective.
Hence the map is a bijection, so |H| = |gH|.
Lemma 2.4. If H 6 G, then left cosets form a partition of G; i.e. if aH ∩ bH 6= ∅,then aH = bH (any two cosets overlap totally or not at all)
Proof. Let c ∈ aH , then c = ah1 for some h1 ∈ H . For all h ∈ H ,
ch = (ah1)h
= a(h1h) ∈ aH,
so cH ⊆ aH . But c = ah1 ⇒ a = ch−11 ∈ H , and aH = ch−1
1 h1 = c, so aH ⊆ cH and
thus cH = aH and if c ∈ aH ∩ bH , then aH = cH = bH .
Now we can prove Lagrange’s theorem.
Proof. (Lagrange) Consider a finite group G and a subgroup H . From (2.4), G is the
union of a finite number of pairwise disjoint cosets, say
G =r⋃
i=1
giH and hence |G| =r∑
i=1
|giH| .
Now by lemma 2.3, H and gH have the same number of elements, namely |H|, so that
G = r |H|.
Definition. If H 6 G, the index of H in G is the number of distinct cosets of H ,
denoted by |G : H|.
Corollary 2.5. We have proved above that |G| = |H| · |G : H|, so |G : H| = |G| / |H|for finite G, H 6 G.
Remark. If H 6 G and g ∈ G, the set Hg = {gh|h ∈ H}, a right coset of H in G. The
number of distinct right cosets of H in G equals |G : H|. If H 6 G, precisely one of
the cosets is a subgroup, because only one can contain the identity.
– 16 –
Part IA: Groups
Remark 2.6. e.g. A4 has 12 elements but has no subgroup of order 6. A5 has order
60, but no subgroups of index 2, 3 or 4.
Lemma 2.7 (Lagrange). If G is a finite group and g ∈ G, then
|G| = k · o(g)
where k ∈ N.
Proof. Consider the subgroup 〈g〉 generated by g in G, then |〈g〉| = o(g), whence
o(g) = |〈g〉| which divides |G|.
Corollary 2.8. If G is a finite group and g ∈ G, then g|G| = e. (see 1.21)
Proposition 2.9. If G has prime order, so |G| = p, then G is cyclic. (In fact, if g is
any element in G \{e} , then G is generated by g. )
Proof. Let g ∈ G, g 6= e, then o(g) 6= 1 and o(g) divides |G| = p ⇒ o(g) = p by
Lagrange’s theorem. Hence G = 〈g〉.
An alternative definition of cosets. Let G be a group and H 6 G, define an equiva-
lence relation
a ∼ b ⇐⇒ a−1b ∈ HThe equivalence classes are the left cosets of H .
Proof. Reflexivity. a ∼ a since a−1a = e ∈ H for any a ∈ G;Symmetry, a ∼ b⇒ b ∼ a since b−1a = (a−1b)
−1 ∈ H ;
Transitivity, a ∼ b, b ∼ c⇒ a ∼ c since a−1c = (a−1b) (b−1c) ∈ H .
Example. Left cosets of H in G = S3.
(a) take H = 〈(123)〉, the left cosets are H, (12)H ;
(b) take H = 〈(12)〉 = {e, (12)}, then we have
Left cosets Right cosets
(123)H = {(123) , (13)} H(123) = {(123) , (23)}(132)H = {(132) , (23)} H(132) = {(132) , (13)}.
– 17 –
Part IA: Groups
2.2 Euler-Fermat Theorem
ConsiderZn = {[0] , [1] , . . . , [n− 1]}Z∗
n = {[a] ∈ Zn : hcf (a, n) = 1}(where Z∗
n is the units mod n)
Definition. Define multiplication modulo n to be
[a]⊗n [b] = [ab] .
This defines a well-defined operation on Z∗n.
Proof. If [a], [b] ∈ Z∗n, so is [ab]: for if (a, n) = 1 = (b, n), then hcf(ab, n) = 1. And
it is also well defined, i.e. [a1] = [a2], [b1] = [b2], then [a1b1] = [a2b2] since if a2 =
a1 + rn, b2 = b1 + sn with r, s ∈ Z, then a2b2 = a1b1 + n(...) so [a1b1] = [a2b2].
Lemma 2.10. Z∗n is an abelian group under ⊗n, multiplication mod n.
Proof. The operation ⊗n is well defined on Z∗n.
Closure. if hcf(a, n) = 1 = hcf(b, n), then (ab, n) = 1⇒ (a⊗n b) ∈ Z∗n;
Associativity. multiplication in Z is associative;
Identity. a⊗n 1 = a∀a ∈ Z∗n;
Abelian. a⊗n b = b⊗n a;
Inverse: since (a, n) = 1, there exists integers r, s such that ar + ns = 1, thus [a]−1 =
[r] ∈ Z∗n.
Definition (Euler’s totient function). For n ∈ N, define
φ(n) = |Z∗n| = |{a ∈ Z : 1 6 a 6 n, (a, n) = 1}|
Example. φ(p) = p− 1 for any prime p, φ(4) = 2 and
φ(n) = n∏
p|n
(
1− 1
p
)
.
Theorem 2.11 (Euler-Fermat). If n ∈ N, and a ∈ Z with (a, n) = 1, then
aφ(n) ≡ 1 (mod n).
Proof. Working in Z∗n, and let [a] ∈ Z∗
n, and φ(n) = |Z∗n|, so [a]φ(n) = [1], so aφ(n) ≡
1 (mod n).
Example. 27100 ≡ 1 (mod 101)
– 18 –
Part IA: Groups
2.3 Direct Product of Groups
Definition. Let H,K be groups (under ∗H , ∗K), then H×K = {(h, k) : h ∈ H, k ∈ K}is a group under the operation
(h1, k1) ∗ (h2, k2) = (h1 ∗H h2, k1 ∗K k2)
for hi ∈ H, ki ∈ K. This is the direct product of H and K
Proof. The groupH×K is closed under the given operation as h1h2 ∈ H and k1k2 ∈ K;
it is associative since multiplication in H and K is associative, the identity element is
e = (eH , eK) and (h, k)−1 = (h−1, k−1).
Example (Klein 4-group). C2×C2 = {(0, 0) , (0, 1) , (1, 0) , (1, 1)}, it is not isomorphic
to C4 because all elements other than (0, 0) have order 2.
Note also that (C2 × C2)×C2 ≃ C2×(C2 × C2) ≃ C2×C2×C2, and thus in the general
case
(H ×K)× L ≃ H × (K × L) ≃ H ×K × L.
Remarks. (a) H ×K is abelian ⇐⇒ both H and K are;
(b) |H ×K| = |H| · |K|;(c)H×K contains two ’special’ subgroups {(h, eK) |h ∈ H} ≃ H and {(eH , K) |k ∈ K} ≃K with trivial intersection.
Example. C2 × C3 = {(0, 0) , (0, 1) , (0, 2) , (1, 0) , (1, 1) , (1, 2)} ≃ C6
Exercise. Cr × Cs is cyclic ⇐⇒ (r, s) = 1.
Lemma 2.12 (Isomorphism theorem). Let G be a group with subgroups H and K
such that:
(a) any g ∈ G is g = hk for some h ∈ H, k ∈ K;
(b) H ∩K = {e};
(c) hk = kh for all h ∈ H, k ∈ K.
Then G ≃ H ×K.
– 19 –
Part IA: Groups
Proof. If h1k1 = h2k2 for some hi ∈ H , ki ∈ K, then h−12 h1 = k2k
−11 ∈ H ∩K = {e} ,
so h1 = h2, k1 = k2, so the expression g = hk with h ∈ H, k ∈ K is unique for each
g ∈ G.
Consider the well-defined Isomorphism
θ : G→ H ×K defined by θ(g) = (h, k).
(note we’ve just proved that g could be written as hk uniquely) This is well-defined,
Isomorphism because
θ(g1g2) = θ (h1k1h2k2)
= θ [(h1h2) (k1k2)]
= (h1h2, k1k2)
= (h1, k1) (h2, k2)
= θ (g1) θ (g2) .
2.4 Groups of Small Order
Lemma 2.13. The Groups of small order are (upto isomorphism):
Order Number of groups Groups
1 1 {e}2 1 C2
3 1 C3
4 2 C4, C2 × C2
5 1 C5
6 2 C6, D6 (≃ S3)
7 1 C7
8 5 C8, C4 × C2, C2 × C2 × C2, D8, Q8
9 2 C9, C3 × C3
10 2 C10, D10
11 1 C11
12 5 C12, C2 × C6, A4, D12, Q12
– 20 –
Part IA: Groups
Proof. For the cases where G has prime order, then G ≃ Cp. Now let’s examine the
non-prime cases:
Groups of Order 4. G ≃ C4, C2 × C2. (Groups of order 4 are abelian.)
Non-identity elements have order 2 or 4. If G contains an element of order 4, then G
is cyclic, so G ≃ C4, otherwise all elements of G \ {e} have order 2. Let a ∈ G \ {e},b ∈ G \ 〈a〉 , then G = {e, a, b, ab} , and ba = ab. So G ≃ 〈a〉 × 〈b〉 ≃ C2 × C2.
Remark. In general for elements of order 2, the group is abelian: ba = (ba)−1 =
a−1b−1 = ab
Groups of Order 6. G ≃ C6, D6.
In Examples Sheet 1 Q10/11, we’ve proved that if G be a group in which every element
other than the identity has order two. G is abelian and that if G is finite, the order
of G is a power of 2. And also that if G be a group of even order, then G contains an
element of order two.
In any case, G contains an element a of order 3 and an element b of order 2. Then
G = {e, a, a2, b, ba, ba2} and if ab = ba, then o(ab) = 6⇒ G ≃ C6. Finally, if ab = ba−1,
then G = 〈a, b|a3 = e, b2 = e, ab = ba−1〉 ≃ D6.
Groups of Order 8. G ≃ C8, C4 × C2, C2 × C2 × C2, D8, Q8.
G contains an element of order 8, then G ≃ C8.
If all elements have order 2, then G is abelian. (see above remark) Take a ∈ G \ {e} ,b ∈ G\〈a〉 and c ∈ G\〈a, b〉, then 〈a, b〉 ≃ C2×C2, andG ≃ 〈a, b〉×〈c〉 = (C2 × C2)×C2.
Henceforth we assume that G contains an element a of order 4. Now let b ∈ G \ 〈a〉with order 2 or 4 (so G = 〈a〉 ∪ b 〈c〉), hence G = {e, a, a2, a3, b, ab, a2b, a3b} and we
need to consider the cases for b2 and ba:
• b2: since a is order 4, b2 ∈ 〈a〉. If b2 = a, a3, then o(b) = 8, which is impossible.
So b2 = e, a2.
• ba: let ba = akb for some k = 1, 2, 3, so bab−1 = ak. Note that a = b2ab−2 =
bakb−1 =(ak)k
= ak2
, and thus ak2−1 = e, so 4 divides k2 − 1 and k is therefore
odd. So ba = ab or ba = a−1b:
– ba = ab, then G is abelian
∗ If b2 = e, then G ≃ 〈a〉 × 〈b〉 ≃ C4 × C2 by (2.12).
– 21 –
Part IA: Groups
∗ If b2 = a2, then replace b by b′ = ab−1. Then (ab−1)2 = a2b−2 = e, so
G ≃ 〈a〉 × 〈b′〉 ≃ C2 × C4 by (2.12).
– ba = a−1b
∗ If b2 = e, then G = 〈a, b|a4 = e, b2 = e, ab = ba−1〉 ≃ D8;
∗ If b2 = a2, then G = 〈a, b|a4 = e, b2 = a2, ab = ba−1〉 ≃ Q8 which is the
definition of a Quaternion group (due to Hamilton).
Definition (Quaternion Group). The Quaternion group, Q2n, is defined as
⟨a, b|a4 = e, b2 = a2, ab = ba−1
⟩.
In particular, we can express Q8 in terms of vectors ±1,±i,±j,±k ∈ R4, with (−1)2 =1, i2 = j2 = k2 = −1, and ij = k = −ji, etc.
Remark. Matrix representations of the quaternion group
(a) Q = {±1,±i,±j,±k} → GL2(C)
1 7→(1 0
0 1
)
, j 7→(i 0
0 −i
)
, j 7→(
0 1
−1 0
)
, k 7→(0 i
i 0
)
(b) Q = {±1,±i,±j,±k} → SL2(F3)
1 7→(1 0
0 1
)
, j 7→(1 1
1 −1
)
, j 7→(−1 1
1 1
)
, k 7→(0 −11 0
)
Note that Q8 6≃ D8, if we look at the number of elements of order 2 and 4:
order 2 order 4
D8 5 2
Q8 1 6
*Q8 < H (Hamiltonian), which in 4-dimensional real algebra, form a basis over
R4, i.e. H = {α1+ βi+ γj + δk} where α, β, γ, δ ∈ R.
– 22 –
Part IA: Groups
3 | Group Actions
“If you think it’s an action, it’s an action.”
J. Saxl
Definition. Let G be a group, X be a non-empty set, we say that G acts on X if there
is a mapping ρ : G×X → X which takes (g, x) 7→ g(x) that satisfies
(a) g(x) ∈ X ;
(b) (g1 ∗ g2)(x) = g1(g2(x));
(c) e(x) = x for all x ∈ X and all g, g1, g2 ∈ G.
Lemma 3.1. If G acts on X , for g ∈ G, the mapping θg : X → X which takes x 7→ g(x)
is a permutation of X .
Proof. θg is a map X → X , bijective since θg−1 is its inverse:
(θg−1 ◦ θg) (x) = θg−1 (θg(x))
= g−1 (g(x))
=(g−1g
)(x)
= e(x) = e
and similarly θg ◦ θg−1 = e.
Lemma 3.2. If G acts on X , the map θ : G → Sym(X) which takes g 7→ θg is a
homomorphism (a permutation representation of G).
Proof. For g ∈ G, we have θ(g) ∈ Sym(X), so θ : G→ Sym(X). It is a homomorphism:
for x ∈ X ,
θ(g1g2)(x) = θ(g1g2)(x)
= (g1g2)(x)
= g1(g2(x))
= g1(θg2(x))
= θg1(θg2(x))
= (θ (g1) θ (g2)) (x)
so θ(g1g2) = θ(g1)θ(g2).
– 23 –
Part IA: Groups
Notation. Let G be a group acting on X , for each g ∈ G, we have θg = gX (g acting
on X), a permutation of X , so that (g1g2)X = gX1 g
X2 .
Remark. Note that eX = e, but there may be other g ∈ G with gX = e. In fact the
kernel of the action is{g ∈ G : gX = e
}.
3.1 Orbits and transitivity
Definition. If G acts on X , we say that G is transitive on X , if for any x1, x2 ∈ X ,
there exists g ∈ G with g(x1) = x2.
If G acts on X , and x ∈ X , the orbit G(x) of x in X is G(x) = {g(x) : g ∈ G}.
Remark. G is transitive on X if and only if for some x, X = G(x). In general, X splits
into G-orbits:
Define relation ∼ on X : x1 ∼ x2 if x2 = g(x1) for some g ∈ G, ∼ is an equivalence
relation onX with orbits being the equivalence classes. Each orbit G(x) is G-invariant,
(so, for g ∈ G, g : G(x)→ G(x), so G acts on G(x)), and G is transitive on each G(x).
Definition. If G acts on X , define
Gx = {g ∈ G : g(x) = x} ,
the stabiliser of x in G.
Lemma 3.3. If G acts on X , and x ∈ X, then Gx ⊆ G.
Proof. e ∈ GX , if g1, g2 ∈ G, so does g1g2 ∈ Gx, so also g ∈ Gx ⇒ g−1 ∈ GX
(g(x) = x, g−1 : x = g−1(x)).
We can also now define kernel of the action more formally.
Definition (kernel). Let G act on X . Then the kernel of the action is⋂
x∈X
Gx
i.e. the set consisting of all the elements of G which act trivially on X . An action is
faithful if its kernel is trivial = {eG}.
Example. G = S4 and X = set of all partitions of {1, 2, 3, 4} into two parts of size 2
= {12|34, 13|24, 14|23}. Then G acts on X in the way you would expect. What is the
kernel? Investigate this action.
– 24 –
Part IA: Groups
Theorem 3.4 (Orbit-Stabiliser). If the finite group G acts on the set X , and x ∈ X ,
then
|G| = |Gx| × |G(x)| .In particular, if G is transitive on X , then |G| = |Gx| × |X| for any x ∈ X .
Proof. First show that Gx 6 G: Obviously e ∈ Gx, now g ∈ Gx =⇒ g−1 ∈ Gx:
gx = x =⇒ x = ex = g−1gx = g−1x.
g1, g2 ∈ Gx ⇒ g1g2 ∈ Gx :(g1 ∗ g2)(x) = g1(g2(x)) = g1(x) = x. By Lagrange’s theorem,
|G| / |Gx| = |G : Gx|, the index of Gx in G. We’ll show that |G : Gx| = |G(x)|.Define (G : Gx) to be the set of left cosets of Gx in G. Then all that is needed is to
prove that the mapping
G(x)→ (G : Gx), g(x) 7→ gGx
is a well-defined bijection.
Well-defined (⇒) and Injective (⇐).
g1Gx = g2Gx ⇐⇒ g−11 g2 ∈ Gx
⇐⇒ g−11 g2x = x
⇐⇒ g1x = g2x
Surjectivity: any coset of Gx is gGx for some g ∈ G, so gGx is the image of g(x).
3.2 Groups of symmetries of a regular n-gon
4
3
1
2
b
b
b
b
Figure 3.1. Tetrahedron
– 25 –
Part IA: Groups
Theorem (Tetrahedron). Let G be the group of all symmetries of a tetrahedron and
G+ the group of all rotational symmetries, then G ≃ S4 and G+ ≃ A4.
Proof. Let X be the set of vertices where |X| = 4, since G is transitive on X , so
|G| = |X| · |G1| = 4 |G1|
by the Orbit-Stabiliser theorem. Now G acts on {2, 3, 4} and is transitive, so
applying the theorem again gives |G1| = 3 |G12| = 3× 2.
So |G| = 24, and in fact G ≃ Sym(X). [In fact, GX contains any transposition, so is
S4.]
Now let’s consider the group G+ of the rotational symmetries (i.e. rigid motions).
G+ 6 G, and∣∣G+
∣∣ = 4
∣∣G+
1
∣∣ = 4× 3 = 12.
In fact, G+ = A4 contains all 3-cycles and these generate A4, so A4 6 G+, which are
the same size, so is equal.
1
1′
2
34
4′
2′
3′
Figure 3.2. Cube
Theorem (Cube). Let G be the group of all symmetries of a cube and G+ the group
of all rotational symmetries, then G ≃ S4 × C2 and G+ ≃ S4.
Proof. V = {vertices : |V | = 8}, F = {faces: |F | = 6} and D = {diagonals: |D| = 4}.Let G act on F : |G+| = 6 ·
∣∣G+
f
∣∣ = 6 · 4 = 24, and |G| = 6 · |Gf | = 6 · 4 · 2. Now let G
act on D, consider the mapping θ : G→ Sym(D) = S4.
There is a non-trivial kernel K of the action (K = 〈(11′) (22′) (33′) (44′)〉 where |K| =2). If g ∈ K fixes 1, it foxes 1′, but also 2, 2′ and also 3, 3′ and 4, 4′. If g ∈ K swaps 1
and 1′, then it also swaps 2, 2′, 3, 3′ and 4, 4′.
– 26 –
Part IA: Groups
G+∩K = {e} so the action of G+ on D is faithful, i.e. there exist trivial kernel. Hence
the mapping θ : G+ → S4 = Sym(D) is injective, which follows that θ(G+) 6 S4, of
order 24, so G+ ≃ S4.
Finally, G ≃ G+ ×K as kh = hk for all h ∈ G+, k ∈ K. [Later we will see that two
normal subgroups, intersecting trivially, commute element wise.]
Figure 3.3. Left: Dodecahedron, Right: Icosahedron
*Theorem (Dodecahedron and Icosahedron). G+ ≃ A5
Let V = {vertices : |V | = 20}, F = {faces: |F | = 12} and E = {edges: |D| = 30} andconsider the action G+ on F (which is transitive). Now for f ∈ F , |G+| = 12 ·
∣∣G+
f
∣∣ =
12 · 5 = 60. (for all symmetries, |G| = 60× 2 = 120).
Lemma (Klein). There are 5 cubes embedded in a dodecahedron.
Proof. These cubes are permuted by G+ (and by G), so we have θ : G+ → S5, a
homomorphism.
So θ(G+) < S5, and it contains all 5-cycles, so A5 6 θ(G+), so they have the same
order. Hence G+ ≃ A5, and G ≃ A5 × C2.
3.3 Actions from groups and Cayley’s theorem
Definition. Let G be a group and let X = G, then the map
ρ : G×X → X defined by ρ(g, x) 7→ gx = g(x)
is an action. This is the left regular action.
– 27 –
Part IA: Groups
Theorem 4.7 (Cayley). Any group G is isomorphic to a subgroup of a symmetry
group.
Proof. Consider the injective homomorphism
φ : G→ Sym(X)
arising from the left regular action.
Injective. If, for some x ∈ X, g1(x) = g2(x), then g1x = g2x so g1 = g2.
Hence G is isomorphic to the subgroup θ(G) = {θ(G) : g ∈ G} 6 Sym(G) and since
the action is faithful, so G ≃ θ(G) 6 Sym(X).
Remark. Let θ : G → H be a homomorphism. Then θ(G) = {θ(G) : g ∈ G} 6 H .
This is because eH = θ(eG) ∈ θ(G), and if a, b ∈ θ(G), say a = θ(g) and b = θ(h) with
g, h ∈ G, then we have a−1b = ((θ(g))−1 θ(h) = θ(g−1h) ∈ θ(G).
Example 4.8. Let (G, ∗) be a group, H 6 G and X = (G : H) the set of left cosets of
H in G.
Then G acts on (G : H) by the mapping
g : (G : H)→ (G : H) which takes xH 7→ (gx)H = g(xH).
Well-defined: x1H = x2H → x−11 x2 ∈ H , so
(gx1)−1(gx2) = x−1g−1gx2
= x−1x2 ∈ H.
So g(x1H) = g(x2H)
Action: e(xH) = (ex)H = xH and
(g1g2)(xH) = (g1g2x)H
= g1(g2(xH)).
Action is transitive: if we let x1H, x2H ∈ X , then (x2x−11 )(x1H) = x2H .
The stabiliser of the ’point’ H in X is the subgroup H . Also, the stabiliser of xH is
xHx−1 which is all the elements of the form {xhx−1|h ∈ H}.
Lemma 3.9. IfG acts onX , and x ∈ X, g ∈ G, thenGg(x) = gGxg−1 = {ghg−1|h ∈ Gx}.
Proof. For, (ghg−1)(g(x)) = (gh)(x) = g(x), and conversely.
– 28 –
Part IA: Groups
Lemma 3.10 (Conjugation Action). Let G be a group, and X = G. Define the action
of G on X as G × X → X which takes (g, x) 7→ gxg−1, so gX : X → X which takes
x 7→ gxg−1. This is the conjugation action of G.
(a) The orbit of x ∈ X is the conjugation class
cclG(x) ={gxg−1 : g ∈ G
}
Then G is the disjoint union of the conjugacy classes.
(b) The stabiliser of x ∈ G is the centraliser of x, i.e.
CG(x) = {g : gx = xg} 6 G and |cclG(x)| = |G : CG(x)|
(c) The kernel of the conjugation action is the centre ofG, Z(G) = {g : gx = xg∀x ∈ G}.
Proof. This all follows from previous work on actions, once we have checked that this is
an action, i.e. (g1 ∗ g2)(x) = g1 ∗ g2 ∗x∗ (g1 ∗ g2)−1 = g1 ∗ (g2 ∗x∗ g−12 )g−1
1 = g1(g2(x)) =
gX1 ◦ gX2 (x).
Example. G = D8 = 〈a, b|a4 = e, b2 = e, bab−1 = a−1〉
cclsG(x): {e} {a, a−1} {a2} {b, a2b} {ab, a−1b}|cclsG(x)|: 1 2 1 2 2
CG(x) : D8 〈a〉 D8 〈a2, b〉 〈ab, a2〉|CG(x)|: 8 4 8 4 4
Remarks. (a) Conjugate elements have the same order (e.g. in 3.5, groups of order 8,
bab 6= a2, since (gxg−1)n = (gxg−1)(gxg−1) · · · (gxg−1) = gxng−1, so o(x) = o(gxg−1).
(b) cclG(z) = {z} ⇐⇒ z ∈ Z(G).(c) There is also a conjugation action of G on the set of all subgroups of G: If H 6 G,
define gHg−1 = {ghg−1|h ∈ H} and g : H 7→ gHg−1 gives an action. Then the
stabiliser of the ’point’ H is {g ∈ G|gH = Hg} = NG(H) the normaliser of H .
Note: gHg−1 = H for all g ∈ G ⇐⇒ gH = Hg ∀g ∈ G ⇐⇒ NG(H) = G.
Definition. We say H is normal in G if NG(H) = G. (Note this is a provisional
definition.)
– 29 –
Part IA: Groups
3.4 Conjugacy classes in Sn and An
Definition. If π ∈ Sn, the cycle type of π is (n1, n2, . . . , nk) if the disjoint cycles in π
have length n1 > n2 > · · · > nk > 0. (Note: n1 + n2 + · · ·+ nk = n)
Example. In S3, we have cycle type 13 : (1, 1, 1) = {e} , 21 : (2, 1) = {(12) , (13) (23)}and 3 : (3) = {(123) , (132)} and etc.
Theorem 3.11. Two permutations in Sn are conjugate (in Sn) if and only if they have
the same cycle type.
Proof. If π = (a11a12 . . . a1n1) (a21a22 . . . a2n2
) · · · then for σ ∈ Sn, we have
σπσ−1 = (σ (a11)σ (a12) · · ·σ (a1n1)) (σ(a21)σ(a22) · · ·σ(a2n2)) · · · (⋆)
such that
σπσ−1 : σ(a11) 7→ σ(a12) 7→ σ(a13) 7→ · · · 7→ σ(a1n1) 7→ σ(a11).
Hence conjugate elements have the same cycle type.
Conversely, if π is as before, and τ = (b11b12 . . . b1n1) (b21b22 . . . b2n2
) · · · , same cycle
type, let σ be the permutation in Sn taking aij to bij , so σ(aij) = bij , then by (⋆)
τ = σπσ−1, so τ, π are conjugate in Sn.
Example. (2k) = (12) (1k) (12)−1 for k > 2.
Corollary 3.12. Sn has p(n) conjugacy classes, where p(n) is the partition function,
p(n) is the number of expressions n = n1 + n2 + · · ·+nk, with n1 > n2 > · · · > nk > 0,
ni, k ∈ N.
Example. S4 has 5 conjugacy classes:
example member cycle type size sgn centraliser |CS4|
e 14 1 + S4 24
(12) (3)(4) 212 6 - 〈(12) , (34)〉 4
(123)(4) 22 3 + D8 8
(12)(34) 31 8 + self-centralising 3
(1234) 4 6 - 〈(1234)〉 4
– 30 –
Part IA: Groups
Theorem 3.13. Let x ∈ An, then cclAn(x) ⊆ cclSn
(x) with equality ⇐⇒ CSn(x)
contains an odd permutation (then half are odd and half even). If cclAn(x) $ cclSn
(x),
then cclSn(x) splits into 2 cclAn
(x), of the same size 12|cclSn
(x)|.
Proof. Consider the conjugacy classes of A4:
example member cycle type size centraliser |CA4|
e 14 1 A4 12
(123)(4) 22 3 〈(12) (34) , (13) (24)〉 4
(12)(34) 31 4 〈(123)〉 3
(12)(34) 31 4 〈(123)〉 3
Let x ∈ An, cclAn(x) ⊆ cclSn
(x), thus
|cclSn(x)| = |Sn : CSn
(x)||cclAn
(x)| = |An : CAn(x)|
where CAn(x) 6 CSn
(x) (i.e. the index is 1 or 2), recall that if H < Sn contains an odd
permutation then |H ∩An| = 2.
Now Sn > An index 2,
CAn(x) = An ∩ CSn
(x) 6 CSn(x) index 1 or 2
Exercise. Show that A5 has 5 conjugacy classes, of size 1, 12, 12, 15, 20.
3.5 Cauchy’s theorem
Theorem 3.14 (Cauchy). Let G be a finite group, let p be a prime with p| |G|. ThenG contains an element of order p.
*Remark. The generalisation of Cauchy’s theorem is Sylow’s first theorem, which
implies that if pn is any prime power dividing the order of G, then G has a subgroup
of order pn. See IB Groups, Rings and Modules.
Proof. Let X = {x = (x1, x2, . . . , xp) : x1x2 · · ·xp = e, xi ∈ G} and thus |X| = |G|p−1
so p| |X|. Then let Cp = 〈g : gp = e〉 act onX such that gjx = (x1+j , . . . , xp+j) (mod p).
Consider that x1 · · ·xp = e ⇒ x1+j · · ·xp+j = e so gj is a homomorphism from G to
Sym(X) so is an action.
– 31 –
Part IA: Groups
The orbits of this action is Gx = {gx = x : g ∈ Cp} and thus
|Gx| ={
p if x = {xi, . . . , xi}1 if x 6= {xi, . . . , xi}.
By the Orbit-Stabiliser theorem, |Cp| = |Gx| |Gx| = p so take x1 = (e, . . . , e) then
|Gx1| = p/p = 1 and hence
|X| = |Gx1|+m∑
j=2
|Gxj | = 1 +m∑
j=2
|Gxj| = |G|p−1
and since p| |X| so there must at least p such xjs with orbit 1 so xj = (xj , . . . , xj) ⇒xpj = e and thus there exists an element of order p.
*Elementary proof. Consider all possible p-tuples (a1, a2, . . . , ap) of elements of G
whose product a1a2 · · · ap = e. The number of such p-tuples is np since if we selected
a1, . . . , ap−1 at random, there is a unique ap = a−1p−1 · · · a−1
1 ∈ G, with a1a2 · · · ap−1ap = e.
Now call two p-tuples equivalent if one is merely a cyclic permutation of the other. So
(a1, a2, . . . , ap) is equivalent to exactly p p-tuples except when a p-tuple is of the form
(a, a, . . . , a)
with all of its components equal. It is equivalent only to itself. Thus the equivalence
class of any p-tuple of the form (a, a, . . . , a) has a single member, while all the other
equivalence classes have p members.
Now, if there aren’t any equivalence classes other than {(e, e, . . . , e)} with a single
member, then p|(np − 1) ⇒ np ≡ 1 (mod p). But we’ve assumed at the start that
p|n⇒ np ≡ 0 (mod p) so there must be a p-tuple (a, a, . . . , a) 6= (e, e, . . . , e) such that
ap = e. Thus, there is an element a ∈ G of order p.
Remark. If m| |G| for G finite, it does not imply the existence of H 6 G with |H| = m.
Example. A4 has no subgroups of order 6.
– 32 –
Part IA: Groups
4 | Homomorphisms, Normal subgroups and Quotients
“We must cure this homomorphobia.”
J. Saxl
4.1 Homomorphisms and Normal subgroups
Recall. Let (G, ∗G), (H, ∗H) be groups, the mapping θ : G→ H is a homomorphism if
θ(g1 ∗G g2) = θ(g1) ∗H θ(g2) ∀gi ∈ G. It is an isomorphism if it is also bijective. The
image θ(G) = {θ(g)|g ∈ G} and the kernel ker(θ) = {k ∈ G|θ(k) = eH}.
Example. Let C∗ = (C \ {0} ,×) and H = {z ∈ C : |z| = 1} . Then φ : z 7→ z/ |z| isa homomorphism in which case the image is obviously the unit circle H- in fact, φ does
nothing to the unit circle. The kernel of φ is those elements which map to the identity
1, i.e. the positive half of the real line, i.e. (R > 0,×).
Remark. If G acts on a set X , the kernel of the action consists of elements of G acting
trivially. This is then the kernel kerθ of the corresponding permutation representation
θ : G→ SymX which takes g 7→ gX .
Lemma 4.1. If θ : G→ H is a homomorphism, then θ(eG) = eH , and θ(g)−1 = θ(g−1).
Proof. θ(eG)θ(eG) = θ(eG), so θ(eG) = eH , and θ(g)θ(g−1) = θ(gg−1) = θ(eG) = eH =
θ(g−1)θ(g) so θ(g−1) = θ(g)−1.
Lemma 4.2. If θ : G → H is a homomorphism, then θ is injective if and only if
kerθ = {eG}. (We can think of ker θ as the “obstruction to injectivity” of θ.)
Proof. If kerθ = {e}, and if θ(g1) = θ(g2), then θ(g−11 g2) = e, so g−1
1 g2 ∈ kerθ = {e} ,so g1 = g2. Conversely, if θ is injective and g ∈ kerθ then θ(g) = θ(eG) so g = eG, so
kerθ = {eG}.
Definition. The subgroup K 6 G is normal in G if
gKg−1 = K∀g ∈ G.We write K ⊳ G.
Equivalently, for K 6 G, K ⊳ G ⇐⇒ gK = Kg for any g ∈ G; or ⇐⇒ for any
g ∈ G, k ∈ K, gkg−1 ∈ K; or ⇐⇒ K is the union of some G ccls.
Remark. If G is abelian, then any subgroup is normal, because gkg−1 = k ∈ K.
Lemma 4.3. If K < G of index 2, then K ⊳ G.
– 33 –
Part IA: Groups
Proof. If g ∈ G, then either g ∈ K, in which case gK = K = Kg, or g ∈ G \ K, in
which case gK = G \K = Kg. Either way, K ⊳ G.
Lemma 4.4. Let G1, G2 ⊳ G, with G = 〈G1, G2〉 and G1 ∩ G2 = {e}. Then G ≃G1 ×G2.
Proof. By (2.12), it is enough to show that g1g2 = g2g1 for any g1, g2 ∈ G. This is truebecause g−1
1 g−12 g1g2 ∈ G1 ∩G2 = {e} so g1g2 = g2g1. [[g1, g2], i.e. g
−11 g−1
2 g1 and g−12 g1g2
are the commuters of G.]
Remark. This was used when G was the group of all symmetries of a cube, G1 was
G+ ⊳ G, and G2 was the kernel of the action of G on D, thus G2 ⊳ G.
4.2 Quotient groups
Theorem 4.5 (Quotient group). Let K ⊳G, a normal subgroup of G. Then the set of
all cosets of K in G forms a group G/K under multiplication. (g1K) ∗ (g2K) = g1g2K,
where g1g2K is the quotient group G by K.
Proof. Closure. the multiplication on G/K is well defined: if g1, g2 ∈ G, then g1g2K
is a coset in G/K, which is also well defined, i.e. if g1K = h1K, g2K = h2K, then
g1g2K = h1h2K:
h−12 h−1
1 g1︸ ︷︷ ︸
∈K,say k1
g2 = h−12 k1︸ ︷︷ ︸
let h−1
2k1h2=k∈K
g2 = k h−12 g2︸ ︷︷ ︸
∈K, say k2
= k · k2 ∈ K
Associativity. g1K(g2Kg3K) = g1Kg2g3K = g1(g2g3)K = (g1Kg2K)g3K = g1g2Kg3K =
(g1g2g3)K since multiplication in G is associative.
Identity. the identity is K: eKgK = gK = gKeK for any gK.
Inverse. the inverse of gK is g−1K since gKg−1K = gg−1K = K, and g−1KgK =
K.
Remark: Normality of K is necessary.
Example. Let G = S3, K = 〈(123)〉, then G/K = {K, (12)K} ≃ C2.
Example. D8 = 〈a, b|a4 = e, b2 = e, bab = a−1〉, Z = 〈a2〉⊳G, G/Z = {Z, aZ, bZ, abZ}.All elements of G/Z other than the identity have order 2. (aZ)2 = a2Z = Z, so
G/Z ≃ C2 × C2. Note that 〈b〉 is not normal, but 〈a2〉 and all subgroups of order 4
are normal. Note also that if G is finite and K ⊳ G, then |G/K| = |G| / |K| < |G| ifk 6= {e}.
– 34 –
Part IA: Groups
Example. If G is a p-group, so |G| = pa, a prime power, > 1, by 4.18, Z(G) 6= {e} ,now G/Z(G) = pb for some b < a.
Theorem 4.6 (First Isomorphism theorem). Let G,H be groups and θ : G → H be
a homomorphism, then θ(G) 6 H, ker θ ⊳ and
G/ ker θ ≃ θ(G).
Remark. Hence, the homomorphic images of G are just the quotient of G. In general,
the way to show that two groups are isomorphic is to construct an isomorphism between
them.
Proof. θ(G) 6 H : eH ∈ θ(G) as eH = θ(eG). Let h1, h2 ∈ θ(G), say hi = θ(gi), then
h−11 h2 = θ(g1)
−1θ(g2) = θ(g−11 g2) ∈ θ(G).
kerθ ⊳ G: Let K = kerθ, we note that eG ∈ K, since θ(eG) = eH , and if g1, g2 ∈ K, thenθ(g−1
1 g2) = θ(g−11 )θ(g2) = eH · eH = eH , so g
−11 g2 ∈ K, so K 6 G. To show that K ⊳G:
let k ∈ K and g ∈ G, then gkg−1 ∈ K; θ(gkg−1) = θ(g−1)θ(k)θ(g) = θ(g−1)θ(g) = e
(note that θ(k) = eH).
G/ ker θ ≃ θ(G): Let K = kerθ, and define ψ : G/K → θ(G) where ψ(gK) = θ(g)
where g ∈ G, it is well-defined (⇒) and injective (⇐),
g1K = g2K ⇐⇒ g−11 g2 ∈ K
⇐⇒ θ(g−11 g2) = eH
⇐⇒ θ(g1) = θ(g2)
⇐⇒ ψ(g1K) = ψ(g2K).
Surjectivity: any element in θ(G) is θ(g) for some g ∈ G, so ψ(gK) = θ(g);
Homomorphism:
ψ(g1Kg2K) = ψ(g1g2K) = θ(g1g2) = θ(g1)θ(g2)
= ψ(g1K)ψ(g2K).
Hence ψ is a well-defined isomorphism.
Example. θ : (R,+)→ C∗ = {C \ {0} ,×} which takes t 7→ eti. θ(G) = S1, the unit
circle. kerθ = {2πn|n ∈ Z} = K and R/K ≃ S1.
– 35 –
Part IA: Groups
An alternative proof of multiplication in G/K being well defined if K ⊳ G.
Let A,B ⊆ G, where A×B = {ab|a ∈ A, b ∈ B}.Claim: If K ⊳ G, the product of the sets g1K and g2K is g1g2K:
(g1K) (g2K) = g1 (Kg2)K = g1g2KK ⊆ g1g2K.
Warning! Quotient groups are not in general isomorphic to subgroups of G.
Example. The homomorphic images of (Z,+) are Z and Z/nZ (integers modulo n
under addition).
Let θ : Z → H (surjective), K = ker θ is cyclic (as subgroups of cyclic groups are
cyclic), say K = 〈n〉 , n ∈ Z > 0. Then θ(G) ≃ Z/ 〈n〉, (i) if n > 0, Z/ 〈n〉 is finite, itis integers modulo n; (ii) if n = 0, then θ(G) ≃ Z.
Example. G = Q8 = 〈a, b : a4 = e, b2 = a2, bab−1 = a−1〉, Z = 〈a2〉 and G/Z =
{Z, aZ, bZ, abZ} ≃ C2 × C2 since all elements have order 2.
Theorem 4.7. Let K ⊳ G. Then K is the kernel of the natural surjective homomor-
phism θ : G→ G/K, which takes g 7→ gK.
Proof. Homomorphism:
θ(g1g2) = g1g2K
= g1Kg2K
= θ(g1)θ(g2).
Surjective: just note that gK = θ(g). Now
ker θ = {g ∈ G : θ(g) = eK}= {g ∈ G : gK = K} = K
as required.
Remark. Homomorphic images of G are equivalent to quotients G/K with K ⊳ G.
There are more complicated isomorphism theorems to come in IB Groups, Rings and
Modules. E.g. subgroups of G containing K ⇐⇒ subgroups of G/K.
Corollary 4.8. Let G act on a set by g : x 7→ g(x). Let θ : G → Sym(X) be a
homomorphism, with θg ∈ Sym(X), taking g 7→ θ(g). The kernel of the action is⋂
x∈X Gx = ker θ = K.
– 36 –
Part IA: Groups
By the isomorphism theorem, G/K ≃ θ(G) 6 Sym(X). So it follows that G has a
normal subgroup of index dividing n!, where |X| = n so that |Sym(X)| = n!.
Example. Let H 6 G of index n. Then G has a normal subgroup K of index dividing
n!, because G acts on the set X of all left cosets of H in G, where |X| = n.
4.3 Simplicity of A5
Definition. A group G is simple if {e} and G are the only normal subgroups. e.g. Cp
and A5.
Conjugacy classes in S5 :
cycle type example member size centraliser sign
15 e 1 S5 +
213 (12) 10 〈(12)〉 × S3 -
221 (12) (34) 15 D8 +
312 (123) 20 〈(123)〉 × S2 +
32 (123) (45) 20 〈(123)〉 × 〈(45)〉 -
41 (1234) 20 〈(1234)〉 -
5 (12345) 24 〈x〉 +
The normal subgroups of S5 are {e} , A5, S5 and no others, since any normal subgroup
must be a union of conjugacy classes. The only other possibility would be, in A5, to
replace the conjugacy class 312 by the conjugacy class 32, but if x is of cycle type 32,
then x3 is of type 213 which would not be present.
Conjugacy classes in A5:
cycle type size centraliser
15 1 A5
221 15 V4
312 20 〈x〉5 12 〈x〉5 12 〈x′〉
– 37 –
Part IA: Groups
Note; (12)(12345)(12) = (21345), so the two are conjugate in S5 but not in A5; all
conjugating elements in S5 are odd. Any element in S5 conjugating (12345) to (21345)
is in the coset (45)C with C = 〈(12345)〉 = Cn(x), so S5 \ A5.
Therefore A5 is simple, because its only normal subgroups are {e} and A5.
*More homomorphisms with applications
Let’s take a further look at θ : G→ H , K = ker θ
θ : G θ(G)
G/K
πψ
Theorem 4.9. If G is a p-group of size pa, then G has a subgroup of order pb for each
0 6 b 6 a. Moreover, there is a chain of subgroups, each normal in the preceding one.
{e} = Ga ⊳ · · · ⊳ G2 ⊳ G1 ⊳ G0 = G
with Gi/Gi+1 cyclic of order p, so |Gi| = pa−1. They are called a subnormal series (the
Gi not normal in G in general).
Definition (Galois). G is soluble if there is a subnormal chain {Gi} of subgroups suchthat Gi/Gi+1 is cyclic of prime order.
Examples of soluble groups. Finite p-groups, A4, S4.
Proof. (4.9) It is clear if a = 1, so let a > 1, we argue by induction. We know that
Z = Z(G) is not {e}, i.e. non-trivial centre. Let x ∈ Z of order p, put K = 〈x〉, thenK ⊳ G, let G = G/K, then
∣∣G∣∣ = pa−1, so we can again use an inductive argument.
Let H be a subgroup of G of order pb−1, and H ={g ∈ G|π(g) ∈ H
}, where π is the
natural map π : G→ G/K. Then |H| = pb. (Check!)
’Moreover’: choose G1 ⊳ G of order pa−2, and G1 = {g ∈ G|π(g) ∈ G1} and check
normal in G and continue in G1, then use induction.
Definition. G is a simple group if its only normal subgroup are {e} and G, i.e. its onlyhomomorphic images are G, and {e} .
Example. G is abelian and simple ⇒ G is cyclic of prime order.
Example. A5 is simple of order 60.
– 38 –
Part IA: Groups
5 | Matrix and Orthogonal Groups
5.1 Matrix Groups
Let F be a field - usually F = R or F = C, Mn(F) be the set of all n×n matrices with
entries in F (here usually n = 2, or n = 3).
Recall from Vectors and Matrices. A,B ∈ Mn(F), A = aij and B = bij , then
(AB)ij = aikbkj, thus AB ∈Mn(F). We note that multiplication is associative:
((AB)C)rs =∑
k
(AB)rk (C)ks
=∑
k
∑
j
(arjbjk) cks
=∑
j
arj
(∑
k
bjkcks
)
= (A (BC))rs ,
the determinant of A is
detA =∑
σ∈Sn
(sgnσ) a1σ(1)a2σ(2) · · · anσ(n)
then detAB = detA detB and detA 6= 0 ⇐⇒ A is invertible. In fact, A−1 =1
detAadjA.
Example. n = 2, detA = a11a22 − a12a21, then A−1 =1
detA
(a22 −a12−a21 a11
)
Proposition 5.1. The set GLn(F) = {A ∈Mn(F) : detA 6= 0} is a group under multi-
plication (GLn is called the general linear group).
Proof. If A,B ∈ GLn(F), then AB ∈ GLn(F) since detA 6= 0 6= detB implies detAB 6=0.
Multiplication is associative and the identity is
In =
1 0 . . . 0
0. . .
......
. . . 0
0 · · · 0 1
The inverse of A exists since detA 6= 0.
– 39 –
Part IA: Groups
Lemma 5.2. det : GLn(F) → FX which takes A 7→ detA is a homomorphism,
surjective with ker det = SLn(F) where SLn(F) is the special linear group defined as
SLn(F) = {A ∈Mn(F) : detA = 1}.
Proof. detAB = detA detB thus it is a homomorphism, it is surjective: if λ 6= 0, λ ∈ F,
det
λ 0 · · · 00 1
......
. . . 0
0 · · · 0 1
= λ.
Lemma 5.3. SLn(F) ⊳ GLn(F), GLn(F)/SLn(F) ≃ FX .
Proof. See (5.2) and the isomorphism theorem.
Remark. Z ={λIn : λ ∈ FX
}6 GLn (F) lies in the centre (proof as exercise).
Z ⊳ GLn(F), PGLn(F) = GLn(F)/Z
where PGLn is the projective general group, and PSLn(F) = SLn(F)/ {λI : λn = 1}where PSLn is the projective special group.
In fact, PSLn(F) is almost always simple. We often assume n > 1 since GL1(F) ≃ FX .
5.4 An important action of GLn(F) on V = Fn. A ∈ GLn(F), A : V → V which
takes v 7→ Av.
This is an action of GLn(F) on Fn. It is faithful, i.e. trivial kernel (only In fixes all
elements of Fn). It is transitive on Fn/ {0} , in fact it is also transitive on the set of
basis of Fn.
5.5 Another important action of GLn(F) on Mn(F), which extends the conjugation
action of GLn(F) on its elements. For P ∈ GLn(F),X ∈Mn(X)
P : X 7→ PXP−1
the GLn(F) orbit of X the similarity class of X.
Remark. Two matrices in Mn(F) are conjugate ⇐⇒ they represent the same linear
transformation on the vector space Fn. A goal of Linear Algebra is to find a ’good’
representative for each conjugacy orbit, e.g. for n = 2.
– 40 –
Part IA: Groups
Lemma 5.6 (Change of basis). Let V be a vector space of dimension n over F, let
α : V → V be a linear transformation. Fix a basis B of V , say B = {v1, v2, . . . , vn},then
v ∈ V ←→ [v]B ∈ Fn
and
α : V → V ←→ [α (v)]B = [α]B [v]B
where [α]B is the unique matrix for which this works:
[α]B = ([α (v1)]B|[α (v2)]B| · · · ) .
If we start with basis B′ rather than B. Then [α]B = P [α]B P−1 for some invertible
matrix, it is unique and called the change of basis matrix.
A ’simple’ form of matrices such that each similarity class contains a unique matrix in
canonical form, e.g. Jordan normal form.
Theorem 5.7 (canonical form). The orbit of X ∈M2(C) contains precisely one of
(a)
(λ1 0
0 λ2
)
λ1 6= λ2 ∈ F, distinct eigenvalues
(b)
(λ 0
0 λ
)
λ ∈ F, the class is {λI}
(c)
(λ a
0 λ
)
λ ∈ F, double eigenvalue λ, not diagonalisable
5.2 Orthogonal Groups
Here working over R (or over Fq)
Definition. On ={A ∈ GLn(R) : AAT = I
}n > 2 and SOn = {A ∈ On : detA = 1} .
We use the isomorphism theorem to show this is a subgroup ofO(n); the homomorphism
detO(n)→ {±1} < R∗ is surjective, so
SO(n) = ker (detO(n)/SO(n)) ≃ C2
Lemma 5.8. On 6 GLn(R)
Proof. I ∈ On; A,B ∈ On ⇒ A−1B ∈ On since (A−1B) (A−1B)T= ATBBTA = I.
Remark. A ∈ GLn is orthogonal ⇐⇒ its rows and columns are orthonormal.
Lemma 5.9. SOn ⊳ On of index 2.
– 41 –
Part IA: Groups
Proof. A ∈ On ⇒ detA = ±1
−1 0 · · · 00 1
. . ....
.... . .
. . . 0
0 · · · 0 1
∈ On \ SOn
And, if A,B ∈ On \ SOn, then A−1B ∈ SOn, so On \ SOn is just one left coset.
Recall. x, y ∈ R, x.y =∑n
1 xiyi, A ∈ GLn(R), Ax.y = x.Ay.
Lemma 5.10. Let A ∈ On, and x,y ∈ Rn, then Ax.Ay = x.y and |Ax| = |x|. So
orthogonal maps are isometries
Proof. Consider that
Ax.Ay =∑
i
(∑
k
(aikxk)∑
j
(aijyj)
)
=∑
i,j,k
aikaijxkyj =∑
i,j,k
δkjyj
=∑
j
xjyj = x.y
and that |Ax|2 = Ax.Ax = x.x = |x|2.
Note. Any isometry on Rn, fixing 0, is an orthogonal map. See IB Geometry
Remark 5.11. Consider the group O2 and A =
(a b
c d
)
∈ O2, (which may be identified
as the group of isometries of R2 fixing the origin).Then AAT = I = ATA. Hence
a2 + b2 = 1 = c2 + d2 = a2 + c2 = b2 + d2 and ac + bd = ab+ cd = 0.
So, for a unique θ ∈ [0, 2π), (a, c) = (cos θ, sin θ), (b, d) = ±(− sin θ, cos θ), so A is
either A =
(cos θ − sin θ
sin θ cos θ
)
, or A=
(cos θ sin θ
sin θ − cos θ
)
detA = +1 detA = −1SO2 O2 \ SO2
– 42 –
Part IA: Groups
The former is a rotation
z 7→ eiθz
where (cos θ+ i sin θ)(x+ iy) = (x cos θ − y sin θ) + i(x sin θ+ y sin θ), and the latter is
a reflection in the liney
x=
sin 12θ
cos 12θ
represented as z 7→ eiθz.
Theorem 5.12. Every matrix A ∈ SO3 has eigenvalue 1, the corresponding eigenspace
is the axis of rotation.
Proof. Let λ ∈ R be an eigenvalue, it exists, since real cubics have a real root - so
λ = ±1. The other eigenvalues are either both real or form a complex conjugate pair.
• If the former, all eigenvalues are ±1, product 1, so some eigenvalue +1.
• If the latter, say λ, α, α are the eigenvalues so λ |α|2 = 1, so λ = +1.
Theorem 5.13. Any matrix in SO3 is conjugate to
1 0 0
0 cos θ − sin θ
0 sin θ cos θ
for some θ ∈ [0, 2π).
Proof. Let A ∈ SO3 and v ∈ V = R3 with |v| = 1 and Av = v. Let P ∈ SO3 taking
v to be e1, (where {e1, e2, e3} is the standard basis). Then PAP−1 fixes e, so keeps
e⊥1 = 〈e2, e3〉 invariant, so is a rotation.
Theorem 5.14. O3 ≃ SO3 × {±I3}.
Proof. SO3 ⊳ O3 of index 2, since {±I} ⊳ O3 (since centred) and {±I} ∩ SO3 = {I}.Hence O3 ≃ SO3 × {±I} by (5.7).
Reflections. Let U be a plane in V = R3, let u ∈ V with U⊥ = 〈u〉 and |u| = 1.
Reflection in the plane U is defined as:
rn : v → v − 2(v.n)n
– 43 –
Part IA: Groups
then the matrix of rn with respect to a suitable basis is
1 0 · · · 0
0. . .
. . ....
.... . . 1 0
0 · · · 0 −1
Theorem 5.15. Any matrix in O3 is the product of at most 3 reflections.
Proof. Since
O3 = SO3 ∪
−1 0 0
0 1 0
0 0 1
SO3
it is enough to prove that any element of SO3 is the product of two reflections. Let
A ∈ SO3, let v be its axis of rotation, so v is fixed and v⊥ is invariant under A.
It is enough to show that any element in SO2 is the product of two reflections, which
is clear from the fact that
(cos θ − sin θ
sin θ cos θ
)
=
(cos θ sin θ
sin θ − cos θ
)(1 0
0 −1
)
Remark. We can generalise this to any element of On is a product of at most n
reflections. See IB Geometry.
– 44 –
Part IA: Groups
6 | Mobius transformations
6.1 Mobius group
Definition. A Mobius transformation on C∞ = C∪{∞} (the extended complex plane)
is a mapping
f(z) =az + b
cz + d
with a, b, c, d ∈ C, ad − bc 6= 0 (This restriction makes the mapping injective.), and if
c 6= 0, then define f (−d/c) = ∞, f(∞) = a/c; if c = 0, then define f(∞) = ∞. We
think of ∞ as just a new point in the plane, and it fits our intuition.
Lemma 6.1. f is a permutation of C∞.
Proof. f is injective if we consider
f(z)− f(w) = (z − w) ad− bc(cz + d)(cw + d)
,
so f(z) 6= f(w) in C with cz + d 6= 0 6= cw + d. If z = −d/c 6= w, then f(z) = ∞,f(w) 6=∞. And, if z ∈ C, then f(z) 6= a/c, so ∞ is the unique pre-image. To show f
is surjective, we reverse the argument with the inverse mapping, see below.
Theorem 6.2. The setM of all Mobius transformations is a group under composition,
M < Sym(C∞).
Proof. If f(z) =az + b
cz + dwith ad − bc 6= 0, and g(z) =
αz + β
γz + δwith αδ − βγ 6= 0 for
f, g ∈M.
Closure.
f ◦ g(z) =a
(αz + β
γz + δ
)
+ b
c
(αz + β
γz + δ
)
+ d
=(aα + bγ) z + (aβ + bδ)
(cα + dγ) z + (cβ + dδ)
with (aα + bγ) (cβ + dδ)−(aβ + bδ) (cα + dγ) = (ad− bc) (αδ − βγ) 6= 0, so f ◦g ∈M.
Associative. matrix composition is associative.
Identity. i(z) = z for all z ∈ C∞, and we have z =1 · z1⇒ a = 1 = d, b = 0 = c.
Inverse. f(z) =az + b
cz + dwith ad− bc 6= 0 has inverse g(z) =
dz − b−cz + a
∈M: f ◦ g = e =
g ◦ f on C∞:
– 45 –
Part IA: Groups
• If c = 0, so f(z) = (az + b)/d, g(z) = (dz − b)/a for z ∈ C, and we check
f ◦ g = e = g ◦ f on C. Also f(∞) =∞ = g(∞) so f ◦ g = e = g ◦ f on C∞.
• If c 6= 0, first assume z ∈ C \ {−d/c} we have the mappings
f : C \{
−dc
}
⇋ C \{a
c
}
: g
with fg(z) = z for z ∈ C \ {a/c} and fg(z) = z for z ∈ C \ {−d/c}. Sec-
ondly, gf(−d/c) = g(∞) = −d/c and similarly fg(a/c) = a/c. Also, fg(∞) =
f(−d/c) =∞ and gf (∞) =∞.
Theorem 6.3. There is a surjective homomorphism
θ : GL2 (C) →M(a b
c d
)
7→ f : z 7→ az + b
cz + d
with kernel {λI2|λ ∈ C∗}. Thus GL2 (C) / {λI2|λ ∈ C∗}.
Proof. Let A =
(a b
c d
)
∈ GL2 (C) so detA = ad − bc 6= 0, so θ(A) ∈ M. θ is a
homomorphism:
θ
[(a b
c d
)(α β
γ δ
)]
= θ
[(a b
c d
)]
◦ θ[(
α β
γ δ
)]
(see before). θ is surjective follows from the definition of M. For
(a b
c d
)
∈ ker θ, if
z =az + b
cz + dfor all z ∈ C∞
cz2 + (d− a) z − b = 0 ∀z ∈ C∞
so c = 0 = b, a = d 6= 0, so
(a b
c d
)
= λI, λ = a.
Remark 6.4. SL2(C)/ {±I2} ≃ M, for, consider the restriction θ′ of θ above to SL2(C)
θ′ : SL2(C)→M
– 46 –
Part IA: Groups
θ′ is a homomorphism with kernel {±I2}. θ′ is surjective ontoM; let
f(z) =az + b
cz + d, ad− bd = D 6= 0
then θ′[(
a′ b′
c′ d′
)]
= f with a′ =a√D, b′ =
b√D, etc, thus det
(a′ b′
c′ d′
)
= 1.
Theorem 6.5. The action of M on C∞ is triply transitive if z∞, z0, z1 are distinct
points in C∞, and ω∞, ω0, ω1 are distinct points in C∞. Then there exists f ∈M with
f(zi) = ωi, i ∈ {∞, 0, 1}. Moreover, this f is unique.
Proof. To take z∞, z0, z1 to ∞, 0, 1, let
g(z) =
(z1 − z∞z1 − z0
)z − z0z − z∞
then
z∞ =∞⇒ z 7→ z − z0z1 − z0
z1 =∞⇒ z 7→ z − z0z − z∞
z0 =∞⇒ z 7→ z1 − z∞z − z∞
Similarly, let h ∈ M with h(ωi) = i for i ∈ {∞, 0, 1}, put f = h−1g which takes zi to
ωi. To prove its uniqueness, we note the following
Note 6.6 (Stabilisers of M). M∞01 = {e}, M∞ =
{az + b
d: ad 6= 0
}
, M∞0 ={az
d: ad 6= 0
}
andM∞01 = {z} = {e}.Hence Mz∞z0z1 = g−1M∞01g = {e}.1 Thus, if f1 : zi 7→ ωi, i ∈ {∞, 0, 1} , f1 ∈ M,
then f−1f ∈Mz∞z0z1 = {e}. So f1 = f .
Conjugacy classes in M and fixed points on C∞
Recall from 5.7. Jordan Normal form. Any matrix in GL2(C) is conjugate to one of
(a)
(λ1 0
0 λ2
)
, λ1 6= λ2 6= 0;
1Recall that Gg(x) = gGxg−1.
– 47 –
Part IA: Groups
(b)
(λ 1
0 λ
)
, λ ∈ C \ {0}, which cannot be diagonalised;
(c)
(λ 0
0 λ
)
= λI
Remark. If PAP−1 = B for A,B,P ∈ GL2(C) then θ(P)θ(A)θ(P)−1 = θ(B).
Theorem 6.7. Any non-identity Mobius transformation is conjugate to one of
(a) z 7→ az, a 6= 0, 1;
(b) z 7→ z + 1.
Corollary 6.8. Any non-identity Mobius transformation fixes 1 or 2 points in C∞. 2
Proof. (6.7) Let f ∈M and A ∈ GL2(C) with θ(A) = f ∈M.
(a) If A is conjugate in GL2(C) to
(λ1 0
0 λ2
)
, λ1 6= λ2 6= 0, then f(z) is conjugate to
z 7→ az (with a = λ1/λ2).
(b) If A is conjugate to
(λ 1
0 λ
)
, then f(z) is conjugate to f : z 7→ λz + 1
λ= z + λ−1,
which can be conjugated further to z 7→ z + 1; if g = θ
(λ 0
0 1
)
, then g(z 7→
z + λ−1)g−1 = z 7→ z + 1 since
(λ 0
0 1
)(1 λ−1
0 1
)(λ−1 0
0 1
)
=
(1 1
0 1
)
.
Remark. If A =
(λ 0
0 λ
)
, then A ∈ ker θ, so f = e, which we’ve assumed not so.
An alternate proof without using the Jordan Normal form:
f(z) =az + b
cz + d, with f 6= e
2No fix-point-free Mobius transformation, in contrast with the finite case (example 3 Q10)
– 48 –
Part IA: Groups
consider fixed points of f on C∞, that involves solving the equation z =az + b
cz + d, i.e.
cz2 + (d− a) z − b = 0.
This is quadratic, collinear and non-trivial (as f 6= e).
(a) If there are two fixed points: z1 and z2, then let g ∈M with g(z1) =∞, g(z2) = 0,
then gfg−1 fixes ∞ and 0, so is z 7→ az/δ, thus f is conjugate to z 7→ λz.
(b) If there is only one fixed point, say z∞, then let g ∈ M with g(z∞) = ∞, thengfg−1 fixes precisely ∞. So gfg−1 : z 7→ αz + β, with α = 1. (Otherwise there’ll
be another fixed point β/(1− α).)So gfg−1(z) = z + β, now we can conjugate further to z 7→ z + 1 as in (6.7).
�
Theorem 6.9. If f ∈M, f 6= e, then for some g ∈M, gfg−1 is either z 7→ az, 0 6= a 6=1, here f fixes precisely the points g−1(0), g−1(∞), or z 7→ z + 1, here f fixes precisely
g−1(∞).
Remarks. If g is a finite group acting transitively on a set X , there exists an element
g ∈ G with no fixed points on X :
For, let x ∈ X and H = Gx, for any y ∈ X , choose g ∈ G with g(x) = y; then
Gy = gGxg−1 = gHg−1. Now Ex.3/10 shows there exist f ∈ G \ ⋃g∈G gHg
−1, so f
fixes nothing.
Example 6.10. Iterations of Mobius transformations
Assume f ∈M, fixing a unique point of C∞, what happens to fn(z), where z ∈ C∞?
There exists g ∈ M with h = gfg−1 : z 7→ z + 1, hn(z) = z + n, so hn(z) = ∞,
a fixed point of h, as h → ∞ for any z. Take z ∈ C∞, then hn(g(z)) = ∞, so
fn(z) = g−1(hn(g(z)))→ g−1(∞).
Proposition 6.11. Any Mobius transformation can be written as a composition of
Mobius transformations.
(a) z 7→ αz (a 6=) dilation, rotation;
(b) z 7→ z + β, translation;
(c) z 7→ z−1, inversion.
– 49 –
Part IA: Groups
Proof. Let f(z) = (az + b) / (cz + d) with ad− bc 6= 0. If c = 0, then f(z) = az/d+ b/d,
so f = f2f1 with f1(z) = az/d and f2(z) = z + b/d.
If c 6= 0, then
f(z) =acz + b
d
z + dc
= A +B
z + dc
with A = a/c, B = −(ad − bc)/c2 ∈ C. Hence f = f4f3f2f1 with f1(z) = z + d/c,
f2(z) = z−1, f3(z) = Bz and f4(z) = z + A.
Example. Circles and straight lines in C are circles in C.
Recall. General equation of a line/straight line in C is Azz + Bz + Bz + C = 0 with
A,C ∈ R and |B|2 > AC. If we put z = x+ iy and B = b1 + ib2, then
A(x2 + y2) + 2 (b1x+ b2y)︸ ︷︷ ︸
2Re(Bz)
+C = 0
which is a circle iff A 6= 0. Note that a straight line is a ’circle’ in C∞; it is L ∪ {∞}with L a line in C, and called the ’circle through infinity’.
Theorem 6.12. Any Mobius transformation takes circles/real lines to circles/real lines
in C. 3
Proof. Let f ∈M, we need to check the image under f of a circle in C∞ is a circle in
C∞.
By 6.11, it is enough to check that this is so for f(z) one of z 7→ z−1, z 7→ az and
z 7→ z+ b. This is clear for the latter, and if w = z−1, if one circle in C∞ is Azz+Bz+
Bz + C = 0 with A,C real etc.
Putting w = z−1, we get
Cww +Bw +Bw + A = 0
with A,C, real etc. So the w form a circle in C∞.
Example. Find the image of the real axis under f(z) =z − iz + i
.
Solution: Consider that f(∞) = 1, f(0) = −1 and f(1) = −i, so the image of the real
axis is the circle containing 1,−1,−i, so is the unit circle. Moreover, the upper half
plane goes to the inside of the circle, e.g. i 7→ 0.
Example. Show there is a necklace of circles touching both of these circles.
3circles → circles in C∞
– 50 –
Part IA: Groups
21
1
Consider the transformation w = z/(2 − z).
6.2 Cross ratios
Definition. Let z1, z2, z3, z4 be distinct points of C∞. The cross ratio is the element x
of C∞ such that the f ∈M with
z1 z2 z3 z4f : ∞ 0 1 x = [z1, z2, z3, z4]
so
[z1, z2, z3, z4] =z3 − z1z3 − z2
· z4 − z2z4 − z1
.
If a zj is infinity:
[∞, z2, z3, z4] =z4 − z2z3 − z2
,
[z1,∞, z3, z4] =z3 − z1z4 − z1
,
[z1, z2,∞, z4] =z4 − z2z4 − z1
,
[z1, z2, z3,∞] =z3 − z1z3 − z2
.
Warning! Some people use different permutations of the zj .
– 51 –
Part IA: Groups
Theorem 6.13. If z1, z2, z3, z4 are distinct points in C∞ and w1, w2, w3, w4 are also
distinct points in C∞. Then there exist f ∈ M with f(zj) = wj for j = 1, 2, 3, 4 if and
only if [z1, z2, z3, z4] = [w1, w2, w3, w4].
Proof. (⇒)
Lemma 6.14. If z1, z2, z3, z4 are distinct inC∞ and f ∈ M then [f(z1), f(z2), f(z3), f(z4)] =
[z1, z2, z3, z4] and thusM preserves cross ratios.
Proof of Lemma.
g ∈M :↓ f(z1) f(z2) f(z3) f(z4)∞ 0 1 LHS
gf :↓ z1 z2 z3 z4∞ 0 1 LHS
but from definition gf(z4) = RHS.
(⇐) Assume [z1, z2, z3, z4] = [w1, w2, w3, w4] = x. Let
g ∈M ↓: z1 z2 z3 z4∞ 0 1 x
h ∈M ↑: w1 w2 w3 w4
Now take f = h−1g ∈M as desired.
Example (A geometric application). z1, z2, z3, z4 lie on a common circle in C∞ ⇐⇒[z1, z2, z3, z4] ∈ R.
Proof. Let f ∈M withz1 z2 z3 z4
f :7→ ∞ 0 1 x
where x = [z1, z2, z3, z4]. The circle on ∞, 0, 1 is the extended real line, and M sends
circles to circles in C∞.
*An action of GL2(C) on P′(C)
GL2(C) acts on C2, and so it acts on P′(C) which is the set of 1-dimensional subspace
of C2. The kernel of the action on P′(C) is Z = {λI2 : λ ∈ C∗}.
Look at 〈(x, y)〉 , called x, y, which is an element of C2 if y 6= 0, and ∞ if y = 0.
– 52 –
Part IA: Groups
• If y 6= 0:
(a b
c d
)(x
1
)
=
(ax+ b
cx+ d
)
, x 7→ ax+bcx+d
• If y = 0:
(a b
c d
)(1
0
)
=
(a
c
)
, ∞ 7→ ac
*A geometric view of ∞ in C∞ - The Riemann sphere
Let S be the unit sphere in R3, C the plane {(x, y, 0)|x, y ∈ R} and ζ the north pole
of S (0, 0, 1). Consider the map
φ : C 7→ S \ {ζ}z 7→ z;
which is the stereographic projection from C, where z′ is the unique point of S \ {ζ}which lies on the line zζ .
ζ
φ(z)
x
b
b
b
We have
φ(z) =
(
2x
|z|2 + 1,
2y
|z|2 + 1,|z|2 − 1
|z|2 + 1
)
for z = (x, y, 0)
let φ(∞) = ζ, so
φ : C∞ 7→ S
a bijection. For f ∈M, we can consider the action φfφ−1 on S
circles ←→ circles
↓ ↓∞ ←→ ζ
In fact, the stereographic projection is the one-point compactification of C. (See IB
Geometry)
– 53 –
