COMPUTING CHARACTER TABLES OF FINITE GROUPS

Jay Taylor(Università degli Studi di Padova)

Symmetry

Symmetry

Chemistry

Symmetry

Chemistry Biology

Symmetry

Chemistry Biology Physics

Symmetry

Chemistry Biology Physics

Groups

Symmetry

Chemistry Biology Physics

Groups

NumberTheory

Symmetry

Chemistry Biology Physics

Groups

NumberTheory Topology

Symmetry

Chemistry Biology Physics

Groups

NumberTheory Topology Algebraic

Geometry

DEFINITION

DEFINITIONA group is a pair with a set and a binary operation such that:

(G, ?) G ? : G⇥G ! G

DEFINITIONA group is a pair with a set and a binary operation such that:1. there exists an element such that for

all x 2 G

(G, ?) G ? : G⇥G ! G

e 2 Gx ? e = e ? x = x

DEFINITIONA group is a pair with a set and a binary operation such that:1. there exists an element such that for

all2. for every there exists an element such that

.

x 2 G

(G, ?) G ? : G⇥G ! G

e 2 Gx ? e = e ? x = x

g 2 G g-1 2 Gg ? g-1 = g-1 ? g = e

DEFINITIONA group is a pair with a set and a binary operation such that:1. there exists an element such that for

all2. for every there exists an element such that

.3. for all

x 2 G

(G, ?) G ? : G⇥G ! G

e 2 Gx ? e = e ? x = x

g 2 G g-1 2 Gg ? g-1 = g-1 ? g = e

a ? (b ? c) = (a ? b) ? c a, b, c 2 G.

DEFINITIONA group is a pair with a set and a binary operation such that:1. there exists an element such that for

all2. for every there exists an element such that

.3. for all

x 2 G

(G, ?) G ? : G⇥G ! G

e 2 Gx ? e = e ? x = x

g 2 G g-1 2 Gg ? g-1 = g-1 ? g = e

a ? (b ? c) = (a ? b) ? c a, b, c 2 G.

Examples

• with ,(Z,+) Z = {. . . ,-2,-1, 0, 1, 2, . . . }

DEFINITIONA group is a pair with a set and a binary operation such that:1. there exists an element such that for

all2. for every there exists an element such that

.3. for all

x 2 G

(G, ?) G ? : G⇥G ! G

e 2 Gx ? e = e ? x = x

g 2 G g-1 2 Gg ? g-1 = g-1 ? g = e

a ? (b ? c) = (a ? b) ? c a, b, c 2 G.

Examples

• with ,• with where denote the real numbers.

(Z,+) Z = {. . . ,-2,-1, 0, 1, 2, . . . }

(R⇥,⇥) R⇥ = R \ {0} R

DIHEDRAL GROUPS

DIHEDRAL GROUPS

DIHEDRAL GROUPS1 2

34

DIHEDRAL GROUPS

1

23

4

DIHEDRAL GROUPS1 2

34

DIHEDRAL GROUPS1 2

34

DIHEDRAL GROUPS1 2

34

DIHEDRAL GROUPS

1 2

34

DIHEDRAL GROUPS1 2

34

DIHEDRAL GROUPS1 2

34

DIHEDRAL GROUPS1 2

34

DIHEDRAL GROUPS1

23

4

DIHEDRAL GROUPS1

23

4

90o

DIHEDRAL GROUPS4 1

23

DIHEDRAL GROUPS4 1

23

90o

DIHEDRAL GROUPS

4

12

3

90o

DIHEDRAL GROUPS

4

12

3

90o

DIHEDRAL GROUPS2 1

43

90o

DIHEDRAL GROUPS2 1

43

90o

DIHEDRAL GROUPS

2 1

43

90o

DIHEDRAL GROUPS

2 1

43

180o

90o

DIHEDRAL GROUPS3 4

12

90o

DIHEDRAL GROUPS3 4

12

90o 180o

DIHEDRAL GROUPS

3

41

2

90o 180o

DIHEDRAL GROUPS

3

41

2

90o 180o

DIHEDRAL GROUPS1 4

32

90o 180o

DIHEDRAL GROUPS1 4

32

90o 180o

DIHEDRAL GROUPS

1 4

32

90o 180o

DIHEDRAL GROUPS

1 4

32

90o 180o

270o

DIHEDRAL GROUPS

90o 180o

DIHEDRAL GROUPS

90o 180o 270o

DIHEDRAL GROUPS

90o

180o 270o

DIHEDRAL GROUPS

90o

180o 270o

a b

DIHEDRAL GROUPS

90o

180o 270o

a b ab

DIHEDRAL GROUPS

90o

180o 270o

a b ab aba

DIHEDRAL GROUPS

90o

180o 270o

a b ab aba

abab

DIHEDRAL GROUPS

90o

180o 270o

a b ab aba

abab ababa

DIHEDRAL GROUPS

90o

180o 270o

a b ab aba

abab ababa ababab

DIHEDRAL GROUPS

90o

180o 270o

a b ab aba

abab ababa ababab e

DIHEDRAL GROUPS

90o

180o 270o

a b ab aba

abab ababa ababab e

a2 = e,

DIHEDRAL GROUPS

90o

180o 270o

a b ab aba

abab ababa ababab e

a2 = e, b2 = e,

DIHEDRAL GROUPS

90o

180o 270o

a b ab aba

abab ababa ababab e

a2 = e, b2 = e, (ab)4 = e

DIHEDRAL GROUPS

90o

180o 270o

a b ab aba

abab ababa ababab e

a2 = e, b2 = e, (ab)4 = eiI2(4) = ha, b |

DIHEDRAL GROUPS

DIHEDRAL GROUPS

72o 144o

216o 288o

a b ab aba

(ab)3

(ab)2

(ab)2a (ab)3a (ab)4 e

DIHEDRAL GROUPS

72o 144o

216o 288o

a b ab aba

(ab)3

(ab)2

(ab)2a (ab)3a (ab)4 e

I2(5) = ha, b | a2, b2, (ab)5i

DIHEDRAL GROUPS

72o 144o

216o 288o

a b ab aba

(ab)3

(ab)2

(ab)2a (ab)3a (ab)4 e

I2(m) = ha, b | a2, b2, (ab)mi

MATRIX REPRESENTATION

MATRIX REPRESENTATION

v1

v2

MATRIX REPRESENTATION

v1

v2

MATRIX REPRESENTATION

v1

v2 v1

v2

MATRIX REPRESENTATION

v1

v2 v1

v2

0 11 0

�

a

MATRIX REPRESENTATION

v1

v2

0 11 0

�

a

MATRIX REPRESENTATION

v1

v2

0 11 0

�

a

MATRIX REPRESENTATION

v1

v2

v1v2

0 11 0

�

a

MATRIX REPRESENTATION

v1

v2

v1v2

0 11 0

� 1 00 -1

�

a b

MATRIX REPRESENTATION

v1

v2

0 11 0

� 1 00 -1

�

a b

MATRIX REPRESENTATION

v1

v2

0 11 0

� 1 00 -1

�

a b

MATRIX REPRESENTATION

v1

v2

v1

v2

0 11 0

� 1 00 -1

�

a b

MATRIX REPRESENTATION

v1

v2

v1

v2

0 11 0

� 1 00 -1

� 0 -11 0

�

a b ab

MATRIX REPRESENTATION0 11 0

� 1 00 -1

� 0 -11 0

�

a b ab

-1 00 1

�

aba

-1 00 -1

� 0 -1-1 0

� 0 1-1 0

� 1 00 1

�

abab ababa ababab e

REPRESENTATIONS

REPRESENTATIONS

• an -dimensional -vector space.V n C

REPRESENTATIONS

• an -dimensional -vector space.

• the group of all invertible linear transformations .

V n C

GL(V) ⇠= GLn(C)f : V ! V

REPRESENTATIONS

• an -dimensional -vector space.

• the group of all invertible linear transformations .

• A representation of a group is a map such that for all we have

(G, ?) ⇢ : G ! GL(V)

a, b 2 G

V n C

GL(V) ⇠= GLn(C)f : V ! V

⇢(a ? b) = ⇢(a) � ⇢(b)

CHARACTERS• Let be a representation of a group . The

function defined by

is called the character of .

⇢ : G ! GLn(C) (G, ?)

�⇢ : G ! C

�⇢(a) = Tr(⇢(a))

⇢

CHARACTERS• Let be a representation of a group . The

function defined by

is called the character of .

e a b ab aba abab ababa ababab

0 11 0

�1 00 1

� 1 00 -1

� 0 -11 0

� -1 00 1

� -1 00 -1

�0 -1-1 0

� 0 1-1 0

�

2 0 0 0 0 0 0-2

⇢ : G ! GLn(C) (G, ?)

�⇢ : G ! C

�⇢(a) = Tr(⇢(a))

⇢

CONJUGACY

CONJUGACY• For any matrices recall thatA, B 2 GLn(C)

Tr(ABA-1) = Tr(B)

CONJUGACY• For any matrices recall that

• Hence for any two elements and any character we have

A, B 2 GLn(C)

Tr(ABA-1) = Tr(B)

a, b 2 G � : G ! C

�(a ? b ? a-1) = �(b)

CONJUGACY• For any matrices recall that

• Hence for any two elements and any character we have

• We say are conjugate if there exists an element such that .

A, B 2 GLn(C)

Tr(ABA-1) = Tr(B)

a, b 2 G � : G ! C

�(a ? b ? a-1) = �(b)

a, b 2 G x 2 G

x ? a ? x-1 = b

CONJUGACY• For any matrices recall that

• Hence for any two elements and any character we have

• We say are conjugate if there exists an element such that .

• This defines an equivalence relation on . The resulting equivalence classes are called conjugacy classes.

A, B 2 GLn(C)

Tr(ABA-1) = Tr(B)

a, b 2 G � : G ! C

�(a ? b ? a-1) = �(b)

a, b 2 G x 2 G

x ? a ? x-1 = b

G

CONJUGACY

90o 180o270o

a b

ab

aba

abab

ababa

ababab e

IRREDUCIBLE CHARACTERS

IRREDUCIBLE CHARACTERS• A representation is irreducible if there is no proper

subspace which is invariant under . By this we mean that for all we have .

⇢ : G ! GL(V)

W ✓ V G

⇢(g)W ✓ Wg 2 G

IRREDUCIBLE CHARACTERS• A representation is irreducible if there is no proper

subspace which is invariant under . By this we mean that for all we have .

• We have is irreducible if and only if

⇢ : G ! GL(V)

W ✓ V G

⇢(g)W ✓ W

1

|G|

X

g2G

�⇢(g)�⇢(g) = 1

g 2 G

⇢ : G ! GL(V)

IRREDUCIBLE CHARACTERS• A representation is irreducible if there is no proper

subspace which is invariant under . By this we mean that for all we have .

• We have is irreducible if and only if

• A character with this property is also called irreducible.

⇢ : G ! GL(V)

W ✓ V G

⇢(g)W ✓ W

1

|G|

X

g2G

�⇢(g)�⇢(g) = 1

g 2 G

⇢ : G ! GL(V)

IRREDUCIBLE CHARACTERS

a b

IRREDUCIBLE CHARACTERS

a b

1

8(22 + 0+ 0+ 0+ 0+ (-2)2 + 0+ 0) = 1

CHARACTER TABLES

Theorem

The number of distinct irreducible characters of a finite group is equal to the number of conjugacy classes.

CHARACTER TABLES

Theorem

The number of distinct irreducible characters of a finite group is equal to the number of conjugacy classes.

• Let be representatives for the conjugacy classes and let be the irreducible characters of . The square matrix

is called the character table of .

g1, . . . , gn 2 G

G�1, . . . ,�n

G

(�i(gj))16i,j6n

CHARACTER TABLES

�1

e a b ab (ab)2

�2

�3

�4

1 1 1 1 1

1 1 1

1 1 1

1 1 1

2 0 0 0 -2

-1 -1

-1

-1 -1

I2(4)

-1

CHARACTER TABLES

�1

e a b

�2

�3

�4

1 1 1 1 1

1 1

1 1

1 1 1

2 0 0

-1

-1

-1 -1

I2(2m) (ab)m(ab)r

(-1)r

(-1)r

(-1)m

(-1)m

j 2(-1)j"jr + "-jr

CHARACTER TABLES

�1

e a b

�2

�3

�4

1 1 1 1 1

1 1

1 1

1 1 1

2 0 0

-1

-1

-1 -1

I2(2m) (ab)m(ab)r

(-1)r

(-1)r

(-1)m

(-1)m

j 2(-1)j"jr + "-jr

1 6 j, r 6 m- 1 " = e⇡i/m

SYMMETRIC GROUPS

SYMMETRIC GROUPS

• is the group of all bijective functions .Sn f : {1, . . . , n} ! {1, . . . , n}

SYMMETRIC GROUPS

• is the group of all bijective functions .

• We call the symmetric group on points.

Sn f : {1, . . . , n} ! {1, . . . , n}

Sn n

SYMMETRIC GROUPS

• is the group of all bijective functions .

• We call the symmetric group on points.

• which can be very large even for small . For example

Sn f : {1, . . . , n} ! {1, . . . , n}

Sn n

|Sn| = n! n

|S20| = 2432902008176640000

SYMMETRIC GROUPS

SYMMETRIC GROUPS

1

2

3

1

2

3

1

2

3

1

2

3

1

2

3

1

2

3

1

2

3

1

2

3

1

2

3

1

2

3

1

2

3

1

2

3

Example ( )n = 3

SYMMETRIC GROUPS

�

1

2

3

1

2

3

1

2

3

1

2

3

1

2

3

1

2

3

1

2

3

1

2

3

1

2

3

1

2

3

1

2

3

1

2

3

1

2

3

1

2

3

1

2

3

1

2

3

Example ( )n = 3

SYMMETRIC GROUPS

� =

1

2

3

1

2

3

1

2

3

1

2

3

1

2

3

1

2

3

1

2

3

1

2

3

1

2

3

1

2

3

1

2

3

1

2

3

1

2

3

1

2

3

1

2

3

1

2

3

1

2

3

1

2

3

1

2

3

Example ( )n = 3

SYMMETRIC GROUPS

� = =

1

2

3

1

2

3

1

2

3

1

2

3

1

2

3

1

2

3

1

2

3

1

2

3

1

2

3

1

2

3

1

2

3

1

2

3

1

2

3

1

2

3

1

2

3

1

2

3

1

2

3

1

2

3

1

2

3

1

2

3

1

2

3

Example ( )n = 3

SYMMETRIC GROUPS• A function is called a cycle of length if there exists a

subset such that for any integer and acts on the elements of in the following wayi 62 X

x1

xk

xk-1

x4

x3

x2

f

f

f

f

f

f

f

f 2 Sn k

X = {x1, . . . , xk} ✓ {1, . . . , n} f(i) = i

f X

SYMMETRIC GROUPS

Lemma

Every element of is a product of disjoint cycles.Sn

SYMMETRIC GROUPS

Lemma

Every element of is a product of disjoint cycles.Sn

=

1

2

3

4

1

2

3

4

1

2

3

4

1

2

3

4

�

1

2

3

4

1

2

3

4

SYMMETRIC GROUPS

SYMMETRIC GROUPS• A partition of is a sequence of integers such

that and .n µ = (µ1, . . . , µk)

µ1 + · · ·+ µk = nµ1 > · · · > µk > 1

SYMMETRIC GROUPS• A partition of is a sequence of integers such

that and .

• For example the partitions of are

n µ = (µ1, . . . , µk)

µ1 + · · ·+ µk = nµ1 > · · · > µk > 1

(5), (4, 1), (3, 2), (3, 1, 1), (2, 2, 1), (2, 1, 1, 1), (1, 1, 1, 1, 1)

5

SYMMETRIC GROUPS• A partition of is a sequence of integers such

that and .

• For example the partitions of are

• Given let be a decomposition of into a product of disjoint cycles. If denotes the length of the cycle then the sequence is a partition of , after possibly reordering the entries. We call the cycle type of .

n µ = (µ1, . . . , µk)

µ1 + · · ·+ µk = nµ1 > · · · > µk > 1

(5), (4, 1), (3, 2), (3, 1, 1), (2, 2, 1), (2, 1, 1, 1), (1, 1, 1, 1, 1)

f 2 Sn f1 � · · · � fk f

µi fiµ(f) = (µ1, . . . , µk) n

µ(f) f

5

SYMMETRIC GROUPS

Theorem

Two elements of the symmetric group are conjugate if and only if they have the same cycle type.

SYMMETRIC GROUPS

Theorem

Two elements of the symmetric group are conjugate if and only if they have the same cycle type.

• We will write for the set of all partitions of .P(n) n

SYMMETRIC GROUPS

Theorem

Two elements of the symmetric group are conjugate if and only if they have the same cycle type.

• We will write for the set of all partitions of .

• For any partition we denote by an irreducible character of .

P(n) n

� 2 P(n) ��

Sn

SYMMETRIC GROUPS

Theorem

Two elements of the symmetric group are conjugate if and only if they have the same cycle type.

• We will write for the set of all partitions of .

• For any partition we denote by an irreducible character of .

P(n) n

|P(20)| = 627

� 2 P(n) ��

Sn

HOOKS OF PARTITIONS• Consider the partition .µ = (5, 5, 4, 3, 1, 1) 2 P(19)

HOOKS OF PARTITIONS• Consider the partition .µ = (5, 5, 4, 3, 1, 1) 2 P(19)

HOOKS OF PARTITIONS• Consider the partition .µ = (5, 5, 4, 3, 1, 1) 2 P(19)

HOOKS OF PARTITIONS• Consider the partition .µ = (5, 5, 4, 3, 1, 1) 2 P(19)

HOOKS OF PARTITIONS• Consider the partition .µ = (5, 5, 4, 3, 1, 1) 2 P(19)

H22

HOOKS OF PARTITIONS• Consider the partition .µ = (5, 5, 4, 3, 1, 1) 2 P(19)

h22 = 6

H22

HOOKS OF PARTITIONS• Consider the partition .µ = (5, 5, 4, 3, 1, 1) 2 P(19)

h22 = 6

l22 = 6

H22

HOOKS OF PARTITIONS• Consider the partition .µ = (5, 5, 4, 3, 1, 1) 2 P(19)

h22 = 6

l22 = 6

HOOKS OF PARTITIONS• Consider the partition .µ = (5, 5, 4, 3, 1, 1) 2 P(19)

h22 = 6

l22 = 6

R22

HOOKS OF PARTITIONS• Consider the partition .µ = (5, 5, 4, 3, 1, 1) 2 P(19)

h22 = 6

l22 = 6

R22

µ \ R22

CHARACTER TABLE

Theorem (Murnaghan–Nakayama Formula)

Write as a product of disjoint cycles. Assume is a cycle of length then the element

is contained in the symmetric group . For any partition we have

f 2 Sn

� 2 P(n)

f1 � · · · � fkfk m

g = f1 � · · · � fk-1

Sn-m

��(f) =X

hij=m

(-1)lij��\Rij(g)

CHARACTER TABLE11111 2111 221 311 32 41 5

5

41

32

311

221

2111

11111

1 1 1 1 1 1 1

1

1

1

1

1

1

1

1

1

1

2 1

1

1

0 0 0

0

0

0

0

0

0

4

4

5

5

6

S5

-1 -1

-1 -1

0 -2

-1 -1 -1

-2 -1

-1 -1 -1