Character tables

for some small groups

P R Hewitt

U of Toledo

12 Feb 07

References:

1. P Neumann, On a lemma which is notBurnside’s, Mathematical Scientist 4 (1979),133-141.

2. JH Conway et al., Atlas of Finite Groups,Oxford University Press, 1985.

3. J-P Serre, Linear Representations of FiniteGroups, Springer-Verlag (GTM 42), 1977.

4. GAP: Groups, Algorithms, Programming.http://www.gap-system.org/, especially.../Manuals/doc/htm/tut/chapters.htm.

5. MAGMA: High performance software forAlgebra, Number Theory, and Geometry.http://magma.maths.usyd.edu.au/, especially.../magma/htmlhelp/MAGMA.htm.

Cyclic group:

Cn = 〈a | an = 1〉

Any Cn-module: direct sum of eigenspaces.

Eigenvalues: n-th roots of 1.

Simple modules determined by

λk : a 7→ ζkn := exp(2kiπ/n), k = 0 . . . n− 1.

Cn 1 a a2 a3 · · ·1 1 1 1 1 · · ·λ1 1 ζn ζ2n ζ3n · · ·λ2 1 ζ2n ζ4n ζ6n · · ·λ3 1 ζ3n ζ6n ζ9n · · ·... ... ... ... ...ρ n 0 0 0 · · ·

Degree-weighted row sum: regular character.∑φ∈Irr

φ(1)φ(g) = ρ(g) :=

character of representation on CG

=

|G| , if g = 1

0, otherwise

Each irreducible appears in CG exactly as many

times as its degree. Each nonidentity g is a fpf

permutation of the basis G. In particular:∑φ∈Irr

φ(1)2 = |G|

This is a special case of column orthogonality.

Klein’s “Vierergruppe”:

V :=⟨a, b | a2 = b2 = [a, b] = 1

⟩.

Commuting set of matrices has simultaneous

Jordan form. In particular: common eigenspace.

Simple modules of abelian group are 1-dimensional,

given by a linear character:

λ : G→ C×.

Image of a linear character is cyclic. More

generally, for any simple module (any field,

any group!) the image of the center is cyclic

(Schur’s lemma).

V 1 a b ab1 1 1 1 1λa 1 1 −1 −1λb 1 −1 1 −1λab 1 −1 −1 1ρ 4 0 0 0

First nonabelian group:

D3 = Sym3

Conjugacy classes (representative × class length):

1, (12)× 3, (123)× 2.

[D3, D3] = Alt3 = 〈(123)〉 .

Linear characters: 1, sgn.

R : G→ GLn =⇒ R : [G,G] → SLn .

Remaining character degree:√6− 12 − 12 = 2.

Dihedral representation:

∆(12) = reflection; eigenvalues 1, −1.

∆(123) = rotation; eigenvalues ζk3, ζ−k3 6= 1.

Sym3 1 3 21 (12) (123)

1 1 1 1sgn 1 −1 1δ 2 0 −1ρ 6 0 0

Note how easy it is to compute character inner

products here. For example,

(sgn, δ)D3= 1 ·1 ·2+3 ·(−1) ·0+2 ·1 ·(−1) = 0

More generally: X = character table, with

rows indexed by φ ∈ Irr, columns indexed by

conjugacy class representatives g, entries

Xφ,g = φ(g).

Let L = diagonal matrix with entries

Lg,g = class length = |G| / |C(g)| .

Row orthogonality:

1

|G|XLXH = I.

Row orthogonality yields column orthogonality:

1

|G|XLXH = I =⇒ XHX = |G|L−1 = C,

where the diagonal matrix C has entries

Cg,g = |C(g)| .

In other words the columns are orthogonal vec-

tors, whose squared length is the order of the

corresponding centralizers.

Compare this to the earlier observation about

the weighted row sums: that result said that

the column for the identity has squared length

|G| and is orthogonal to all of the other columns.

Dihedral group of order 8:

D4 =⟨s, t | s2 = t2 = (st)4 = 1

⟩Classes: 1, s× 2, t× 2, r × 2, z × 2,

where r = st and z = r2.

[D4, D4] = Z(D4) = 〈z〉 , D4/[D4, D4] = V.

D4 1 1 2 2 21 z r s t

1 1 1 1 1 1λr 1 1 1 −1 −1λs 1 1 −1 1 −1λt 1 1 −1 −1 1δ 2 −2 0 0 0ρ 8 0 0 0 0

Quaternion group of order 8:

Q := {±1,±i,±j,±k} ⊂ H×.

Classes: 1, −1, i× 2, j × 2, k × 2.

[Q,Q] = Z(Q) = 〈−1〉 , Q/[Q,Q] = V.

Unique nonlinear character ε can be computed

either from the orthogonality relations or from

viewing the quaternions as a 2-dimensional space

over C.

Q 1 1 2 2 21 −1 i j k

1 1 1 1 1 1λi 1 1 1 −1 −1λj 1 1 −1 1 −1λk 1 1 −1 −1 1ε 2 −2 0 0 0ρ 8 0 0 0 0

A := Alt4 = PSL2(3)

Classes: 1, (12)(34)× 3, (123)× 4, (132)× 4.

[A,A] = V, A/[A,A] = C3.

Permutation character = 1+ irreducible.

Alt4 1 3 4 41 (12)(34) (123) (132)

1 1 1 1 1λ1 1 1 ζ3 ζ23λ2 1 1 ζ23 ζ3π 3 −1 0 0

π is irreducible because A is doubly transitive.

More generally, if G has permutation rank r on

X then the permutation character equals 1+π

where (π, π)G = r − 1. We prove this after we

discuss tensor products.

A := SL2(3)

Since SL2(3)/±1 = Alt4, the character table of

Alt4 is embedded in the one for SL2(3). Each

element in Alt4 pulls back to two elements in

SL2(3), the negatives of one another. The

traces of these elements differ unless the trace

is 0. The only elements of trace 0 are square

roots of −I, the unique element of order 2.

Thus, the Sylow 2-subgroups of SL2(3) are

quaternion. The elements of order 2 in Alt4pull back to a single class of elements qL, of

order 4, whereas every other element pulls back

to a pair of nonconjugate elements, on of order

twice the other. Say (123) pulls back to t and

−t, of orders 3 and 6, respectively.

In particular, there are exactly 3 characters

which are nontrivial on −1, and the sum of the

squares of their degrees is 12. Hence each has

degree 2. Now if R : G → GLn and λ : G → C×

are homomorphisms then so is λ ·R : G→ GLn.

Moreover,

tr(λ(g)Rg) = λ(g) tr(Rg).

Thus linear characters act by multiplication

on the rows of the character table. We have

not seen this effect before because whenever

λ(g) 6= 1 it has happened that φ(g) = 0. This

cannot happen with our degree-2 characters,

because the character values of the elements

of order 3 cannot be 0.

SL2(3) 1 1 6 4 4 4 41 −1 q t −t t2 −t2

1 1 1 1 1 1 1 1λ1 1 1 1 ζ3 ζ3 ζ23 ζ23λ2 1 1 1 ζ23 ζ23 ζ3 ζ3π 3 3 −1 0 0 0 0φ0 2 −2 0 −1 1 −1 1φ1 2 −2 0 −ζ23 ζ23 −ζ3 ζ3φ2 2 −2 0 −ζ3 ζ3 −ζ23 ζ23

More generally, if φ and ψ are characters, then

so is φψ — altho only rarely will the product

be irreducible. For suppose that φ = tr(R)

and ψ = tr(S), where R have degrees m and n,

respectively. Consider the action on the tensor

product R⊗S : G→ GLmn. Since one basis for

the tensor product is the tensor product of any

bases for the factors, it is straightforward to

check that

tr(R⊗ S) = tr(R) tr(S).

Actually, there are two tensor product con-

structions. The one above is called the internal

tensor product. The external tensor product is

applied when R is a representation of one group

G, and S is a representation of another H. In

this case R⊗S is a representation of G×H, and

we write φ⊗ψ for the character. The exercise

above shows that

(φ⊗ ψ)(g, h) = φ(g)ψ(h).

One can also check that

(φ1 ⊗ ψ1, φ2 ⊗ ψ2)G×H = (φ1, φ2)G · (ψ1, ψ2)H .

So, the external tensor product of irreducible

characters is irreducible for the product group.

Why isn’t the internal tensor product of irre-

ducibles also irreducible? We can recover the

internal tensor product from the external when

G = H, by the diagonal embedding G→ G×G.

Only rarely does an irreducible character re-

main irreducible when restricted to a subgroup.

Let’s return to the character computation of

permutation rank. A computation similar to

the one for the dimension of the class alge-

bra shows that if G acts on a set X then the

number of G-orbits equals the dimension of the

fixed point subspace of G on CX. That is, if

πX(g) = tr(g on CX)

= #{fixed points for g on X}

then

#{G-orbits on X} = (1, πX)G .

This result is sometimes erroneously referred

to as “Burnside’s Lemma”, but is in fact due to

Cauchy and Frobenius. (See Peter Neumann’s

account, listed in the references.)

Permutation rank on X

= #{orbits of G on X ×X}=

(1, πX×X

)G

=(1, π2

X

)G

= (πX , πX)G= |πX |2 .

The third equality comes from the identity

C(X ×X) = CX ⊗ CX.

(Consider bases.) In particular, if G is doubly

transitive on X then |πX |2 = 2. This implies

that πX is the sum of exactly 2 irreducibles,

one of which must be the trivial character.

One last point about tensor products: if V and

W are G-modules, then so is V ⊗W :

gf := g ◦ f ◦ g−1.

There is a natural isomorphism

V ∗ ⊗W → Hom(V,W ),

where f ⊗ w is sent to the homomorphism

x 7→ f(x) · w.

The character on V ∗ is the complex conjugate

of the character on V . (Look at eigenvalues.)

Either using these observations or by direct

computation we find that if φ and ψ are the

characters on V and W then the character on

Hom(V,W ) is φ · ψ.

G := Alt5 = SL2(4) = PSL2(5).

Classes: 1, (12)(34)× 15, (123)× 20,

(12345)× 12, (13524)× 12.

Doubly transitive permutation representations

of degrees 5 and 6. (Projective lines over F4

and F5, respectively.)

Remaining 2 characters must have degree 3.

Alt5 1 15 20 12 121 (12)(34) (123) (12345) (13245)

1 1 1 1 1 1θ 3 −1 0 τ τ ′

θ′ 3 −1 0 τ ′ τπ 4 0 1 −1 −1σ 5 1 −1 0 0

To find τ and τ ′ (above) we can either use the

orthogonality relations or use the representa-

tion of G as the rotations of a dodecahedron. If

we use the orthogonality relations we find that

both τ and τ ′ satisfy the “Fibonacci” equation

x2 = x + 1. The Galois group of this equa-

tion acts on the character values. If τ 7→ τ ′

generates this group then its composition with

the homomorphism G→ GL3 produces another

homomorphism.

There is also the outer automorphism of G

(conjugation by anything in Sym5−Alt5). Pre-

composition of this with any representation

produces another, swapping the two conjugacy

classes of 5-cycles. In this case the outer au-

tomorphism has the same effect as the Galois

automorphism. (Not always true.)

The character tables for both D4 and Q also

admit both outer and Galois automorphisms.

Compute these.

H := Sym5 = PGL2(5).

Sym5 1 10 15 30 20 20 241 (12) (12)(34) (1234) (123) (123)(45) (12345)

1 1 1 1 1 1 1 1sgn 1 −1 1 −1 1 −1 1π 4 2 0 0 1 −1 −1

sgn ·π 4 −2 0 0 −1 −1 −1σ 5 −1 1 1 −1 −1 0

sgn ·σ 5 1 1 −1 −1 1 0τ 6 0 −2 0 0 0 1

Note:

τ ↓Alt5= θ+ θ′.

Exercises:

• SL2(5).

• SL3(2) = PSL2(7): Note that there is a

doubly transitive representation of degree

8 (projective line over F7).

• Alt6 = Sp4(2): Note that there are two

doubly transitive representations (swapped

by the outer automorphism) and a rank 3

representation of degree 15 (projective 3-

space over F2).

• Sym6.

• Alt8 = SL4(2).