Character tables for some small groups P R Hewitt U of Toledo 12 Feb 07
References: 1. P Neumann, On a lemma which is not Burnside 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 Finite Groups, 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 for Algebra, Number Theory, and Geometry. http://magma.maths.usyd.edu.au/, especially.../magma/htmlhelp/magma.htm.
Cyclic group: C n = a a n = 1 Any C n -module: direct sum of eigenspaces. Eigenvalues: n-th roots of 1. Simple modules determined by λ k : a ζ k n := exp(2kiπ/n), k = 0... n 1.
C n 1 a a 2 a 3 1 1 1 1 1 λ 1 1 ζ n ζn 2 ζn 3 λ 2 1 ζn 2 ζ4 n ζ6 n λ 3 1 ζn 3 ζn 6 ζn 9..... 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 a 2 = b 2 = [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 ab 1 1 1 1 1 λ a 1 1 1 1 λ b 1 1 1 1 λ ab 1 1 1 1 ρ 4 0 0 0
First nonabelian group: D 3 = Sym 3 Conjugacy classes (representative class length): 1, (12) 3, (123) 2. [D 3, D 3 ] = Alt 3 = (123). Linear characters: 1, sgn. R: G GL n = R: [G, G] SL n. Remaining character degree: 6 1 2 1 2 = 2. Dihedral representation: (12) = reflection; eigenvalues 1, 1. (123) = rotation; eigenvalues ζ k 3, ζ k 3 1.
Sym 3 1 3 2 1 (12) (123) 1 1 1 1 sgn 1 1 1 δ 2 0 1 ρ 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 X φ,g = φ(g). Let L = diagonal matrix with entries L g,g = class length = G / C(g). Row orthogonality: 1 G XLXH = I.
Row orthogonality yields column orthogonality: 1 G XLXH = I = X H X = G L 1 = C, where the diagonal matrix C has entries C g,g = C(g). In other words the columns are orthogonal vectors, 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: D 4 = s, t s 2 = t 2 = (st) 4 = 1 Classes: 1, s 2, t 2, r 2, z 2, where r = st and z = r 2. [D 4, D 4 ] = Z(D 4 ) = z, D 4 /[D 4, D 4 ] = V.
D 4 1 1 2 2 2 1 z r s t 1 1 1 1 1 1 λ r 1 1 1 1 1 λ s 1 1 1 1 1 λ t 1 1 1 1 1 δ 2 2 0 0 0 ρ 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 ɛ 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 2 1 1 i j k 1 1 1 1 1 1 λ i 1 1 1 1 1 λ j 1 1 1 1 1 λ k 1 1 1 1 1 ɛ 2 2 0 0 0 ρ 8 0 0 0 0
A := Alt 4 = PSL 2 (3) Classes: 1, (12)(34) 3, (123) 4, (132) 4. [A, A] = V, A/[A, A] = C 3. Permutation character = 1+ irreducible. Alt 4 1 3 4 4 1 (12)(34) (123) (132) 1 1 1 1 1 λ 1 1 1 ζ 3 ζ3 2 λ 2 1 1 ζ3 2 ζ 3 π 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.
 := SL 2 (3) Since SL 2 (3)/±1 = Alt 4, the character table of Alt 4 is embedded in the one for SL 2 (3). Each element in Alt 4 pulls back to two elements in SL 2 (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 SL 2 (3) are quaternion. The elements of order 2 in Alt 4 pull back to a single class of elements q L, 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 GL n and λ: G C are homomorphisms then so is λ R: G GL n. Moreover, tr(λ(g)r g ) = λ(g) tr(r g ). Thus linear characters act by multiplication on the rows of the character table. We have not seen this effect before because whenever λ(g) 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.
SL 2 (3) 1 1 6 4 4 4 4 1 1 q t t t 2 t 2 1 1 1 1 1 1 1 1 λ 1 1 1 1 ζ 3 ζ 3 ζ3 2 ζ3 2 λ 2 1 1 1 ζ3 2 ζ3 2 ζ 3 ζ 3 π 3 3 1 0 0 0 0 φ 0 2 2 0 1 1 1 1 φ 1 2 2 0 ζ3 2 ζ2 3 ζ 3 ζ 3 φ 2 2 2 0 ζ 3 ζ 3 ζ3 2 ζ3 2
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 GL mn. 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 constructions. 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 (φ 1 ψ 1, φ 2 ψ 2 ) G H = (φ 1, φ 2 ) G (ψ 1, ψ 2 ) H. So, the external tensor product of irreducible characters is irreducible for the product group. Why isn t the internal tensor product of irreducibles 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 remain 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 algebra 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 then π X (g) = tr(g on CX) = #{fixed points for g on X} #{G-orbits on X} = (1, π X ) 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, π X X = ( 1, π 2 X ) G G = (π X, π X ) G = π X 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 π X 2 = 2. This implies that π X 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 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 := Alt 5 = SL 2 (4) = PSL 2 (5). Classes: 1, (12)(34) 15, (123) 20, (12345) 12, (13524) 12. Doubly transitive permutation representations of degrees 5 and 6. (Projective lines over F 4 and F 5, respectively.) Remaining 2 characters must have degree 3. Alt 5 1 15 20 12 12 1 (12)(34) (123) (12345) (13245) 1 1 1 1 1 1 θ 3 1 0 τ τ θ 3 1 0 τ τ π 4 0 1 1 1 σ 5 1 1 0 0
To find τ and τ (above) we can either use the orthogonality relations or use the representation 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 x 2 = x + 1. The Galois group of this equation acts on the character values. If τ τ generates this group then its composition with the homomorphism G GL 3 produces another homomorphism.
There is also the outer automorphism of G (conjugation by anything in Sym 5 Alt 5 ). Precomposition of this with any representation produces another, swapping the two conjugacy classes of 5-cycles. In this case the outer automorphism has the same effect as the Galois automorphism. (Not always true.) The character tables for both D 4 and Q also admit both outer and Galois automorphisms. Compute these.
H := Sym 5 = PGL 2 (5). Sym 5 1 10 15 30 20 20 24 1 (12) (12)(34) (1234) (123) (123)(45) (12345) 1 1 1 1 1 1 1 1 sgn 1 1 1 1 1 1 1 π 4 2 0 0 1 1 1 sgn π 4 2 0 0 1 1 1 σ 5 1 1 1 1 1 0 sgn σ 5 1 1 1 1 1 0 τ 6 0 2 0 0 0 1 Note: τ Alt5 = θ + θ.
Exercises: SL 2 (5). SL 3 (2) = PSL 2 (7): Note that there is a doubly transitive representation of degree 8 (projective line over F 7 ). Alt 6 = Sp 4 (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 F 2 ). Sym 6. Alt 8 = SL 4 (2).