Representations of finite groups

Abstract A micro-introduction to the theory of representations of finite groups.

1 Representations

Let G be a finite group. A linear representation of G is a homomorphism ρ : G → GL(V) where GL(V) is the group of invertable linear transformations of the vector space V. We shall restrict ourselves to finite-dimensional V. The dimension dim V = n is called the degree of the representation. In order to make life easy, we only consider vector spaces over C, the field of complex numbers.

(The theory is easy for finite groups because we can average over the group to get something that is invariant for the group action. In the averaging process we divide by the order of the group, and the theory (of modular representations) is more difficult when the characteristic of the field divides the order of G. For Schur’s Lemma we need an eigenvalue, and life is a bit easier for algebraically closed fields.)

Two representations ρ_{1} : G → GL(V_{1}) and ρ_{2} : G → GL(V_{2}) are called equivalent when they are not really different: V_{1} and V_{2} have the same dimension, and for a suitable choice of bases in V_{1} and V_{2} the matrices of ρ_{1}(g) and ρ_{2}(g) are the same, for all g ∈ G. (Equivalently, ρ_{1} and ρ_{2} are equivalent when there is a linear isomorphism f : V_{1} → V_{2} such that fρ_{1}(g) = ρ_{2}(g)f for all g ∈ G.)

A subspace W of V is called ρ(G)-invariant if ρ(g)W ⊆ W for all g ∈ G. The first example of averaging is to get a ρ(G)-invariant complement of a ρ(G)- invariant subspace.

Theorem 1.1 (Maschke) Let G be a finite group, let V be a vector space over the field F, and let ρ : G → GL(V) be a linear representation of G on V. If the subspace W of V is ρ(G)-invariant and |G| is nonzero in F, then there is a ρ(G)-invariant subspace U of V such that V = U ⊕ W.

Proof: Let U_{0} be a complement of W in V, so that V = U_{0} ⊕ W. Let P_{0} be the projection onto W along U_{0}, that is, P_{0}v = w when v = u_{0} + w with u_{0} ∈ U_{0} and w ∈ W. Put

P = \frac{1}{|G|} ∑_{g∈G} ρ(g)P_{0}ρ(g)^{−1}.

Then P ρ(g) = ρ(g)P for all g ∈ G, and P is a projection onto W. Now the kernel U of P is ρ(G)-invariant. □

1.1 Direct sum

Given two representations ρ_{1} : G → GL(V_{1}) and ρ_{2} : G → GL(V_{2}), their direct sum ρ_{1} ⊕ ρ_{2} is the representation ρ : G → GL(V), where V = V_{1} ⊕ V_{2}, defined by ρ(g)v = ρ_{1}(g)v_{1} + ρ_{2}(g)v_{2} for v = v_{1} + v_{2} with v_{1} ∈ V_{1} and v_{2} ∈ V_{2}. In matrix form, this says that the matrix R(g) of ρ(g) is given by

R(g) = ( \begin{matrix}R_{1}(g) & 0 \\ 0 & R_{2}(g) \end{matrix} ) .

The degree of a direct sum is the sum of the degrees of the summands. A representation ρ is called irreducible if it is not the direct sum ρ_{1} ⊕ ρ_{2} of two representations of nonzero degree.

By induction on dim V we see that each representation is the direct sum ρ_{1} ⊕ ⋯ ⊕ ρ_{m} of (zero or more) irreducible representations.

1.2 Tensor product

Given two representations ρ_{1} : G → GL(V_{1}) and ρ_{2} : G → GL(V_{2}), their tensor product ρ_{1} ⊗ ρ_{2} is the representation ρ : G → GL(V), where V = V_{1} ⊗ V_{2}, defined by ρ(g)v = ρ_{1}(g)v_{1} ⊗ ρ_{2}(g)v_{2} for v = v_{1} ⊗ v_{2} with v_{1} ∈ V_{1} and v_{2} ∈ V_{2}. In matrix form, this says that the matrix R(g) of ρ(g) is R(g) = R_{1}(g) ⊗ R_{2}(g). The degree of a tensor product is the product of the degrees of the factors.

1.3 Schur’s Lemma

Theorem 1.2 (Schur) Let ρ_{1} : G → GL(V_{1}) and ρ_{2} : G → GL(V_{2}) be irreducible representations. Let f : V_{1} → V_{2} be a linear map with ρ_{2}(g)f = fρ_{1}(g) for all g ∈ G.

(i) If ρ_{1} and ρ_{2} are not equivalent, then f = 0.

(ii) If V_{1} = V_{2} = V and ρ_{1} = ρ_{2}, then f = λI, where I is the identity on V.

Proof: (i) We may suppose f ≠ 0. The subspace W_{1} = ker f of V_{1} is ρ_{1}(G)- invariant, so W_{1} = 0 or W_{1} = V_{1}, but f ≠ 0, so W_{1} = 0. The subspace W_{2} = im f of V_{2} is ρ_{2}(G)-invariant, so W_{2} = 0 or W_{2} = V_{2}, but f ≠ 0, so W_{2} = V_{2}. Now f is an isomorphism, and ρ_{1} and ρ_{2} are equivalent.

(ii) Let λ be an eigenvalue of f (there is one, since C is algebraically closed). Now apply the previous argument to f − λI instead of f. Since f − λI is not an isomorphism, is must be 0. □

By averaging over G we can turn a linear map f into one that satisfies the hypothesis of Theorem 1.2. This yields:

Corollary 1.3 Let ρ_{1} : G → GL(V_{1}) and ρ_{2} : G → GL(V_{2}) be irreducible representations. Let f : V_{1} → V_{2} be linear and put f̃ = \frac{1}{|G|} ∑_{g∈G} ρ_{2}(g)^{−1}fρ_{1}(g).

(i) If ρ_{1} and ρ_{2} are not equivalent, then f̃ = 0.

(ii) If V_{1} = V_{2} = V and ρ_{1} = ρ_{2}, then f̃ = λI, where λ = \frac{1}{n}tr f.

Proof: In case (ii), tr f̃ = tr f and tr I = n. □

Now choose bases, and let f = E_{jk} be the linear map with a matrix that is zero everywhere except at the (j, k)-entry where it is 1. We obtain orthogonality relations for the matrix entries of representations.

Corollary 1.4 Let ρ_{1} : G → GL(V_{1}) and ρ_{2} : G → GL(V_{2}) be irreducible representations.

(i) If ρ_{1} and ρ_{2} are not equivalent, then

\frac{1}{|G|} ∑_{g∈G} R_{2}(g)_{ij}^{−1}R_{1}(g)_{kl} = 0 for all i, j, k, l

where R_{i}(g) denotes the matrix of ρ_{i}(g) for i = 1, 2 and g ∈ G.

(ii) If V_{1} = V_{2} = V and ρ_{1} = ρ_{2} = ρ, then

\frac{1}{|G|} ∑_{g∈G} R(g)_{ij}^{−1}R(g)_{kl} = \{ \begin{matrix}1/n & if i = l and j = k \\ 0 & otherwise \end{matrix}

where R(g) denotes the matrix of ρ(g) for g ∈ G.

Proof: Apply the previous corollary, and take the (i, l)-matrix entry of f̃ where f = E_{jk}. In case (ii), tr f equals 0 if j ≠ k and 1 if j = k. □

2 Characters

Given a representation ρ : G → GL(V), let its character be the map χ : G → C defined by χ(g) = tr ρ(g). It will turn out that ρ is determined up to equivalence by its character χ.

Lemma 2.1 Let χ = χ_{ρ} denote the character of ρ. Then (i) χ_{ρ_{1}⊕ρ_{2}} = χ_{ρ_{1}} + χ_{ρ_{2}}, (ii) χ_{ρ_{1}⊗ρ_{2}} = χ_{ρ_{1}}χ_{ρ_{2}}, (iii) χ_{ρ}(1) = n if ρ has degree n, (iv) χ(g^{−1}) = \overline{χ(g)} for all g ∈ G,

(v) χ(h^{−1}gh) = χ(g) for all g, h ∈ G.

Proof: Only part (iv) requires comment. Since G is finite, g has finite order, so ρ(g) has finite order, and its eigenvalues are roots of unity. If ρ(g) has eigenvalue ζ, then ρ(g^{−1}) has eigenvalue ζ^{−1} = \overline{ζ}. And the trace of ρ(g^{−1}) is the sum of its eigenvalues. □

2.1 Inner product

For functions ϕ, ψ : G → C, put

⟨ϕ, ψ⟩ = \frac{1}{|G|} ∑_{g∈G} \overline{ϕ(g)} ψ(g).

Theorem 2.2 If the representation ρ : G → GL(V) is irreducible, then its character χ satisfies ⟨χ, χ⟩ = 1. If the irreducible representations ρ, ρ′ are inequivalent, then their characters χ, χ′ satisfy ⟨χ, χ′⟩ = 0.

Proof: Since χ(g) = tr ρ(g) = ∑_{i} R(g)_{ii} and \overline{χ(g)} = χ(g^{−1}), we have

⟨χ, χ′⟩ = \frac{1}{|G|} ∑_{g∈G} χ(g^{−1})χ′(g) = \frac{1}{|G|} ∑_{g∈G} ∑_{i,j} R(g)_{ii}^{−1}R′(g)_{jj}.

Now apply Corollary 1.4. □

2.2 The character determines the representation

Theorem 2.3 Let σ : G → GL(V) be a representation with character ϕ, and let σ = σ_{1} ⊕ ⋯ ⊕ σ_{m} be a decomposition of σ into irreducible representations. Let ρ : G → GL(W) be an irreducible representation with character χ. Then the number of σ_{i} equivalent to ρ equals ⟨ϕ, χ⟩.

Proof: Let σ_{i} have character ϕ_{i}, so that ϕ = ϕ_{1} + ⋯ + ϕ_{m}. Then ⟨ϕ, χ⟩ = ⟨ϕ_{1}, χ⟩ + ⋯ + ⟨ϕ_{m}, χ⟩. Now apply Theorem 2.2. □

It follows that although the decomposition into irreducible subspaces is not unique in general, the number of subspaces of given type is determined. (In fact, also the sum of all subspaces of given type is determined uniquely.)

Corollary 2.4 Two representations are equivalent if and only if they have the same character. □

We saw that if ρ is irreducible, then ⟨χ, χ⟩ = 1. But the opposite is also true: if ⟨χ, χ⟩ = 1 then ρ is irreducible. Indeed, if χ = ∑_{i} m_{i}χ_{i} where the χ_{i} are distinct irreducible characters (that is, characters of irreducible representations), then ⟨χ, χ⟩ = ∑_{i} m_{i}^{2}, and this equals 1 only when there is only one summand and m_{1} = 1.

2.3 The regular representation

So far, we have not constructed any representations. Let G be a finite group. The regular representation of G is the representation ρ on the vector space C^{G} (with basis G) defined by ρ(g)h = gh for g, h ∈ G. This representation has degree n = |G|, and character χ satisfying

χ(g) = tr ρ(g) = \{ \begin{matrix}|G| & if g = 1, \\ 0 & otherwise. \end{matrix}

If χ_{i} is any irreducible character, of degree n_{i}, then

⟨χ, χ_{i}⟩ = \frac{1}{|G|} ∑_{g∈G} \overline{χ(g)}χ_{i}(g) = χ_{i}(1) = n_{i},

so that χ = ∑ n_{i}χ_{i} and |G| = ∑_{i} n_{i}^{2}.

It follows that there are only finitely many distinct irreducible characters, all found in the character of the regular representation.

2.4 Class functions

For g ∈ G, the conjugacy class C(g) is the set {h^{−1}gh | h ∈ G}. Lemma 2.1(v) says that a character is a class function, that is, is constant on conjugacy classes. We shall see that conversely any class function is a linear combination of characters.

For a class function ϕ and a representation ρ, let f_{ϕ,ρ} = \frac{1}{|G|} ∑_{g∈G} \overline{ϕ(g)}ρ(g), a linear transformation of the vector space V, weighted average of the ρ(g).

Lemma 2.5 If ρ is irreducible, then f_{ϕ,ρ} = λI, where nλ = tr f_{ϕ,ρ} = ⟨ϕ, χ⟩.

Proof: We have

ρ(g)^{−1}f_{ϕ,ρ}ρ(g) = \frac{1}{|G|} ∑_{h∈G} \overline{ϕ(h)}ρ(g^{−1}hg) = \frac{1}{|G|} ∑_{h∈G} \overline{ϕ(ghg^{−1})}ρ(h) = f_{ϕ,ρ}.

Now by Schur’s Lemma (Theorem 1.2) f_{ϕ,ρ} = λI for some constant λ, and λ is found by taking traces. □

Theorem 2.6 The irreducible characters form an orthonormal basis for the vector space of class functions. In particular, the number of irreducible characters equals the number of conjugacy classes of the group G.

Proof: The ‘orthonormal’ part is the content of Theorem 2.2. Remains ‘basis’. If ϕ is a class function orthogonal to all irreducible characters χ_{i}, then consider the linear transformation f_{ϕ,ρ} for various ρ. The above lemma says that f_{ϕ,ρ} = 0 when ρ is irreducible. For arbitrary ρ the function f_{ϕ,ρ} is a direct sum of the functions f_{ϕ,ρ_{j}} for the irreducible constituents ρ_{j} of ρ, hence f_{ϕ,ρ} = 0 for all ρ. Now let ρ be the regular representation and compute the image of f_{ϕ,ρ} on the basis vector 1. Since ρ(g)1 = g, we find 0 = f_{ϕ,ρ}1 = \frac{1}{|G|} ∑_{g∈G} \overline{ϕ(g)}g, so that all

coefficients \overline{ϕ(g)} vanish, and ϕ = 0. □

2.5 Character table

The square matrix X with rows indexed by the irreducible characters χ_{i} and columns by the conjugacy classes of G and entries X_{χ,C} = χ(g) for g ∈ C, is called the character table of G.

Let D be the diagonal matrix with rows and columns indexed by the conjugacy classes of G, where D_{CC} = \frac{|C|}{|G|}, so that tr D = 1.

The fact that different characters are orthogonal is expressed by XD\overline{X}^{⊤} = I. But if AB = I then also BA = I, so it follows that \overline{X}^{⊤}X = D^{−1}. This shows that also the columns of X are orthogonal, and that the sizes of the conjugacy classes can be seen from X.

3 Example: Sym(5)

Let G be the symmetric group Sym(5). Its character table is

1 (12)(34) (123) (12345) (12) (1234) (12)(345)
χ_{1} 1  1  1  1  1  1  1
χ_{2} 1  1  1  1  −1  −1  −1
χ_{3} 4  0  1  −1  2  0  −1
χ_{4} 4  0  1  −1  −2  0  1
χ_{5} 5  1  −1  0  1  −1  1
χ_{6} 5  1  −1  0  −1  1  −1
χ_{7} 6  −2  0  1  0  0  0

The table is square, with 7 rows and columns. The columns are labeled by representatives of the conjugacy classes. The conjugacy classes have sizes 1, 15, 20, 24, 10, 30, 20, respectively. The sum of the squares of the character degrees 1^{2} + 1^{2} + 4^{2} + 4^{2} + 5^{2} + 5^{2} + 6^{2} = 120 equals |G|.

How was this table constructed? By finding some easy characters and decomposing those into irreducibles.

1. The first is the trivial character, the character of the trivial representation that maps every g ∈ G to the identity I = (1) of order 1. This gives χ_{1}.

2. The second is the sign character. A permutation can be even or odd, and the sign character is 1 on even and −1 on odd permutations. This gives χ_{2}.

3. The third construction is that of a permutation character. If G acts as a group of permutations on a set Ω, we find a representation in C^{Ω}. (The regular representation is the example where G acts on itself.) Now χ(g) = tr ρ(g) is the number of fixpoints of g.

The group Sym(5) has an obvious action on the set S = {1, 2, 3, 4, 5}, and the permutation character is π = (5, 1, 2, 0, 3, 1, 0) with entries in the order of the columns of the table. Now ⟨π, π⟩ = 2, so this is the sum of two irreducible characters. And ⟨π, χ_{1}⟩ = 1, so χ_{1} is one of them. Then χ_{3} = π − χ_{1} must be the other. This gives χ_{3}.

The group Sym(5) also has an action on the ten pairs from S. The corresponding permutation character is π_{2} = (10, 2, 1, 0, 4, 0, 1). Now ⟨π_{2}, π_{2}⟩ = 3, and ⟨π_{2}, χ_{1}⟩ = 1, and ⟨π_{2}, χ_{3}⟩ = 1, so π_{2} decomposes into three irreducibles, namely χ_{1} and χ_{3} and χ_{5} = π_{2} − χ_{1} − χ_{3}. This gives χ_{5}.

4. The fourth construction is that of taking tensor products. We find irreducible characters χ_{4} = χ_{2}χ_{3} and χ_{6} = χ_{2}χ_{5}. Now only χ_{7} is left, and we can write it down using the orthogonality relations of the columns. But we can also compute the product χ_{3}^{2} and find that it decomposes as χ_{3}^{2} = χ_{1} + χ_{3} + χ_{5} + χ_{7}. That completes the table.

3.1 Alt(5)

The even permutations in Sym(5) form the alternating group Alt(5). It has character table

1 (12)(34) (123) (12345) (12354)
χ_{1} 1  1  1  1  1
χ_{2} 3  −1  0  s  t
χ_{3} 3  −1  0  t  s
χ_{4} 4  0  1  −1  −1
χ_{5} 5  1  −1  0  0

where s = (1 − \sqrt{5})/2 and t = (1 + \sqrt{5})/2.

We lose the classes of odd permutations, but the class of 5-cycles now splits into two, since (12345) and (12354) are no longer conjugate. (In Sym(5) one had (12354) = (45)(12345)(45), but (45) is odd.) The restrictions of the characters of Sym(5) to Alt(5) give characters again, and we find χ_{1}, χ_{4}, χ_{5} and χ_{2} + χ_{3}. (The formula for the inner product of two characters involves a factor \frac{1}{|G|}, so if the irreducible character χ of Sym(5) vanishes outside Alt(5), and χ′ is the restriction of χ to Alt(5), then ⟨χ′, χ′⟩ = 2.) If x is an element of Sym(5)\Alt(5), and χ a character of Alt(5), then also χ′ defined by χ′(g) = χ(x^{−1}gx) is a character of Alt(5). The characters χ_{2} and χ_{3} must be related this way, so have the same value on the classes that do not split, and all that remains is to find s and t. The orthogonality relations give s^{2} +t^{2} = 3 and st = −1 (and we already knew s + t = 1), and this determines s, t. That completes the table.

Exercise Construct matrices for a representation with character χ_{2}.

4 Some additional material

4.1 Frobenius reciprocity

Let G be a group and H a subgroup and let ϕ : H → C be a class function on H. The function ϕ^{G} obtained by inducing ϕ up to G is by definition ϕ^{G}(g) = \frac{1}{|H|} ∑_{x∈G} ϕ̂(x^{−1}gx), where ϕ̂ : G → C is defined by ϕ̂(h) = ϕ(h) for h ∈ H,

and ϕ̂(x) = 0 for x ∈ G \ H. Now ϕ^{G} is a class function on G. Conversely, if ψ : G → C is a class function on G, then the restriction ψ|_{H} of ψ to H is a class function on H.

Proposition 4.1 Let ϕ : H → C be a class function on H and ψ : G → C a class function on G. Then ⟨ϕ^{G}, ψ⟩ = ⟨ϕ, ψ|_{H}⟩. In particular, if ϕ is a character of H, then ϕ^{G} is a character of G.

Proof:

\begin{aligned}⟨ψ, ϕ^{G}⟩ & = & \frac{1}{|G|.|H|} ∑_{g,x∈G} \overline{ψ(g)}ϕ̂(x^{−1}gx) = \frac{1}{|G|.|H|} ∑_{g,x∈G} \overline{ψ(x^{−1}gx)}ϕ̂(g) \\ & = & \frac{1}{|H|} ∑_{g∈H} \overline{ψ(g)}ϕ(g) = ⟨ψ|_{H}, ϕ⟩. \end{aligned}

Necessary and sufficient in order to be a character is that the inner product with all irreducibles is a nonnegative integer. □

Another useful identity is (ϕ.ψ|_{H})^{G} = ϕ^{G}.ψ.

4.2 Permutation characters

Recall that if G acts as a group of permutations on a set Ω, then the corresponding permutation character is the function π : G → N defined by π(g) = #{ω ∈ Ω | gω = ω}, the number of fixpoints of g in this action.

If G acts transitively on Ω, then the rank of the action is the number of orbits of a point stabilizer, or, what is the same, the number of orbits of G in the natural action on Ω × Ω.

For a group G, let 1_{G} be the trivial character on G (that is identically 1).

Proposition 4.2 Let π be the permutation character of a permutation representation of the group G on the set Ω. Let G have orbits Ω_{1}, …, Ω_{m}. Let, for 1 ≤ i ≤ m, the group H_{i} be the stabilizer in G of some element in Ω_{i}. Then π = ∑_{i=1}^{m}(1_{H_{i}})^{G}. In particular,

(i) The number of orbits m of G equals ⟨1, π⟩.

(ii) If G is transitive, then it has rank ⟨π, π⟩.

(iii) For m = 2, the number of orbits of G on Ω_{1} × Ω_{2} (which equals the number of orbits of H_{1} on Ω_{2} and the number of orbits of H_{2} on Ω_{1}) equals ⟨π_{1}, π_{2}⟩.

Proof: The permutation representation on Ω is the direct sum of the representations on the Ω_{i}, so we may assume that G is transitive, i.e., m = 1. Put H = H_{1}. Now Ω can be identified with the set of left cosets gH of H, with G acting by left multiplication. A left coset xH is fixed by multiplication by g when gxH = xH, i.e., when x^{−1}gx ∈ H (and each left coset xH has |H| representatives x). Now by definition (1_{H})^{G}(g) = \frac{1}{|H|}#\{x ∈ G | x^{−1}gx ∈ H\} = π(g), so that π = (1_{H})^{G}.

For (i), ⟨1_{G}, π⟩ = ⟨1_{H}, 1_{H}⟩ = 1.

For (ii): the action of G on Ω × Ω (via g(a, b) = (ga, gb)) has character π^{2}. Now the rank is ⟨1, π^{2}⟩ = ⟨π, π⟩.

For (iii): the action of G on Ω_{1} × Ω_{2} (via g(a, b) = (ga, gb)) has character π_{1}π_{2}. Now the number of orbits is ⟨1, π_{1}π_{2}⟩ = ⟨π_{1}, π_{2}⟩. □