5

Classical Representation Theory of Finite Groups

One of the most powerful tools for the study of finite groups is the theory of representations and particularly the theory of characters. These subjects as generalized to locally compact groups constitute a major area of modern analysis that generalizes classical Fourier analysis. The subject of representation theory of finite groups is almost wholly the creation of Frobenius. Notable improvements and simplifications of the theory are due to Schur. During the past fifty years, very deep results have been added to the -theory by Brauer, and representation theory has played an important role in the explosive growth of the structure theory of finite groups, which began with the Feit-Thompson proof of a hundred-year-old conjecture by Burnside that every finite group of odd order is solvable.

In this chapter, we are concerned with the classical theory of representations of finite groups acting on finite dimensional vector spaces over the field of complex numbers or more generally over fields of characteristics not dividing the group order. This restriction on the characteristic implies complete reducibility of the representation or, equivalently, semi-simplicity of the group algebras. This permits the application of the structure and representation theory of finite dimensional semi-simple algebras to the representation theory of finite groups. However, there is considerably more to the story than this, namely, the theory of characters, much of which can be developed without recourse to the theory of algebras.

We shall derive the classical results of Frobenius, Schur, and Burnside as well as Brauer’s results on induced characters and splitting fields. One of the most important contributions of Brauer is his modular theory, which deals with representations over fields whose characteristics do divide the group order and the relation between the representations over such fields and representations over . We shall not consider this theory in our account. A number of applications of character theory to structural results on finite groups will be given.

5.1   REPRESENTATIONS AND MATRIX REPRESENTATIONS OF GROUPS

DEFINITION 5.1.   By a representation ρ of a group G we shall mean a homomorphism of G into the group GL(V) of bijective linear transformations of a finite dimensional vector space V over afield F.

We shall say that ρ is a representation of G over F and that it acts on V/F. The dimensionality of V is called the degree of the representation. The defining conditions for a representation are that ρ is a map of G into GL(V) such that

for g_i ∈ G. These have the immediate consequences that ρ(l) = l_v, ρ(g^–1) = ρ(g)^–1 and in general, ρ(g^m) = ρ(g)^m for m ∈ . We remark also that the conditions that ρ(g)∈ GL(V) and (1) holds can be replaced by the following: ρ(g) ∈ End_FV, (1) holds, and ρ(l) = 1. These immediately imply that ρ(g)∈ GL(V), so p is a representation.

Let B = (u₁, u₂,…, u_n) be a base for V/F. If a ∈ End_F V, we write

and obtain the matrix (α) whose (i, j)-entry is α_ij. The map a (α) is an isomorphism of End_F V onto M_n(F). Hence if ρ is a representation of G acting on V, then the map

where ρ_B(g) denotes the matrix of ρ(g) relative to the base B, is a homomorphism of G into the group GL_n(F) of invertible n × n matrices with entries in F. Such a homomorphism is called a matrix representation of G of degree n. A change of the base B to C = (v₁, …, v_n) where v_i = Σμ_ji u_j and (μ) ∈ GL_n(F) replaces ρ_B by ρ_c where

This matrix representation is said to be similar to ρ_B. It is clear that any matrix representation can be obtained from a representation in the manner indicated above.

A homomorphism of G into the symmetric group S_n or, equivalently, an action of G on the finite set {1, 2, …, n} (BAI, p. 71), gives rise to a representation. Let the action be described by

where π is the homomorphism of G into S_n. Let V be the vector space with base (u₁, u₂, …, u_n) and let ρ(g) be the linear transformation of V such that

Then ρ(g₁g₂) = ρ(g₁)ρ(g₂) and ρ(l) = 1, so ρ is a representation of G. Representations of this type are called permutation representations. They are characterized by the property that they are representations that stabilize some base B of V/F. Of particular interest is the permutation representation obtained from the action of G on itself by left translations (BAI, p. 72). The corresponding representation of G is called the regular representation.

EXAMPLES

1. Let G = , the cyclic group generated by an element g of order n. We have the homomorphism of G into S_n mapping g into the cycle (12…n). This action is equivalent to the action of G on itself by left translations. The associated permutation representation maps g into the linear transformation ρ(g) such that

This gives a matrix representation in which g is represented by the matrix

This is obtained by specializing (2) to (6)

2. Let G = S₃ whose elements are

The identity map is an isomorphism of G onto itself. This gives rise to a permutation representation and an associated matrix representation in which

For a given group G and given field F we now consider the class Σ(G, F) of representations of G acting on vector spaces over F as base field. There is a rich algebraic structure that can be defined on Σ(G, F). First, let ρ₁ and ρ₂ be representations of G acting on the vector spaces V₁/F and V₂/F respectively. Form the vector space V_{l F} V₂ and define ρ₁ ρ₂ by

where, as usual, a₁ a₂ for the linear transformations a_i of V_i is the linear transformation of V₁ V₂ such that

If b_i, i = l, 2, is a second linear transformation of V_i, then we have (a₁ a₂)(b₁ b₂) = a₁b₁ a₂b₂. It follows that if g₁, g₂ ∈ G, then

Hence ρ₁ ρ₂ is a representation. We call this the tensor product of the given representations ρ₁ and ρ₂.

If (u₁, u₂,…, u_n) is a base for V₁/F and (v₁, v₂,…, v_m) is a base for V₂/F, then the mn vectors u_i v_j constitute a base for (V₁ V₂)/F. We order these lexicographically:

Then if a_l ∈ End_F V_l has the matrix (α⁽¹⁾) relative to the base (u_l, …, u_n), and a₂ ∈ End_F V₂ has the matrix (α⁽²⁾) relative to (v₁, …, v_m), we have

and

Hence the matrix of a₁ a₂ relative to the base (11) is

We denote this matrix as ((α⁽¹⁾) (α⁽²⁾)). In particular, we see that if ρ_1B1(g) is the matrix of ρ₁(g) relative to B₁ = (u_l, …, u_n) and ρ_2B2(g) is the matrix of ρ₂(g) relative to B₂ = (v_l, …, v_m), then the matrix of (ρ₁ ρ₂) (g) relative to the base (11) is ρ_1B1(g) ρ_2B2(g).

As usual, we denote the dual space of linear functions on V/F by V*. If (u_l, u₂, …, u_n) is a base for V/F, we have the dual (or complementary) base (u_l^*, u₂^*, …, u_n^*) of V^*/F where u_i^* is the linear function on V such that u_i^*(u_j) = δ_ij, 1 ≤ j ≤ n. If a is a linear transformation in V/F, we have the transposed transformation a* in V* such that

for y ∈ V and x* ∈ V*. If au_i = Σ_jα_jiu_j and a*u*_k = Σβ_lku*_l, then . Hence β_ik = α_ki and so the matrix of a* relative to the dual base (u₁*, …, u_n*) of (u_l, …, u_n) is the transpose of the matrix of a relative to (u_l, …, u_n). The map a a* is an anti-isomorphism of the algebra End_F V onto End_F V* and hence a (a*) ^–1 is an isomorphism of the group GL(V) of bijective linear transformations in V into GL(V*).

Now let ρ be a representation of G acting on V. We compose this with the isomorphism a (a*)^–1 of GL(V) onto GL(V*). This gives a representation g (ρ(g)*)^–1 of G acting on V*. We call this the contragredient representation of ρ and denote it as ρ*. Evidently it has the same degree as ρ. Moreover, if B = (u_l, …, u_n) and B* is the dual base for V*, then we have the following relation for the matrix representations g ρ_B(g) and g ρ*_B*(g): ρ*_B*(g) = (^tρ_B(g))^–1

The map ρ ρ* may be regarded as a unary composition in Σ(G, F). We also have an important nullary composition. This is the unit representation 1 or 1_G for which V is one-dimensional and 1(g) = 1_v for all g. The corresponding matrix representation (determined by any base) is g (1) where (1) is the l × l matrix with entry 1.

Let ρ_i, i = 1, 2, be a representation of G acting on V_i/F. Then we say that ρ₁ and ρ₂ are equivalent if there exists a bijective linear map η of V₁ onto V₂ such that

EXERCISES

        1. Let ρ be the representation of the cyclic group given in example 1. Let be the base field. Show that ρ is equivalent to the representation ρ' such that ρ'(g)u_i = ζⁱu_i where ζ = e^2πi/n.

        2. Let p_i, i = 1, 2, be a representation of G acting on V_i/F and put V = hom_F(V_l, V₂). If l ∈ V, define ρ(g)l = ρ₂(g)lρ₁(g)^–1. Verify that ρ is a representation and show that ρ is equivalent to ρ₁* ρ₂.

        3. Let l denote the identity map of GL(V). This is a representation of GL(V) acting on V. Consider the representation l* l acting on V* V. Show that the set of vectors c ∈ V* V such that (l* l)(a)c = c for all a ∈ GL(V) is a one-dimensional subspace of V* V. Find a non-zero vector in this space.

        4. Show that if p_i, i = 1, 2, is a representation acting on V_i, then ρ₁ ρ₂ and ρ₂ ρ₁ are equivalent.

5.2   COMPLETE REDUCIBILITY

The study of the representations of a group G can be reduced to the study of the representations of the group algebra of G. Let G be a group, F a field, then the group algebra F[G] is the algebra over F having the set G as base and multiplication defined by

for α_g, β_h ∈ F. Let ρ be a representation of G acting on the vector space V/F. Then the group homomorphism ρ has a unique extension to the algebra homomorphism

of F[G] into End_FV. Conversely, given a homomorphism ρ of F[G] into End_F V, where V is a finite dimensional vector space over F, the restriction of ρ to G is a representation since ρ(l) = 1, which implies that every ρ(g) ∈ GL(V). Now we recall that a representation of F[G] ( = homomorphism of F[G] into End_FV) can be used to make V into an F[G]-module. One simply defines the action of F[G] on V by

for x ∈ V. Again, this can be turned around: Given a module V for F[G], this is a module for F, hence, a vector space over F and, assuming finite dimensionality of V/F, we obtain a representation ρ of G where ρ(g) is the map x gx, which is a linear transformation of V/F. Thus representations of G acting on (finite dimensional) vector spaces over a field F are equivalent to F[G]-modules, which as F-modules are finite dimensional.

The standard concepts of module theory can be carried over to representations of groups via the group algebra. If ρ is a representation of G acting on V, a submodule U of V as F[G] -module is the same thing as a subspace of V that is ρ(G)-invariant in the sense that it is stabilized by every ρ(g), g ∈ G. Then we obtain a representation ρ|U of G acting on U in which (ρ|U) (g) is the restriction of ρ(g) to U. We shall call this a subrepresentation of ρ. We also have the module V/U. The associated representation of G is ρ|V/U where (ρ|V/U) (g) is the linear transformation x + U ρ(g)x + U. This will be called a factor representation of ρ.

Let B = (u_l, …, u_n) be a base for V such that (u_l, …, u_r) is a base for the ρ(G)-invariant subspace U. Consider the matrix representation ρ_B determined by B. Since U is stabilized by every ρ(g), ρ(g)u_i for 1 ≤ i ≤ r is a linear combination of the vectors (u_l, …, u_r). Hence every matrix ρ_B(g), g ∈ G, has the “reduced” form

The matrices in the upper left-hand corner are those of the matrix representation of G acting on U associated with the base (u₁, …, u_r) and those in the lower right-hand corner are the matrices of the representation of G on V/U relative to the base (u_{r + 1} + U, …, u_n + U). It is clear that conversely if there exists a base (u₁, …, u_n) such that the matrices of the ρ(g) relative to this base all have the form (18), then U = Σ₁^rFu_j is a ρ(G)-invariant subspace, hence, an F[G]-submodule of V.

If v = U U' where U and U' are submodules, then we can choose a base B = (u_l, …, u_n) for V such that (u₁, …, u_r) is a base for U and (u_{r + 1},…, u_n) is a base for U'. Then the corresponding matrices ρ_B(g) have the form (18) in which the r × (n – r) blocks in the upper right-hand corner are all 0. If ρ₁ = ρ|U and ρ₂ = ρ|U' then we say that the representation ρ is a direct sum of the subrepresentations ρ₁ and ρ₂ and we write ρ = ρ₁ ρ₂. Let p be the projection on U determined by the decomposition V = U U'. Then if we write any x ∈ V as x = y + y', y ∈ U, y' ∈ U', we have ρ(g)x = ρ(g)y + ρ(g)y' with ρ(g)y ∈ U, ρ(g)y' ∈ U' for all g ∈ G. Since p is the map x y, we have pρ(g)x = ρ(g)y = ρ(g)px. Thus p commutes with every ρ(g). Conversely, suppose U is any ρ(G)-invariant subspace and there exists a projection p of V on U that commutes with every ρ(g). We can write V = pV (l – p)V and pV = U. Also, ρ(g) (1 – p)V = (1 – p)ρ(g)V = (1 – p)V. Hence U' = (1 – p)V is ρ(G)-invariant and we have the decomposition V = U U'.

We shall call a representation ρ of G irreducible (completely reducible) if the corresponding F[G]-module is irreducible (completely reducible). We recall that a module V is completely reducible if and only if it satisfies either one of the following conditions: (1) V = ΣV_i where the V_i are irreducible submodules, (2) V ≠ 0 and for every submodule U there exists a submodule U' such that V = U U' (Theorem 3.10, p. 121). We shall use these conditions to prove two theorems giving sufficient conditions for complete reducibility. The first of these is a fundamental theorem in the representation theory of finite groups. This is

MASCHKE’S THEOREM.   Every representation ρ of a finite group G acting on a vector space V/F such that the characteristic char is completely reducible.

Proof. Let U be a ρ(G)-invariant subspace of V and write V = U U₀ where U₀ is a second subspace (not necessarily invariant). Let p₀ be the projection on U determined by this decomposition. We shall now construct by an averaging process a projection on U that commutes with every ρ(g), g ∈ G. We put

Since char , |G|^–1 exists in F and p is well defined. If g' ∈ G, then

Hence ρ(g')p = pρ(g') for g' ∈ G. Evidently, p ∈ End_FV. If y ∈ U, then p₀y = y and since ρ(g)y ∈ U, p₀ρ(g)y = ρ(g)y. Hence ρ(g)^–1 p₀ρ(g)y = y and

If x ∈ V, then p₀x ∈ U and ρ(g)^–1 p₀ρ(g)x ∈ U. Hence px ∈ U. The two conditions on p ∈ End_F V, py = y for y ∈ U and px ∈ U for x ∈ V, imply that p is a projection on U. Then V = U U' where U' = (l – p)V and since p commutes with every ρ(g), U' is ρ(G)-invariant.

This result has a formulation in terms of matrix representations that should be obvious from the discussion above. The result in its matrix form was proved by H. Maschke for the case in which F = . The validity of the result for any F with char was first noted by L. E. Dickson.

If ρ is a completely reducible representation of G acting on V, then the F[G]-module V decomposes as where the V_i^k are irreducible and for any k, l but . By Theorem 3.12 (p. 123), the submodules are the homogeneous components of V. If ρ_i is any irreducible representation of G equivalent to the subrepresentation determined by the V_i^(k), then we write

We call ρ_i an irreducible constituent of ρ and m_i its multiplicity. By Theorem 3.14, the equivalence classes of the irreducible constituents and their multiplicities are uniquely determined. We remark also that the multiplicity m_i = q_i/n_i where q_i = dim W_i, n_i = dim V_i^(k), which shows also that m_i is independent of the particular decomposition of V as a direct sum of irreducible submodules.

If H is a subgroup of G and ρ is a representation of G acting on V, then the restriction of ρ to H is a representation of H that we shall denote as ρ_H. We shall now show that if and ρ is completely reducible, then ρ_H is completely reducible. It suffices to prove this for ρ irreducible. Moreover, in this case we can say considerably more on the relation between the irreducible constituents and their multiplicities. First, we need to give a definition of conjugacy of representations of a normal subgroup of a group.

Let and let σ be a representation of H acting on U. For any g ∈ G we can define the map

This is the composite of the automorphism h ghg^–1 of H with the homomorphism σ. Hence (22) is a representation of H acting on U, which we shall denote as ^gσ. Any representation equivalent to ^gσ will be called a conjugate of σ or, more precisely, a g-conjugate of σ. Evidently, any σ(H)- invariant subspace is ^gσ(H)-invariant and since σ = ^g–1 (^gσ), it is clear that σ and ^gσ have the same lattices of F[H]-submodules. In particular, if σ is irreducible, then any conjugate of σ is irreducible. Now suppose σ₁ and σ₂ are equivalent representations of H acting on U₁ and U₂ respectively. Then we have a bijective linear map η of U₁ onto U₂ such that σ₂(h) = ησ_l(h)η^–1, h ∈ H. Then also, are equivalent.

We can now prove

(A. H.) CLIFFORD’S THEOREM.   Let and let ρ be an irreducible representation of G. Then ρ_H is completely reducible and all of its irreducible constituents are conjugate and have the same multiplicity.

Proof.   Let ρ act on V and let U be an irreducible F[H]-submodule of V. Evidently, Σ_{g ∈ G}ρ(g)U is a ρ(G)-invariant subspace of V containing U. Hence V = Σρ(g)U. Let h ∈ H, y ∈ U. Then

Thus ρ(g)U is ρ(H)-invariant. Let σ denote the representation of H acting on U and σ' the representation of H acting on ρ(g)U. Now by (23), . Since y ρ(g)y is a bijective linear map of U onto ρ(g)U, we see that σ' is a conjugate of σ. Hence ρ(g)U is an irreducible F[H]- module and V = Σρ(g)U is a decomposition of V as sum of irreducible F[H]- modules such that the corresponding representations are conjugates. Then V is a direct sum of certain ones of the ρ(g)U. Hence ρ_H is completely reducible and its irreducible constituents are conjugate. If U₁ and U₂ are isomorphic irreducible F[H]-submodules, then ρ(g)U₁ and ρ(g)U₂ are isomorphic F[H] submodules. It follows that every ρ(g) permutes the homogeneous components of V as F[H]-module. Thus we have an action of G through ρ(G) on the set S of homogeneous components of V as F[H]-module. Moreover, this action is transitive since if U is any irreducible F[H]-submodule, then any other irreducible F[H]-submodule is isomorphic to a ρ(g)U for some g ∈ G. This implies that all of the irreducible constituents have the same multiplicity.

EXERCISES

        1. Prove the following extension of Maschke’s theorem. Let ρ be a representation of a group G that is not necessarily finite. Suppose G contains a normal subgroup H of finite index [G : H] not divisible by char F. Show that if ρ|H is completely reducible, then ρ is completely reducible.

        2. Let F = and let H(x, y) be a hermitian form on V/ that is positive definite in the sense that H(x, x) > 0 for all x ≠ 0 (BAI, pp. 381–384). Let ρ be a representation of a group G acting on V such that every ρ(g) is unitary with respect to H, (H(ρ(g)x, ρ(g)y) = H(x, y), x, y ∈ V). Such a representation is called unitary. Show that if U is a ρ(G)-invariant subspace, then is ρ(G)-invariant. Use this to prove that ρ is completely reducible.

        3. Same notations as in exercise 2. Let G be finite, ρ a representation of G acting on V. Let H₀(x, y) be any positive definite hermitian form on V/. Put . Show that H(x, y) is positive definite hermitian and that every ρ(g) is unitary relative to H. Use this and exercise 2 to prove Maschke’s theorem for F = .

        4. Define the projective linear group PGL(V) of a finite dimensional vector space V/F as GL(V)/F*1 where F* is the multiplicative group of non-zero elements of F. A homomorphism ρ of a group G into PGL(V) is called a projective representation of G. For each g ∈ G let μ(g) denote a representative in GL(V) of the coset of ρ(g) ∈ PGL(V). Then μ(g₁g₂) = γ_g1,_g2μ(g₁)(g₂) where γ_g1,_g2 ∈ F*. Define ρ(G)-invariant subspaces and complete reducibility as for ordinary representations. Prove the analogue of Maschke’s theorem for projective representations.

        5. Show that ρ is irreducible (completely reducible) if and only if ρ* is irreducible (completely reducible).

        6. Let ψ be a representation of G acting on W, V a ψ(G)-invariant subspace. Put ρ = ψ|V. Show that ρ* ψ has a subrepresentation equivalent to the unit representation.

        7. Let ψ and ρ be representations of G acting on W and V respectively. Assume ρ is irreducible and ρ* ψ has a subrepresentation equivalent to the unit representation. Show that ψ has a subrepresentation equivalent to ρ.

        8. Show that the following are irreducible groups of linear transformations of the finite dimensional vector space V/F: (1) GL(V), (2) O(V, Q) the group of orthogonal linear transformations relative to a non-degenerate quadratic form Q, (3) O⁺ (V, Q) the rotation subgroup of O(V, Q). (See Chapter 6 of BAI.)

5.3   APPLICATION OF THE REPRESENTATION THEORY OF ALGEBRAS

There are two main methods for developing the representation theory of finite groups: the structure and representation theory of finite dimensional algebras applied to group algebras and the theory of characters. Many results can be obtained by both methods, and the choice of the method is often a matter of taste. The first method is introduced in this section and the theory of characters appears in section 5. Our primary concern will be with representations of finite groups over fields whose characteristics do not divide the order of the group. However, the first two theorems that are given do not require these restrictions.

We begin by defining certain algebras of linear transformations associated with a representation ρ of a group G acting on a vector space V/F. First, we have the enveloping algebra Env ρ(G), which is the set of linear transformations of the form . If we extend ρ to a homomorphism ρ of A = F[G] as in (16), then Env ρ(G) = ρ(A). Next we have the algebra A' = End_A V. Evidently, this is the set of linear transformations of V that commute with every a ∈ ρ(A), so End_AV is the centralizer in End_FV of ρ(A). If ρ is irreducible, then by Schur’s lemma, A' is a division algebra. Next we have A′ = End_A, V, which is the algebra of linear transformations that commute with every a' ∈ A'. Evidently, A" ⊃ Env ρ(G) = ρ(A). We remark that if we define A"' = End_A"V, then it is trivial to see that A"' = A'. Hence the process of creating algebras by taking endomorphism algebras breaks off with A". We call this the double centralizer of ρ(A) and we say that ρ has the double centralizer property if A" = ρ(A). We have the following general result on completely reducible representations.

THEOREM 5.1.   Any completely reducible representation ρ of a group G has the double centralizer property.

Proof.   It is easily seen that this is a special case of Theorem 4.10 (p. 222). However, it is somewhat simpler to base the proof directly on the density theorem: Let (u₁, u₂, … ,u_n) be a base for V/F and let l be an element of the double centralizer of ρ(A). By the density theorem, there exists an a ∈ ρ(A) such that au_i = lu_i, 1 ≤ i ≤ n. Hence l = a ∈ ρ(A) and A" = ρ(A).

We state next

THEOREM 5.2.   A finite group has only a finite number of inequivalent irreducible representations.

This follows by applying Theorem 4.5 (p. 208) to the artinian ring A = F[G]. We omit the details, since this result will not play any role in the sequel. We shall now prove

THEOREM 5.3.   The group algebra A = F[G] of a finite group G over a field F is semi-simple if and only if char .

Proof.   Suppose first that char . Then every representation of G is completely reducible. Hence A regarded as (left) A-module is completely reducible. Then A is semi-simple by the structure theorem for semi-primitive artinian rings (p. 203). Now suppose char F| |G|. Consider the element

of A. This is non-zero and we evidently have

for all g' ∈ G. Hence F_Z is an ideal in A. From (25) we obtain , since char F| |G|. Hence F_Z is a non-zero nilpotent ideal in A and A is not semi-simple.

For the remainder of this section, we assume that G is finite and that char . We have

where the A_i are the simple components. By the matrix form of Wedderburn’s theorem,

where Δ_i is a division algebra over F. If I_i is a minimal left ideal of A_i, it is a minimal left ideal of A and hence gives an irreducible representation of G. Moreover, I_i I_j if i ≠ j and every irreducible representation of G is equivalent to one obtained from one of the I_i (Theorem 4.4, p. 208). Hence there are s equivalence classes of irreducible representations of G. We determine next the degree of the irreducible representations provided by I_i. Let {e_ij|1 ≤ i, j ≤ n} be the usual set of matrix units in M_n(Δ), Δ a division algebra. Then

and M_n(Δ)e_ii is a minimal left ideal and every minimal left ideal of M_n(Δ) is isomorphic as M_n(Δ)-left module to M_n(Δ)e_ii. We have the subalgebra Δ1 of matrices d1 = diag{d, d, …, d} and we can regard M_n(Δ) as (left) vector space over Δ by defining d(a) for the matrix (a) to be (d1)(a). Then M_n(Δ) has the base {e_ij} over Δ, so its dimensionality over Δ is n². Similarly, M_n(Δ)e_ii = Δe_1i …Δe_ni, so the dimensionality of M_n(Δ)e_ii over Δ is n. Then the dimensionality of M_n(Δ)e_ii over F is

where d = [Δ : F]. It now follows that [I_i : F] = n_id_i where d_i = [Δ_i : F] and this is the degree of the irreducible representation provided by I_i. We can summarize these results as

THEOREM 5.4.   Let A = F[G] be the group algebra of a finite group G over a field F such that char and let A = A₁ …A_s where the A_i are the simple components of the semi-simple algebra A. Assume that A_i M_ni(Δ_i) where Δ_i is a division algebra and let [Δ_i : F] = d_i. If I_i is a minimal left ideal of A_i, then I_i provides an irreducible representation of G and the irreducible representations obtained from the I_i for 1 ≤ i ≤ s form a set of representatives of the equivalence classes of irreducible representations of G. Moreover, the degree of the irreducible representation provided by I_i is n_id_i.

Let {ρ_i|l ≤ i ≤ s} be a set of representatives of the equivalence classes of irreducible representations of G. As above, we may take ρ_i to be the irreducible representation provided by a minimal left ideal I_i of A_i At any rate we may assume that ρ_i is equivalent to the irreducible representation determined by I_i. If ρ is any representation of G acting on V, then V = V₁ …V_r where the V_j are ρ(G)-invariant and ρ|V_j is irreducible. Then ρ|V_j ρ_i for some i and the number m_i of j such that ρ|V_j ρ_i for a fixed i is independent of the decomposition of V as V₁ …V_r. As on p. 254, we can write ρ ~ m_lρ_l m₂ρ₂ … ^msρ_s and call m_i the multiplicity of ρ_i in the representation ρ.

Now consider A as (left) A-module. Let ρ denote the representation of G determined by this module. The base G of A is stabilized by every ρ(g): x gx. It is clear that ρ is the regular representation of G (p. 248). The decompositions (26) and (28) show that the multiplicity of the irreducible representation ρ_i determined by I_i in the regular representation is n_i. We therefore have the following result on the regular representation of G.

THEOREM 5.5.   Let the notation be as in Theorem 5.4. Then the multiplicity of ρ_i in the regular representation is n_i.

The number of equivalence classes of irreducible representations of G is the number s of simple components A_i of A = F[G]. Let cent A denote the center of A. Then

where the center, cent A_i, is isomorphic to the center of M_ni(Δ_i) and hence to the center of Δ_i. Hence cent A_i is a field and is a simple component of the semisimple commutative algebra cent A. Then s is the number of simple components of cent A. We now determine a base for this algebra.

PROPOSITION 5.1.   Let G = C₁( = {1}) ∪ C₂ ∪ … ∪ C_r be the decomposition of G into conjugacy classes (BAI, P. 74). Put

Proof.   If g ∈ G, then since the map x g^–1xg permutes the elements of the conjugacy class C_i. Hence c_i commutes with every g ∈ G and with every . Thus c_i ∈ centF[G]. Now let ∈ cent F[G]. If h ∈ G,

so the condition h^–1ch = c gives γ_hgh^–1 = γ_g, h ∈ G. Thus any two elements g and g' in the same conjugacy class have the same coefficient in the expression and hence c is a linear combination of the c_i. It is clear that the c_i are linearly independent and so (c₁, c₂, … , c_r) is a base for cent F[G].

Both the foregoing result and the fact that the number s of equivalence classes of irreducible representations of G is the number of simple components show that s ≤ r, the number of conjugacy classes. More precisely, we have r = Σ₁^s [cent A_i : F]. This shows that s = r if and only if cent A_i = F for all i, that is, the A_i are central simple. Since cent A_i is a finite dimensional field extension of F, we have cent A_i = F for all i if F is algebraically closed. Hence in this important case the number of equivalence classes of irreducible representations is the number of conjugacy classes. In the algebraically closed case we have the following important result.

THEOREM 5.6.   Let G be a finite group, F an algebraically closed field such that char . Let s be the number of conjugacy classes of G. Then the number of equivalence classes of irreducible representations over F (acting on vector spaces V/F) is s and if ρ₁, …, ρ_s are representatives of these classes and n_i is the degree of ρ_i, then

Proof.   The first statement has been proved. To see the second, we use (26) and (27) and the fact that since F is algebraically closed, the only finite dimensional division algebra over F is F itself. Then A = F[G] = A₁ … and A_i M_n(F). Then and Theorem 5.4 shows that n_i is the degree of the irreducible representation ρ_i associated with A_i.    

EXAMPLES

1. Let G = , the cyclic group of order n generated by z and let F = . Then A = [G] is a direct sum of n copies of . Hence we have n inequivalent irreducible representations, all of degree 1. It is clear that in the corresponding matrix representations we have z (e^2πir/n) where ( ) is a 1 × 1 matrix and r = 1, 2, …, n.

2. Let G = D_n, the dihedral group of order 2n generated by two elements r, s such that rⁿ = 1, s² = 1, srs^–1 = r^–1 (BAI, pp. 34, 70). The elements of D_n are r^k, r^ks, 0 ≤ k ≤ n – 1, and we have the relation sr^k = r^–ks. Hence (r^ks)² = 1, so the n elements r^ks are of period two. Using the multiplication table: , it is readily seen that if n is odd = 2v + l, v ≥ 1, then there are v + 2 conjugacy classes with representatives: 1, r^k, 1 ≤ k ≤ v, s, and if n = 2v, v ≥ 1, then there are v + 3 conjugacy classes with representatives 1, r^k, 1 ≤ k ≤ v, s, rs. On the other hand, we can list these numbers of inequivalent irreducible matrix representations over C as follows:

n = 2v +1, v ≥ 1.

ρ₁, the unit representation.

ρ₂, the matrix representation of degree 1 such that r (1), s (–1).

σ_l, 1 ≤ l ≤ v, the matrix representation of degree 2 such that

where ω = e^2πi/n.

n = 2v, v ≥ 1.

ρ_l, the unit representation.

ρ₂, the matrix representation of degree 1 such that r (1), s (– 1).

ρ₃, the matrix representation of degree 1 such that r (–1), s (1).

ρ₄, the matrix representation of degree 1 such that r (–1), s (– 1).

The representations σ_l, 1 ≤ l ≤ v – 1, as above.

It is easy to verify that the representations listed are irreducible and inequivalent. Hence they constitute a set of representatives of the equivalence classes of irreducible representations of D_n. As a check, we can verify the degree relation (32) in the two cases:

We shall consider next a process of extension of the base field of a representation. We need to recall some simple facts about extension of the base field of a vector space. For our purpose, it suffices to restrict our attention to finite dimensional vector spaces. Thus let V/F be an n-dimensional vector space over the field F and let K be an extension field of F. We can form the tensor product V_K = K _F V, which can be regarded as a vector space over K. The injective map v l _v permits us to regard V as contained in V_K as an F-subspace such that KV = V_K. Moreover, F-independent elements of V are K-independent and any base for V/F is a base for V_k/K (see p. 220). A linear transformation l of V/F has a unique extension to a linear transformation of V_K/K. We denote the extension by / also. These extensions span End_K V_K as vector space over K, that is, End_K V_K = K End_F V. Moreover, End_K V_K K _F End_FV.

Now suppose we have a representation ρ of G acting on V/F. Then the defining relations ρ(g₁g₂) = ρ(g₁)ρ(g₂), ρ(l) = 1_V give the relations ρ(g₁g₂) = ρ(g₁)ρ(g₂), ρ(l) = 1_VK for the extensions. Hence ρ_K : g ρ(g) (in V_K) is a representation of G acting on V_K. We call this the representation obtained from ρ by extending the base field to K. It is clear that if A = F[G], then K[G] A_K and . We also have the following useful result.

PROPOSITION 5.2. .

Proof. The elements of End_F[G]V (End_K[G]V_K) are the linear transformations of V (V_K) that commute with every ρ(g), g ∈ G. Hence it is clear that K End_F[G] V ⊂ End_K[G] V_K. To prove the reverse containment, we choose a base (λ_i) for K/F. Then any element of End_K V_K can be written in one and only one way in the form Σλ_il_i where l_i ∈ End_FV. The conditions that this commutes with every ρ(g) imply that the l_i commute with every ρ(g), hence, that l_i ∈ End_F[G] V. Thus Σλ_il_i ∈ End_K[G] V_K and End_K[G] V_K = K End_F[G]V. We have noted that . This implies that if E is any subspace of End_FV/F, then . In particular, we have .    

It is clear that if U is a ρ(G)-invariant subspace of V, then KU U_K is a ρ(G)-invariant subspace of V_K. Hence if ρ_K is irreducible, then ρ is irreducible. The converse need not hold, as the following examples show.

EXAMPLES

1. Let G = , the cyclic group of order 4 and let F = . We have the representation of degree two over in which ρ(g) is the linear transformation with matrix

relative to the base (u, v) of V/. Since the characteristic polynomial of this matrix is λ² + 1, ρ(g) acts irreducibly on V. Now consider ρ. We have the base (z = u + iv, w = u – iv) for V/ and

Hence _Z and _w are ρ(G)-invariant subspaces. The irreducible representations ρ|_Z and ρ | _w are inequivalent.

2. Let G be the quaternion group {± 1, ±i, ±j, ±k}, which is a subgroup of the multiplicative group of Hamilton’s quaternion algebra over . Let ρ be the representation of G such that ρ(g) is the left multiplication x gx, x ∈ , g ∈ G. ρ is irreducible since is a division algebra. On the other hand, M₂() since is a splitting field for (p. 228). Since M₂() is a direct sum of two minimal left ideals, it follows that is a direct sum of two ρ(G)-invariant subspaces I₁ and I₂. The representations ρ|I₁ and ρ|I₂ are irreducible and equivalent.

A representation ρ is called absolutely irreducible if ρ_k is irreducible for every extension field K of the base field F. We have the following criterion for irreducibility and absolute irreducibility.

THEOREM 5.7.   Let G be a finite group, ρ a representation of G acting on V/F where char , A = F[G]. Then ρ is irreducible if and only if A' = End _A V is a division algebra and ρ is absolutely irreducible if and only if A' = F1.

Proof.   If ρ is irreducible, then A′ is a division algebra by Schur’s lemma and if ρ is absolutely irreducible, then is a division algebra for every extension field K of F. This implies that A' = F1. For, suppose that A' ≠ F1 and let c ∈ A', F1. Then the minimum polynomial f(λ) of c over F has degree > 1 and this is irreducible since A' is a division algebra. Put K = F[λ]/(f(λ)) and consider A'_K. The minimum polynomial of c ∈ A'_K is f(λ) and this is reducible in K[λ]. Hence A'_K is not a division algebra, contrary to what we had proved. Hence A' = F1.

Next assume ρ is reducible, so we have a ρ(G)-invariant subspace U ≠ V, 0. By Maschke’s theorem, there exists a projection p on U that commutes with every ρ(g). Then p is an idempotent ≠ 0, 1 in A' and A′ is not a division algebra. Thus if A' is a division algebra, then ρ is irreducible. Since A′ = F1 implies A'_K = K1, it follows that if A' = F1, then ρ is absolutely irreducible.    

A field F is called a splitting field for the group G if every irreducible representation of G over F is absolutely irreducible. We have the following

THEOREM 5.8.   Let G be a finite group, F a field with char . Then F is a splitting field for G if and only if F[G] is a direct sum of matrix algebras M_n(F).

Proof.   Let A = F[G] = A₁…A_S where the A_i are the simple components of A and let I_i be a minimal left ideal of A_i The representation ρ_i of G acting on I_i is irreducible and every irreducible representation of G over F is equivalent to one of the ρ_i. Hence F is a splitting field for G if and only if every ρ_i is absolutely irreducible. By Theorem 5.7, this is the case if and only if End_AI_i = F1 for 1 ≤ i ≤ s. Now End_AI_i = Δ'_i is a division algebra and by Theorem 5.1, where and n_i is the dimensionality of I_i as vector space over Δ'_i. On the other hand, since the A_j, j ≠ i, annihilate I_i and A_i is simple, . Thus . Now suppose F is a splitting field. Then Δ'_i = F and . Conversely, suppose for some n'_i. Then . By the isomorphism theorem for simple artinian algebras, this implies that n'_i = n_i and . Then Δ_i = F1, End_A I_i = Δ'_i = F1, and F is a splitting field.    

EXERCISES

        1. Let G be finite, ρ an irreducible representation of G over F where char . Show that if ρ is the unit representation and otherwise.

        2. Let G be a finite group, F = or /(ρ) where . Show that there exists a finite dimensional extension field of F that is a splitting field for G.

        3. Let ρ₁ and ρ₂ be representations of G over an infinite field F and let K be an extension field. Show that if ρ_1K and ρ_2K are equivalent, then ρ₁ and ρ₂ are equivalent. Does this hold for F finite?

5.4   IRREDUCIBLE REPRESENTATIONS OF S_n

It is generally a difficult problem to determine the structure of the group algebra F[G] for a given finite group G. As we saw in the previous section, if char , this amounts to determining a set of representatives for the equivalence classes of irreducible representations of G over F, or, as we shall now say more briefly, determining the irreducible representations of G over F. In this section, we give one important example for which this can be done: G = S_n. We shall determine a set of idempotents generating minimal left ideals of F[S_n] (char ) that give the irreducible representations of S_n over F. These results are due to Frobenius and to A. Young; our exposition follows one due to J. von Neumann as presented in van der Waerden’s Algebra vol. 2, p. 246.

We recall that the number of conjugacy classes of S_n is p(n) the number of (unordered) partitions of n (BAI, p. 75). If {r₁, r₂, …, r_h} is such a partition (that is, r_i ≥ 1 and , then the permutations that are products of disjoint cycles of lengths r₁, r₂, …, r_h form a conjugacy class and every conjugacy class is obtained in this way. For the partition α = {r₁, r₂, …, r_h} we assume and we use this order to order the set of partitions lexicographically. Thus if β = {s₁, s₂, …, s_k} with then α > β if r_i > s_i at the first place where r_i ≠ s_i. Each partition α = {r₁, r₂, …, r_h} defines a (Young) tableau

and each tableau defines a set of diagrams D, E,… obtained by distributing the numbers 1, 2, …, n in the boxes so that no two numbers appear in the same box. For example, if α = {3, 2, 2}, then one of the D is

If D ( = D ) is a diagram, we define the group R(D) of row permutations of D to be the subgroup of S_n of permutations stabilizing the subsets filled in the rows of D. Thus, for D as in (34), R(D) is the set of products of cycles all of whose numbers appear in one of the rows {1, 3,7}, {2, 6), or {4, 5} (so R(D) = {1, (13), (17), (37), (137), (173), (26), (45), (13), (26), etc.}). Similarly, we define the subgroup C(D) of column permutations of D to be the subgroup of S_n of permutations stabilizing the columns of D.

If D_α is a diagram and σ ∈ S_n, then σD_α is the diagram obtained from D_α by applying σ to its entries. Evidently this is an E_α and every E_α is a σD_α for some σ ∈ S_n. It is clear that .

With each diagram D ( = D_α) we shall associate elements of the group algebra F[S_n] as follows: Put

where sg τ denotes the sign of the permutation τ. Evidently, S_D ≠ 0 and A_D ≠ 0 in F[S_n]. But also F_D ≠ 0. To see this, we observe that R(D) ∩ C(D) = 1, since an element common to these subgroups stabilizes every row and every column and hence fixes every element in {1, 2, …,n}. It now follows that if σ₁, σ₂ ∈ R(D) and τ₁, τ₂ ∈ C(D), then τ₁σ₁ = τ₂σ₂ implies and σ₁ = σ₂ and τ₁ = τ₂. Thus the products τσ appearing in F_D are distinct and hence F_D ≠ 0.

Since and for ρ ∈ S_n and , it follows that , and . Also if σ ∈ R(D) and τ ∈ C(D), then the definition (35) gives

The main properties of the elements S_D, A_D, F_D will be derived from the following combinatorial result.

LEMMA 1.   Let α and β partitions such that α ≥ β and let D_α and E_β be associated diagrams. Suppose no two numbers appearing in the same row in D_α are in the same column of E_β. Then α = β and E_α = στD_β for some σ ∈ R(D_α) and τ ∈ C(D_α).

Proof.   The number of entries in the first row of D_α is the same as or exceeds that in the first row of E_β. If greater, then since the number of columns of E_β is the number of entries in the first row of E_β, two entries of the first row of D_α occur in the same column in E_β, contrary to hypothesis. Hence both diagrams have the same number of entries in the first row. Also we have a column permutation τ'₁ of E_β so that the first row of τ'₁ E_β has the same entries as the first row of D_α. Next we note that the entries in the second row of D_α occur in distinct columns of τ'₁ E_β and in rows after the first. It follows that D_α and τ'₁E_β and hence D_α and E_β have the same number of entries in the second row and a column permutation τ'₂ of τ'₁ E_β (and of E_β) brings these into the second row. Continuing in this way, we see that β = α and there exists τ' ∈ C(E_α) such that the entries of each row of τ'E_α and of D_α are the same. Hence there is a σ ∈ R(D_α) such that . Now . Hence and . Then .

Now assume α > β. Then Lemma 1 implies that there exist i, j, i ≠ j, in a row of D_α and in a column of E_β. If π = (ij) then π ∈ R(D_α). Then, by (36), . Hence

Hence . If ρ is any element of S_n, then and since we have . Thus

LEMMA 2.   An element a ∈ F[S_n] satisfies the equations τaσ = (sgτ) a for all σ ∈ R(D) and τ ∈ C(D) if and only if a = γF_D, γ ∈ F, F_D = A_D S_D.

Proof.   We have . Hence any γF_D, γ ∈ F, satisfies the conditions. Next let satisfy the given conditions. Then

for σ ∈ R(D) and τ ∈ C(D). In particular, γ_τσ = γ₁ sg τ, so if we can show that γ_ρ = 0 if ρ is not of the form τσ, τ ∈ C(D), σ ∈ R(D), then we shall have a = γ₁ F_D by (35). Hence suppose . Then . Then ρ^–1D ≠ στD and Lemma 1 implies that there exists a transposition . Then π = ρ^–1π'ρ where π' is a transposition contained in C(D). Then ρ = π'ρπ and by (38). Hence γ_ρ = 0 and the proof is complete.

Now let x ∈ F[S_n] and consider the element F_D x F_D = A_D S_D x A_D S_D. By (36) we have for σ ∈ R_D, τ ∈ C_D. Hence, by Lemma 2, F_D x F_D = γ F_D for γ ∈ F. In particular, F_D² = γF_D. We proceed to show that γ ≠ 0. For this purpose, we consider the map x F_Dx of F[S_n] into itself. If γ = 0, F_D² = 0 and the map x F_Dx is nilpotent and hence has trace 0. On the other hand, if ρ ∈ S_n, then if we look at the matrix of x ρx relative to a base, we see that the trace of this map is 0 if ρ ≠ 1 and is n! if ρ = 1. Since the formula (35) for F_D shows that the coefficient of 1 in the expression for F_D is 1, the trace of x F_Dx is n! ≠ 0 (since char ).

We now put e_D = γ^–1 F_D. Then e_D² = e_D ≠ 0 and e_DF[S_n]e_D = Fe_D. Also, if α and β are distinct partitions, then follows from (37) if D_α is a diagram associated with α and E_β is one associated with β.

We recall that if e and f are idempotents of a ring A, then the additive groups hom_A(Ae, Af) and eAf are isomorphic and the rings End_AAe and eAe are anti-isomorphic (p. 180). If we apply this and Theorem 5.7 to the representations of G acting on F[S_n]e_D , we see that this representation is absolutely irreducible and that if α ≠ β, then the representations provided by F[S_n]e_Dx and F[S_n]e_Eβ are inequivalent. Since the number of conjugacy classes of S_n is p(n), we obtain in this way a full set of representatives of the equivalence classes of irreducible representations. In terms of the group algebra F[S_n] the result we have proved is

THEOREM 5.9.   If F is a field of characteristic 0 or of prime characteristic exceeding n, then

The method of proof is constructive and in theory it can be used to carry out the decomposition of F[S_n] into simple components. The determination of the n_i, which are the degrees of the irreducible representations, can be made by calculating the characters of S_n as defined in the next section for an arbitrary group. There is an extensive literature on the characters of S_n. We shall not consider any of this here. Evidently, Theorem 5.9 has the following consequence.

COROLLARY.   The field and any field /(p) with p > n is a splitting field for S_n.

5.5   CHARACTERS. ORTHOGONALITY RELATIONS

DEFINITION 5.2.   If ρ is a representation of a group G acting on a vector space V/F, then the F-valued function on G defined by

where tr ρ(g) is the trace of the linear transformation ρ(g), is called the character of G afforded by ρ. If p is irreducible, then χ_ρ is called irreducible and if F = , then χ_ρ is called a complex character. The degree of ρ is called the degree of χ_ρ.

As we shall see in a moment, two representations of G over are equivalent if and only if they have the same character. Moreover, a great deal of experience has shown that the characters encapsulate precisely the information on the representations that is useable for the applications. For these reasons, it is fair to say that the central problem of representation theory is that of determining the complex irreducible characters of a given group, or, more precisely, of developing methods for this purpose.

We begin by listing some simple facts about characters.

1. Equivalent representations have the same character. If ρ₁ on V₁ is equivalent to ρ₂ on V₂, then there exists a bijective linear map η of V₁ onto V₂ such that ρ₂(g) = ηρ₁(g)η^–1. This implies that trρ₂(g) = trρ₁(g) and χ_ρ2 = χ_ρ1.

2. Any character is a class function, that is, it is constant on every conjugacy class and hence it defines a map of the set of conjugacy classes into the base field. Let g, h ∈ G. Then . Hence .

3. If char F = 0, then the degree of ρ is χ_ρ(1). This is clear since ρ(l) = l_v, so χ_ρ(1) = trρ(l) = dim V.

4. Let U be a ρ(G)-invariant subspace of the space V on which ρ acts and let ρ|U and ρ|V/U be the corresponding subrepresentation and factor representation. Then .

This follows by choosing a base (u₁, … , u_n) for V such that (u₁, … , u_r) is a base for U (and hence (u_{r + l} + U, … u_n + U) is a base for V/U). Then the matrices of the ρ(g) all have the reduced form (18). The result follows from this.

5. If ρ₁ and ρ₂ are representations of G, then

that is, for every g ∈ G, . To see this, we refer to the matrix (12) for the linear transformation a₁ a₂ in V_l V₂ where a_i is a linear transformation in V_i. It is clear from this that tr (a₁ a₂) = (tr a₁) (tr a₂). Hence .

For the applications we are interested primarily in complex characters. We shall now note some of their properties.

6. If G is finite, then any complex character of G is a sum of mth roots of unity where m is the exponent of G, defined to be the least common multiple of the orders of the elements of G. If g ∈ G, then g^m = 1 and hence ρ(g)^m= 1. Hence the minimum polynomial of ρ(g) is a factor of λ^m – 1, and so it has distinct roots that are mth roots of unity. It follows that ρ(g) has a matrix of the form

where the ω_i are mth roots of unity. Then .

There are several useful consequences of this result. First we have the following

PROPOSITION 5.3.   Let ρ be a complex representation of degree n of a finite group G. Then for any g ∈ G

and

if and only if ρ(g) = ω1 where ω is an mth root of unity, m the exponent of G. In particular, if

then ρ(g) = 1.

Proof.   We have an mth root of unity. Then . Moreover, equality holds if and only if all the ω_i are on the same ray through the origin. Since they are on the unit circle, this holds if and only if they are equal. Hence if and only if ρ(g) = ω1. The last statement is an immediate consequence of this.    

The fact that implies ρ(g) = 1, and the obvious converse of this leads us to define the kernel of the character χ_ρ as the set of g such that (the degree of ρ). Then ker χ_ρ = ker ρ is a normal subgroup of G. Also we define

Then we have shown that Z(χ_ρ) is the set of g such that ρ(g) = ω1. It is clear that these form a normal subgroup of G containing ker χ_ρ and Z(χ_ρ)/ker ρ is isomorphic to a subgroup of the multiplicative group of rath roots of unity. Hence Z(χ_ρ)/ker ρ is cyclic.

Another important consequence of the proof of property 6 is

7. Let ρ be a complex representation of a finite group G, ρ* the contragredient representation. Then

(that is, . To see this, we suppose g ∈ G and we choose a base B in V such that the matrix of ρ(g) relative to this base is (39). Then the matrix of ρ*(g) relative to the dual base B* (p. 250) is . Hence .

EXAMPLES

1. Let 1 be the unit representation: V is one-dimensional and 1(g) = 1_V, g ∈ G. The character afforded by this is the unit character χ₁:g 1 ∈ F.

2. Let G be finite, ρ the regular representation of G. To determine χ_ρ we use the base G = {g₁ = l, g₂, …, g_n} for F[G]. We have χ_ρ(1) = n. On the other hand, if i > 1, then all of the diagonal elements of the matrix of ρ(g_i) relative to the base G are 0 since g_ig_j ≠ g_j. Hence, χ_ρ(g_i) = 0. Thus the character of the regular representation is given by

3. Let G = D_n, the dihedral group of order 2n generated by r, s such that rⁿ = 1, s² = 1, srs^–1 = r^–1. If we refer to the results given in example 2, p. 261, we obtain the following character tables:

In these tables, the representatives of the conjugacy classes are listed in the top row and the rows correspond to the irreducible representations given before. In both cases, 1 ≤ k ≤ v, and 1 ≤ l ≤ v if n = 2v + l, l ≤ l ≤ v – 1 if n = 2v.

We shall derive next some fundamental orthogonality relations connecting the irreducible complex characters of a finite group. We consider first a more general result in the situation in which G is a finite group, F a splitting field for G with char . Let {ρ₁, …, ρ_s} be a set of representatives of the (absolutely) irreducible representations of G over F and suppose ρ_i acts on V_i, 1 ≤ i ≤ s. We assume also that ρ₁ is the unit representation. Let 1 ≤ i, j ≤ s and consider hom_F(V_i, V_j). We have a representation ρ_ij of G acting on hom_F(V_i, V_j) obtained by defining

for l ∈ hom_F(V_i, V_j). It is clear that this gives a representation of G (exercise 2, p. 251). Now let l be any element of hom_F(V_i, V_j) and form the element

We have

This implies that η(l) is a homomorphism of V_i regarded as F[G]-module into V_j regarded as F[G]-module, that is, η(l) ∈ hom_F[G](V_i, V_j). Now if i ≠ j, then V_i and V_j are irreducible and non-isomorphic F[G]-modules. Hence by Schur’s lemma,

if i ≠ j and l is any element of hom_F(V_i V_j). Next suppose i = j. Then η(l) ∈ End_F[G] V_i = F1, so we have

for any l ∈ End_FV_i. We can use (47) and (48) to derive the following result, which is due to Schur.

THEOREM 5.10.   Let G be a finite group, F a splitting field for G with char . Let {ρ₁, …, ρ_s} be a set of representatives of the equivalence classes of irreducible representations of G over F and for each i let ρ⁽ⁱ⁾ be a matrix representation given by ρ_i Then char and we have the following relations:

These relations are called the Schur relations.

Proof.   Let be a base for the space V_i on which ρ_i acts so that and let f_lr be the element of hom_F(V_i, V_j) such that . Then . Then (47) implies the first set of relations (49). Also for j = i, the foregoing relations and (48) give

Hence

Put t = k in these equations and sum on k. This gives

which shows that char and . Substituting this in (50) gives the second set of Schur’s relations.    

We have . Hence if we put l = k and t = r in (49) and sum on k and r, we obtain

Now suppose F = . Then it is clear from the fact that ρ_i(g) has a matrix of the form diag {ω₁, …, ω_n} where the ω’s are roots of unity (see property 6 above) that . Hence we obtain from (51) the basic orthogonality relations for irreducible complex characters:

We now consider the complex vector space ^G of complex valued functions on G. We have the usual definitions of addition and multiplication by complex numbers: If for a ∈ . We define a hermitian form on ^G by

Then

and equality holds if and only if φ = 0. Hence (φ|ψ) is positive definite. We shall now write χ_i for χ_ρi. Then the relations (52) are equivalent to

These state that {χ₁, …, χ_s} is an orthonormal set of vectors in ^G, that is, they are mutually orthogonal and all have length one. It is clear from this that the irreducible complex characters are linearly independent over . We have observed that characters are class functions. The set of class functions forms a subspace cf (G) of ^G whose dimensionality is the number of conjugacy classes. We have seen that this number is the same as the number s of irreducible representations over . Hence, it is clear that the irreducible characters constitute a base for cf (G).

Now let ρ be an arbitrary complex representation of G, χ its character. We have defined the multiplicity m_i of ρ_i in ρ on p. 255. It is clear from the definition and from the fact that the representations are completely reducible that two representations are equivalent if and only if for every i = 1, 2, …, s, the multiplicity of ρ_i in the two representations is the same. It is clear also that we have the formula

where m_i is the multiplicity of ρ_i in ρ. Now, by (54),

Hence the m_i, are determined by χ and consequently complex representations are equivalent if and only if they have the same character. Also, by (55), we have

and this has the value 1 if and only if one of the m_i is 1 and the rest are 0. This shows that a character is irreducible if and only if it has length 1 in the hermitian metric for ^G.

Let C₁ = {1}, C₂, …, C_S be the conjugacy classes of G and let h_k = |C_k|. We now write χ_ik for χ_i(g), g ∈ C_k. Then the orthogonality relations give

Thus we have

Let X = (χ_ij) ∈ M_s() and let H = diag {h_l, …, h_s}. Then the foregoing relations amount to the matrix equation

From this, one deduces , which is equivalent to

We shall call (58) and (60) the row and the column orthogonality relations respectively for the characters. We remark that if we take j = k = 1 in (60) we obtain . This is the relation (32) that we had obtained in a more general situation by using the theory of algebras.

If ρ is a representation of a group G acting on V/F, then ρ has a unique extension to a homomorphism of F[G] into End_FV. This maps the element of the group algebra into . We shall denote this extension by ρ also. Similarly, the character χ_ρ defined on G can be extended uniquely to a linear map χ_ρ of F[G] into F by putting . If c ∈ cent F[G], the center of the group algebra, then ρ(c) is in the centralizer of ρ, so if ρ is absolutely irreducible, then ρ(c) = ω(c)1 where ω(c) ∈ F. If the degree of ρ is n, then χ_ρ(c) = nω(c) so

Since the map c ρ(c) is an algebra homomorphism of cent F[G] into F1, it is clear that c (1/n)χ_ρ(c) is an algebra homomorphism.

We have seen that if we put , then (c₁, …, c_s) is a base for cent F[G]. Evidently, where the n_jkg ∈ , the set of nonnegative integers. Since c_jc_k ∈ cent F[G], we have h^–1c_jc_kh = c_jc_k, h ∈ G. This gives . It follows that we have a multiplication table of the form

where m_jkl ∈ for the base (c_l, …, c_s) of centF[G] over F. It is readily seen that m_jkl is the number of pairs (x, y), x ∈ C_j, y ∈ C_k, such that xy is a given z in C_l.

Now suppose F = and let the notation be as before. We apply the representation ρ_i to (62) to obtain . If n_i = deg ρ_i = χ_i(1), then we have . Hence we have the character relation

If we use the definition of and the fact that characters are class functions, we obtain for any g_j ∈ C_j, so χ_i(c_j) = h_jχ_ij. Hence we have the character relation

or

By a character table for a group G, we mean a table giving the values χij for the irreducible complex characters. A fundamental problem for the study of a given group G is the computation of its character table. As an illustration of this, we consider the following

EXAMPLE

We wish to determine a character table for S₄. The following is a list of representatives for the conjugacy classes: (1, (12), (123), (1234), (12) (34)). We denote the corresponding conjugacy classes as (C_l, C₂, C₃, C₄, C₅). Their cardinalities are respectively (h₁, h₂, h₃, h₄, h₅) = (1, 6,8, 6,3). The alternating group A₄ is a normal subgroup of index 2 in S₄, and V = {1, (12) (34), (13) (24), (14) (23)} is a normal subgroup of index 6 in S₄ (BAI, p. 261, exercise 4). If we let S₃ denote the subgroup of S₄ fixing 4, then S₄ is a semi-direct product of V and S₃: Any element of S₄ can be written in one and only one way as a product sv where s ∈ S₃, v ∈ V. If s_i ∈ S₃ and v_i ∈ V, then (s₁v₁)(s₂v₂) = (s₁s₂)(s₂^–1v₁s₂v₂) and s₂^–1v₁s₂ ∈ V. Hence sv s is a homomorphism η of S₄ onto S₃ with kernel V. If ρ is a representation of S₃, then ρη is a representation of S₄ whose character is χ_ρη.

Now S₃ D₃ under a map such that (12) s, (123) r. Hence the character table for D₃ gives the following character table for S₃:

We denote the characters obtained by composing this with η again by χ_l, χ₂, χ₃ Since. (12) (13) (24) = (1324), the part of the character table obtained from these characters is

If n_i is the degree of χ_i 1 ≤ i ≤ 5, then Σn_i² = 24 gives n₄² + n₅² = 18. Hence n₄ = n₅ = 3, so the missing two irreducible characters are of degree three. Hence the last two rows of the character table have the form (3, α, β, γ, δ) and (3, α', β', γ', δ'). If we use the relation (60) with j = 1 and k = 2, 3,4, 5, we obtain

Hence the last row is (3, –α, –β, –γ, – 2 – δ). If we use (58) with i = 4 and j = 1, 2, 3, we obtain the relations

These equations give β = 0, δ = – 1, γ = –α. The orthogonality relation (60) for j = 2, k = 4 gives = – 1. On the other hand, α = χ₄(12) is a sum of square roots of 1, hence real. Thus = α( – ) = – α² = – 1 and α² = 1, so α = ± 1, γ = 1. Thus the last two rows are either (3, 1,0, – 1, – 1) and (3, – 1, 0,1, – 1) or (3, – 1, 0,1, – 1) and (3, 1,0, – 1, – 1). Both determinations give the same table except for the order of the last two rows.

The foregoing example illustrates how the orthogonality relations plus other information that one can get a hold of can be used to calculate a character table. Further results that we shall derive presently will supply additional information useful for calculating character tables. We should note also that there is a substantial literature on the characters of S_n, beginning with a classical paper published by Frobenius in 1900 that gives formulas for the characters of any S_n. See, for example, Weyl’s Classical Groups, p. 213.

EXERCISES

        1. Determine a character table for A₄.

        2. Determine a character table for the quaternion group.

        3. Let G be a subgroup of S_n, ρ the corresponding permutation representation of G over , χ its character. Show that where r is the multiplicity of the unit representation ρ₁ in ρ. Show that is also the total number of fixed points in {1, 2, …, n} for all g ∈ G.

        4. Let the notations be as in exercise 3. Show that the number of orbits of G in {1, 2, …, n} is the multiplicity r of ρ₁ in ρ.

        5. A permutation group G is called k-fold transitive for k = 1, 2, … if given any two k-tuples (i₁, i₂, …, i_k) and (j₁, j₂, …, j_k) of distinct i’s and j’s, there exists a g ∈ G such that gi₁ = j_i, l ≤ l ≤ k. Let G be doubly ( = 2-fold) transitive, ρ the corresponding permutation representation. Show that ρ ~ ρ₁ ρ_i where ρ_i is an irreducible representation ≠ ρ₁.

        6. Let χ₁, …, χ_s be the irreducible complex characters of G and C₁ = {1}, C₂, …, C_s be conjugacy classes of G, h_i = |C_i|. Put and where χ_ij = χ_i(g) for some g ∈ C_j. Show that the e_i are orthogonal idempotents in the center of [G] and Σe_i = 1.

        7. Show that an element g in a finite group G is conjugate to its inverse g^–1 if and only if χ(g) ∈ for every irreducible complex character χ.

5.6   DIRECT PRODUCTS OF GROUPS. CHARACTERS OF ABELIAN GROUPS

It is easy to see that if G₁ and G₂ are finite groups, then F[G₁ × G₂] F[G₁] _F F[G₂]. If char and F is a splitting field for G₁ and G₂, this can be used to reduce the study of the representations over F of G₁ × G₂ to that of the components G_i. In the most important case, in which F = , we can obtain the results also by using characters. We shall follow this approach.

Let ρ_i, i = 1, 2, be a representation of G_i acting on V_i over any F. If we compose the projection (g₁, g₂) g₁ with ρ₁, we obtain a representation ρ'₁ of G₁ × G₂. Evidently G₂ ⊂ kerρ'₁. Similarly, we obtain a representation ρ'₂ of G₁ × G₂ by composing the projection (g₁, g₂) g₂ with the representation ρ₂ of G₂. We now form ρ'₁ ρ'₂, which we denote as ρ₁ ≠ ρ₂. Then

and hence

We have the canonical imbeddings g₁ (g_l, 1₂) and g₂ (1₁, g₂) of G₁ and G₂ in G₁ × G₂. Then

Similarly,

All of this has an immediate extension to direct products of more than two factors.

Now suppose the G_i are finite, F = , and ρ_i is an irreducible representation of G_i. Then we have

Hence ρ₁ # ρ₂ is an irreducible representation of G₁ × G₂. Next suppose the irreducible characters of G_i are χ₁ ⁽ⁱ⁾, … , χ_s ⁽ⁱ⁾ and let ρ_k⁽ⁱ⁾ be a representation affording χ_k⁽ⁱ⁾. The representations ρ_k⁽¹⁾ # ρ_l⁽²⁾ of G₁ × G₂ are irreducible and (67) and (68) imply that ρ_k⁽¹⁾ # ρ_l⁽²⁾ and ρ_k'⁽¹⁾ # ρ_l'⁽²⁾ have distinct characters and hence are inequivalent if (k', l') ≠ (k, l). Hence we obtain in this way s₁, s₂ inequivalent irreducible representations. Since the degree of ρ_k⁽¹⁾ # ρ_l⁽²⁾ is the product of the degree of ρ_k⁽¹⁾ and the degree of ρ_l⁽²⁾, the sum of the squares of the degrees of the ρ_k⁽¹⁾ # ρ_l⁽²⁾ is the product of the sum of the squares of the degrees of the irreducible representations of G₁ and the sum of the squares of the degrees of the irreducible representations of G₂. This is |G₁||G₂| = |G₁ × G₂|. It follows that the set of representations {ρ_k⁽¹⁾ # ρ_l⁽²⁾} is a set of representatives of the equivalence classes of the irreducible representations of G₁ × G₂. This proves

THEOREM 5.11.   Let G₁ and G₂ be finite groups, {ρ₁⁽ⁱ⁾, …, ρ_s⁽ⁱ⁾} a set of representatives of the equivalence classes of irreducible representations over of G_i. Then every ρ_k⁽¹⁾ # ρ_l⁽²⁾ is an irreducible representation of G₁ × G₂ and {ρ_k⁽¹⁾ # ρ_l⁽²⁾} is a set of representatives of the equivalence classes of irreducible representations of G₁ × G₂.

A character χ of degree one is called linear. Evidently, such a character is irreducible and may be identified with the representation affording it. Thus χ is a homomorphism of the given group into the multiplicative group of a field. Conversely, any homomorphism of a group G into the multiplicative group F* of a field F is a linear character of G. We recall that these characters have played an important role in the Galois theory of fields (BAI, p. 291). If χ and χ' are linear characters of G, then χ₁χ₂ defined by (χχ')(g) = χ(g)χ'(g) is a linear character. In this way the set of linear characters of G form a group with the unit character: g 1 as unit and the inverse of χ as g χ(g)^–1.

If G is an abelian group, every irreducible complex character of G is linear. For, if ρ is an irreducible representation of G acting on V over , then End_[G]V = l and since G is abelian, Env ρ ⊂ End_[G]V so Env ρ = l. Then any subspace is ρ(G)-invariant and since ρ is irreducible, V is onedimensional. Hence χ_ρ is linear. Since G is abelian, every conjugacy class consists of a single element; hence the number of these is |G|. Then G has |G| irreducible complex characters.

The last result can also be seen without using representation theory. In fact, we can easily determine the structure of the group of irreducible (= linear) complex characters of any finite abelian group. If G is any abelian group, the group of irreducible complex characters is called the character group of G. Now let G be finite abelian. Then G is a direct product of cyclic groups (BAI, p. 195): G = G₁ × G₂ × … × G_r where and g_i has order e_i. We may assume the G_i are subgroups and every element of G can be written in one and only one way as where 0 ≤ k_i < e_i. Let , the character group of G. Then so χ(g_i) is an e_ith root of unity. The set of these is a cyclic subgroup Z_ei of the multiplicative group *. We now define a map of into Z_e1 × Z_e2 × … × Z_er by

It is clear that this is a homomorphism. The kernel is the set of χ such that χ(g_i) = l, 1 ≤ i ≤ r. Then and χ = 1. Hence (69) is a monomorphism. Moreover, the map is surjective. For, if (u₁, …, u_r) ∈ Z_e1 × … × Z_er, then u_i^e_i = 1 and hence we have a homomorphism of into Z_ei sending g_i u_i. Then we have a homomorphism χ of G = G₁ × … × G_r into * sending g_i u_i 1 ≤ i ≤ r. Evidently, χ is a character that is mapped into (u_l, …, u_r) by (69). Hence we have an isomorphism of G with its character group .

EXERCISES

        1. Show that the number of complex linear characters of a finite group is the index [G: G'], G' the commutator group of G.

        2. Let G be a finite group, ρ the regular representation of G over the field F. Form the field F(x_g1, …, x_gm) in m-indeterminates where m = |G| and g_i x_gi is 1–1. The determinant

is called the group determinant of G over F. (The study of such determinants was one of the chief motivations for Frobenius’ introduction of the theory of characters.) Show that if F = and G is abelian, then

This result is due to R. Dedekind.

        3. Let Fab denote the category of finite abelian groups with homomorphisms as morphisms. We have a contravariant functor D from Fab to itself such that and if f: A → B, then is defined by ψ χ where χ(x) = ψ(f(x)), x ∈ A. Show that is naturally equivalent to the identity functor (cf. p. 22).

        4. Show that if H is a subgroup of a finite abelian group G, then the subgroup of of χ such that χ_H = 1_H can he identified with and hence its order is |G/H|. Use this to show that the map χ χ_H of into is surjective. Note that this implies that any linear character of H can be extended to a linear character of G.

5.7   SOME ARITHMETICAL CONSIDERATIONS

In this section, G will be finite and all representations and characters are complex. We shall apply some elementary results on integral complex numbers that were given in BAI, pp. 279–281, to obtain important results on the degrees of the irreducible representations and on the characters of S_n.

We recall that a ∈ is called algebraic if a is algebraic over the subfield , that is, a is a root of a non-zero polynomial with coefficients in . The complex number a is called integral (or an integer) if a is a root of a monic polynomial with integer coefficients. A useful criterion to prove integrality is that a is integral if and only if there exists a finitely generated -submodule M of such that 1 ∈ M and aM ⊂ M. The subset A of of algebraic numbers is a subfield and the subset I of integral complex numbers is a subring. Moreover, if a ∈ is a root of a monic polynomial in A[λ] (in I[λ]), then a ∈ A (I). We showed also that a is integral if and only if a is algebraic and its minimum polynomial over has integer coefficients, and that the only rational numbers that are integral in are the elements of .

We have seen that the characters of G are sums of roots of unity (property 6 on p. 270). Since roots of unity are integral, it follows that χ(g) is integral for any character χ and any g ∈ G. We recall the following notations that were introduced in section 5.5.

            1. χ₁, …, χ_s are the irreducible characters. χ₁ is the unit character, that is, the character of the unit representation ρ₁.

            2. C₁ = {1}, C₂, …, C_s are the conjugacy classes.

            3. χ_ij = χ_i(g_j) where g_j ∈ C_j.

            4. n_i = χ_i1. This is the degree of the representation ρ_i such that χ_ρi, = χ_i.

            5. h_j = |C_j|.

We recall also the following character relations (p. 276):

where the m_jkl ∈ . We shall deduce from these equations

PROPOSITION 5.4.   The complex numbers h_jχ_ij/n_i are integral.

Proof. Fix i and put u_k = h_kχ_ij/n_i, 1 ≤ k ≤ s, u₀ = 1. Let . Then (64) shows that u_kM ⊂ M. Hence u_k is integral by the criterion we noted above.

We can now prove

THEOREM 5.12.   n_i||G|.

Proof.   We use the formula (58) for i = j to obtain . Since the and are integral, so is |G|/n_i. Since this is rational, it is contained in . Hence n_i||G|.

The foregoing result is due to Frobenius. We also have the theorem that every n_i = 1 if G is abelian. Both of these results are special cases of the following more general theorem, which is due to Schur.

THEOREM 5.13.   n_i|[G: Z], Z the center of G.

Proof.   (Tate). Let m be a positive integer and let G_m = G × G × … × G, m times. Let ρ_i be an irreducible representation of G over affording the character χ_i, V the vector space on which ρ_i acts. We have the representation ρ of G_m acting in V_m = V … V, m times, such that ρ(g₁, …, g_m) = ρ_i(g₁) … ρ_i(g_m) (see section 5.6). By iterating the result of Theorem 5.11, we see that ρ is irreducible. If c ∈ Z, the irreducibility of ρ_i implies that . Evidently γ_i is a homomorphism of Z into *. By the definition of ρ we have

and hence for c_j ∈ Z,

It follows that the subset D of elements of G_m of the form (c₁, c₂, … ,c_m), c_i ∈ Z, is in the kernel of ρ. Evidently D is a normal subgroup of G_m and ρ defines a representation of Gm/D that is irreducible. We now apply Theorem 5.12 to to obtain . This implies that and this holds for all m. Put u = |G|/n_i|Z|. Then the relation shows that the -submodule Of is contained in |Z|^–1. Since is noetherian, so is |Z|^–1 and hence so is M. Thus M is finitely generated as -module. Since 1 ∈ M and uM ⊂ M, it follows that u ∈ . This proves that (n_i|Z|)||G|. Then n_i[G : Z] as required.    

Let F be a subfield of . Then I ∩ F is a subring of F called the ring of algebraic integers of F. The study of the arithmetic of such rings constitutes the theory of algebraic numbers. We shall give an introduction to this theory in Chapter 10. Now suppose F is a splitting field for the finite group G. Then F[G] = M_n1(F) … M_ns(F) and s is the number of conjugacy classes of G (see p. 280). We have s inequivalent irreducible representations ρ_i over F and these remain irreducible on extension of F to . Thus {ρ₁, …, ρ_s} a set of representatives of the classes of the irreducible complex representations. If ρ_i acts on V_i/F, then to compute χ_ρ1(g), g ∈ G, we choose a base for V_i/F and take the trace of the matrix of χ_ρ1(g) relative to this base. Nothing is changed if we pass to . Thus we see that χ_ρι(g) = χ_ρι(g). This shows that the irreducible complex characters have values in F. We know also that these values are contained in I. Since any character is an integral combination of irreducible characters, this gives the following

THEOREM 5.14.   Let F be a subfield of that is a splitting field for the finite group G. Then any complex character of G has values that are integral algebraic numbers of F.

In section 4, we showed that is a splitting field for S_n. Hence we have the

COROLLARY.   The complex characters of S_n have values in .

5.8   BURNSIDE’S p^aq^b THEOREM

One of the earliest applications of the theory of characters was to the proof of the following beautiful theorem due to Burnside.

THEOREM 5.15.   If p and q are primes, then any group of order p^aq^b is solvable.

Quite recently, John Thompson succeeded in giving a proof of this theorem that does not use representation theory. However, this is considerably more complicated than the original proof with characters, so the original proof—which we shall give here—remains a good illustration of the use of representation theory to obtain results on the structure of finite groups.

We prove first the following

LEMMA.   Let χ be an irreducible complex character of a finite group G, ρ a representation affording χ. Suppose C is a conjugacy class of G such that (|C|, χ(l)) = 1. Then for any g ∈ C, either χ(g) = 0 or ρ(g) ∈ 1.

Proof.   Since (|C|, χ(1)) = 1, there exist integers l and m such that l|C| + mχ(l) = 1. Then

Now χ(g) is an algebraic integer, and by Proposition 5.4 (p. 283), , where n = deg ρ = χ(l), is an algebraic integer. Hence χ(g)/n is an algebraic integer. We recall also that χ(g) ≤ n and this was proved by showing that χ(g) is a sum of n roots of unity (p. 270). Thus χ(g) = ω_l + … + ω_n where the ω_i ∈ W, a cyclotomic field of complex roots of unity (BAI, p. 252). Let H = Gal W/. and let s ∈ H. Then s maps roots of unity into roots of unity. Hence sχ(g) is a sum of n roots of unity. Hence . It is clear that since a = χ(g)/χ(1) is an algebraic integer, so is sa = sχ(g)/χ(1) and |sa| ≤ 1. Then the norm satisfies |N_W/(a)| ≤ 1. Since this is an algebraic integer and a rational number, it follows that |N_W/(a)| ∈ ⁺. Hence either N_W/(a) = 0, in which case a = 0 and χ(g) = 0, or |N_W/(a)| = 1. In the latter case, |a| = 1, |χ(g)| = n, and Proposition 5.3 shows that ρ(g) = ωl, ω a root of unity.    

We recall that the only abelian simple groups are the cyclic groups of prime order. Since this class of simple groups is rather trivial, one generally excludes it from the study of simple groups. We follow this convention in the following

THEOREM 5.16.   Let G be a finite (non-abelian) simple group. Then no conjugacy class of G has cardinality of the form p^a, p a prime, a > 0.

Proof.   Suppose C is a conjugacy class of G such that |C| = p^a, p prime, a > 0. Let ρ₁, …, ρ_s be the irreducible representations of G, χ₁, …, χ_s the corresponding characters. We assume ρ₁ is the unit representation, so χ₁(g) = 1 for all g. Let n_i = χ_i(1), the degree of ρ_i. If then the foregoing lemma shows that either χ_i(g) = 0 or ρ_i(g) ∈ l for every g ∈ . Now the elements g ∈ G such that ρ_i(g) ∈ l form a normal subgroup G_i of G. Hence if χ_i(g) ≠ 0 for some g ∈ C, then G_i ≠ 1. Then G_i = G since G is simple. Also since G is simple and ρ_i is not the unit representation for i > 1, it follows that G ρ_i(G) if i > 1. Since ρ_i(G) is abelian, this is excluded. Hence if and i > 1, then χ_i(g) = 0 for all g ∈ C. We now use the orthogonality relation (60), which gives

for g ∈ C. Since n_l = 1 and χ₁(g) = 1, we have some n_i, i > 1, divisible by p. Let n₂, …, n_t be the n_i, i > 1, divisible by p. Then (73) gives the relation

Since the n_j are divisible by p and the χ_j(g) are algebraic integers, this implies that 1/p is an algebraic integer, contrary to the fact that the only rational numbers that are algebraic integers are the elements of .    

We can now give the

Proof of Theorem 5.15. Let |G| = p^aq^b where p and q are primes. Let P be a Sylow p-subgroup of G. Since P is of prime power order, its center Z ≠ 1. If z ≠ 1 is in Z, then the centralizer C(z) of z contains P, so [G : C(z)] is a power of q. Now [G:C(z)] is the cardinality of the conjugacy class C containing z (BAI, p. 75). Hence if [G:C(z)] > 1, then G is not simple non-abelian by Theorem 5.16. On the other hand, if [G: C(z)] = 1, then z is in the center of G. Then the center of G is not trivial, so again G is not simple non-abelian. It is now clear that unless G is cyclic of prime order, it contains a normal subgroup H ≠ 1, ≠G. Using induction on |G| we can conclude that H and G/H are solvable. Then G is solvable (BAI, p. 247).    

5.9   INDUCED MODULES

Let G be a group, H a subgroup of finite index in G, σ a representation of H acting on the vector space U/F. There is an important process, introduced by Frobenius, for “extending” σ to a representation σ^G of G. We recall that we have an action of G on the set G/H of left cosets of H in G. If G/H = {H₁ = H, H₂, …, H_r} and g ∈ G, then gH_i = H_π(g)i and π(g) is a permutation of {l, …, r}. The map π : g π(g) is a homomorphism of G into the symmetric group S_r (BAI, p. 72). Now put U^G ≡ U^(r), the direct sum of r copies of U. Let {s₁, s₂, …, s_r} be a set of representatives of the left cosets of H, say, H_i = s_iH, 1 ≤ i ≤ r. From now on we shall call such a set a (left) cross section of G relative to H. If g ∈ G, then

where μ_i(g) ∈ H. We can define an action of G on U^G = U^(r) by

where the u_j ∈ U. Using the fact that π is a homomorphism,and that H acts on U, we can verify that (75) does indeed define an action of G on U^G. Since this action is by linear transformations in a finite dimensional vector space, we have a representation σ^G of G acting on U^G. We can verify also that another choice of cross section of G relative to H defines an equivalent representation. All of this will become apparent without calculations by giving an alternative, conceptual definition of σ^G, as we shall now do.

Let B = F[H] the group algebra of H, and A = F[G], the group algebra of G. To be given a representation σ of H amounts to being given a module U for B that is finite dimensional as vector space over F. Now B is a subalgebra of A and hence A is a B-B-bimodule in which the actions of B on A are left and right multiplications by elements of B. We can form A _B U, which is a left ,A-module in which for a₁, a₂ ∈ A and u ∈ U we have

We shall now show that [A _B U : F] < ∞, so the A = F[G]-module A _B U defines a representation of G. Moreover, we shall see that this representation is equivalent to σ^G as defined above.

As before, let {s_l, s₂, …, s_r} be a cross section of G with respect to H. Then any element of G can be written in one and only one way in the form s_ih, h ∈ H. Since the elements of G form a base for A, it follows that A regarded as right B-module is free with (s_l, …, s_r) as base. It follows that every element of A _B U can be written in one and only one way in the form

u_i ∈ U. Hence if (u⁽¹⁾, …, u⁽ⁿ⁾) is a base for U/F, then

is a base for A _B U, so [A _B U : F] = rn < ∞. Thus the module A _B U defines a representation of G.

Now let g ∈ G. Then, by (76) and (74), we have

Comparison of this with (75) shows that the map is an equivalence of the induced representation of G as defined first with the representative of G provided by A _BU.

The module U^G (or A _BU) is called the induced G-module U and the associated representation σ^G is called the induced representation of G. As we have seen, if (u⁽¹⁾, …, u⁽ⁿ⁾) is a base for U/F, then (78) is a base for the induced module. Now suppose

for h ∈ H, so h α(h) = (α_ij(h)) is a matrix representation of H determined by σ and

is the character of σ. By (79), the matrix of σ^G(g) relative to the base (78) is obtained by applying the permutation π(g) to the r rows of n × n blocks in the matrix

Only the non-zero diagonal blocks of the matrix thus obtained contribute to the trace. These blocks occur in the ith row (of blocks) if and only if π(g)i = i, which is equivalent to gs_iH = s_iH and to s_i^–1 gs_i ∈ H.It follows that the character

where the summation Σ' is taken over all i such that s_i^–1 gs_i ∈ H. This can be put in a better form by extending the definition of the function χ_σ from H to F to a function on G defined by = 0 if g ∈ G – H. Using this, we obtain the formula

This can be written in another form if G is a finite group, by noting that . This is clear if g ∈ H since χ_σ is a class function and if , then . Hence

Then (84) gives the formula

EXAMPLES

1. Let G be the dihedral group D_n of order 2n generated by r, s, such that rⁿ = 1, s² = 1, srs^–1 = r^–1. Let H = , σ a representation of degree 1 of H such that r ω1 where ωⁿ = 1. We may take S₁ = 1, s₂ = s as representatives of the cosets of H. Then π(r) = 1, π(s) = (12), and

It follows that a matrix representation determined by σ^G maps

2. Let G be as in example 1, but now let H = . Let σ be the representation of degree 1 such that s –1. We have , so we may take {s_i} where s_i = r^{i – 1}, 1 ≤ i ≤ n, as a set of coset representatives. We have

It follows that we have a matrix representation determined by σ^G such that

3. An interesting special case of induction is obtained by beginning with the unit representation 1 = 1_H of the subgroup H acting on a one-dimensional space U₁ = Fu. Let {s₁, …, s_r} be a cross section of G relative to H and suppose where μ_i(g) ∈ H. The set of elements s_i u, 1 ≤ i ≤ r, is a base for U₁^G and, by definition of the induced representation, the action of g on U₁^G is given by . On the other hand, we have the action of G on the set G/H of left cosets {s_iH|l ≤ i ≤ r} given by . It is clear from this that 1^G is the permutation representation obtained from the action of G on G/H.

We can determine the character χ₁^G for the representation 1^G. It is clear from the formula that the trace of the matrix of 1^G(g) relative to the base (S₁u, …, s_nu) is the number of fixed points of π(g). This is the number of cosets aH such that g(aH) = aH. Hence we have the formula

We note also that the condition gaH = aH is equivalent to ^a–lg = a^–1ga ∈ H.

We recall that the action of G in G/H is transitive and any transitive action of G on a finite set N = {1, 2, …} is equivalent to the action of G on a set of left cosets relative to a subgroup H of index n in G (BAI, p. 75). In fact, if G acts transitively on N, then we can take H to be the stabilizer of any element of N. It is clear from this that the study of induced representations of the form 1_H^G, where 1_H is the unit representation of a subgroup H of finite index in G, is essentially the same thing as the study of transitive actions of G on finite sets.

We shall now give two useful characterizations of induced modules in the general case in which H is a subgroup of G of finite index in G, U a module for H. Let A = F[G], B = F[H], {s₁, …, s_r} a cross section of G relative to H, and consider the induced module U^G of G. We have U^G = A _BU = (s_i U) and [s_i U: F] = [U: F] = n. Hence the map x s_i x of U into s_i U is a linear isomorphism. Since we may take one of the s_i = 1, we see that the map x 1 x is a linear isomorphism of U onto the subspace 1 U = {1 ) x|x ∈ U] of U^G. If b ∈ B, then b( 1 x) = b x = bl x = 1 bx. Hence x 1 x is a B-isomorphism of U onto 1 U, which is a B-submodule of U^G. We shall now identify U with its image 1 U and so regard U as a B-submodule of U^G. It is clear also that we have U^G = s₁U … s_rU, a direct sum of the subspaces s_iU. The properties we have noted give a useful internal characterization of U^G that we state as

PROPOSITION 5.5.   Let V be an A-module. Then V is isomorphic to U^G for some B-module U if and only if V contains U (strictly speaking a B-submodule isomorphic to U) such that V = s₁U … s_rU for some cross section {s₁, …, s_r} of G relative to H.

Proof. We have shown that U^G has the stated properties for the B-submodule U ( = 1 U) and any cross section of G relative to H. Conversely, let V be an A-module having the stated properties. Then V = AU. Since s_i ∈ G, s_i is invertible in A and hence the map x s_ix is a bijection. It follows that [s_iU : F] = [U : F] and since . We have the map (a, x) ax of A × U into V, which is F-bilinear. Since (ab)x = a(bx) for b ∈ B, it follows that we have an F-linear map η of A _BU into V such that η(a x) = ax. Evidently, η is an A-homomorphism. Since V = AU, η is surjective and since [V : F] = [A _BU : F] < ∞, η is injective. Hence V and U^G are isomorphic, A-modules.    

The second characterization of U^G that we shall give is a categorical one that applies to any ring A and subring B. We consider the categories of modules A-mod and B-mod. If M is an A-module, we can regard M as B-module. Evidently, homomorphisms of A-modules are homomorphisms of B-modules. In this way we obtain a functor R from the category A-mod to B-mod sending an A-module M into M regarded as B-module and A-homo-morphisms into the same maps regarded as B-homomorphisms. We call R the restriction of scalars functor from A-mod to B-mod. Now let N be a given B-module. Then we can form A _BN and regard this as an A-module in the usual way. We have a B-homomorphism u of N into A _BN sending y ∈ N into 1 y. We claim that the pair (A N, u) constitutes a universal from N to the functor R (p. 42). This means that if M is a (left) A-module and η is a B-homomorphism of N into RM = M, then there exists a unique homomorphism of A _BN into M (as A-module) such that

is commutative. To see this, we consider the map (a, y) a(ηy) of the product set A × N into M. Evidently this is additive in a and y and if b ∈ B, then (ab)(ηy) = a(b(ηy)) = a(η(by)). Hence (a, y) a(ηy) defines a balanced product from A × N into M (p. 126). Hence we have a group homomorphism of A N into M such that . It is clear that is a homomorphism of A-modules and if y ∈ N, then . Hence (87) is commutative. Now let η' be any A-homomorphism of A _BN into M such that ηy = η'uy = η'(l y). Then η′(a y) = η′(a(l y)) = aη'(l y) = a(ηy). Hence , so is unique.

The fact that for any B-module N, (A _BN, u) is a universal from N to the functor R implies that we have a functor from B-mod to A-mod sending any B-module N into the A-module A _BN and mapping any homomorphism η of B-modules into 1 η, which is a homomorphism of A-modules. This functor is a left adjoint of the functor R (p. 49).

This is applicable in particular to A = F[G] and B = F[H] where H is a subgroup of finite index in G. In this case for a given B-module U such that [U : F] < ∞, we obtain the universal object U^G = A _BU, which is finite dimensional over F and hence provides a representation of G.

EXERCISE

        1. Let G be a finite group, H a subgroup, and let σ be the regular representation of H. Show that σ^G is the regular representation of G.

5.10   PROPERTIES OF INDUCTION. FROBENIUS RECIPROCITY THEOREM

We shall now derive the basic properties of induced representations. We prove first

THEOREM 5.17.   Let K be a subgroup of finite index in G, H a subgroup of finite index in K, σ a representation of H acting on U, ρ a representation of G acting on V, ρ_H the restriction of ρ to H. Then

           (1) H is of finite index in G and σ^G and (σ^K)^G are equivalent.

           (2) If W is a σ(H)-invariant subspace of U, then W^G is a σ^G(G)-invariant subspace of U^G. Moreover, if U = W_l W₂ where W_i is σ-invariant, then U^G = W₁^G W₂^G.

           (3) σ^G ρ and (σ ρ_H)^G are equivalent.

Proof. (1) Write A = F[G], B = F[H], C = F[K]. It is clear that H is of finite index in G. (In fact, [G : H] = [G : K] [K : H].) Now σ^K acts on acts on , and σ^G acts on A _BU. The A-modules and are isomorphic (Proposition 3.7, p. 135) and A _CC and A are isomorphic as left A-modules (Proposition 3.2, p. 130). Hence and A _BU are isomorphic as A-modules, which means that (σ^K)^G and σ^G are equivalent.

(2) If W is σ(H)-invariant subspace of U, then this is a B-submodule of U. Since A is free as right module over B, W^G = A _BW can be identified with its image in U^G = A _BU. Then (σ|W)^G is a subrepresentation of σ^G. The second statement follows in the same way.

(3) The two A-modules we have to consider are and . In the first case the action of g ∈ G is and in the second case we have . Hence the F-isomorphism of onto sending is not an A-isomorphism. We shall define an A-isomorphism of onto such that and hence . For a fixed g ∈ G we consider the map into . Since this is F-bilinear, we have an F-homomorphism τ_g of U _F V into such that . Since G is a base for A/F, for any we can define an F-homomorphism τ_a of U _FV into by . Then . We now have a map

of A × (U _FV) into , which is F-bilinear. Moreover, if h ∈ H, then

It follows that we have a balanced product of A as right B-module and U _FV as left B-module. Hence we have a homomorphism of into sending . This is an F-homomorphism and if g' ∈ G, then and . Hence our map is an A-homomorphism. It is clear also that the map is surjective and since the two modules have the same dimensionality over F, the map is an isomorphism.    

For our next result on contragredience and induction we shall make use of an involution in the group algebra (BAI, p. 112). We note first that if g, h ∈ G, then (gh)^{– 1} = h^{– 1} g^{– 1} and (g^{– 1})^{– 1} = g. Hence if is the linear transformation of A = F[G] into itself such that , then and . Thus is an involution in A. We shall call this the main involution of A and denote it as j. If H is a subgroup of G, then the main involution in B = F[H] is the restriction to B of the main involution in A. Now let H have finite index and let {s_l, …, s_r} be a cross section. Any element of A can be written in one and only one way as , we define

and we have the map (a, c) (a|c) of A × A into B. We list its important properties:

        (i) (a|c) is independent of the choice of the representatives s_i, since any other choice has the form S_ih_i, h_i ∈ H. Using these replaces b_i by h_i^{– 1}b_i, c_i, by h_i^{– l}c_i. Then and (88) is unchanged.

        (ii) (a|c) is F-bilinear. Clear.

        (iii) If b ∈ B, then

This is clear from the definition.

        (iv) If e ∈ A, then

It suffices to prove this for e = g ∈ G. Then where π(g) ∈ S_r and μ_i(g) ∈ H. Then and , so . Hence

Thus (ga|c) = (a|g^{– 1}c) holds for all g ∈ G.

        (v) (s_i|s_j) = δ_ijl.

This is clear from the definition.

We shall now establish commutativity of induction and contragredience. This is given in

THEOREM 5.18   If H is a subgroup of finite index and o is a representation of H, then (σ^G)* and (σ*)^G are equivalent.

Proof. Let U be the vector space on which σ acts and let U* be the dual space of linear functions on U. lf x ∈ U and y* ∈ U*, we write . Then is F-bilinear. The space U* is a left B-module defined by the contragredient representation σ* of H and we can make this a right B-module by defining , b ∈ B. Since for h ∈ H, by definition of σ*, we have , which implies that

for x ∈ U, y* ∈ U*, b ∈ B. We now introduce

This is defined for x ∈ U, a, c ∈ A, y* ∈ U*, and is F-bilinear in c and x for fixed a and y*. Moreover, if b ∈ B, then by property (iii) above,

By definition of the tensor product, this implies that we have an F-linear map of A _BU into F sending . Thus we have an element φ_{a, y*} of the dual space (A _BU)* such that

Now the map (a, y*) φ_{a, y*} is F-bilinear and since

φ_{ab, y*} = φ_{a, by*}. Hence we have a linear map η of A_BU* into (A_BU)* sending ay* into φ_{a, y*}. If g ∈ G, . On the other hand, . Hence by (iv), , which implies that η is an A-homomorphism. To conclude the proof, we shall show that η is injective. Since A_BU* and (A_BU)* have the same finite dimensionality over F, this will imply that η is an isomorphism. Now suppose . We can write and the condition implies that for all c ∈ A, x ∈ U and hence for all c. If we take c = S_j and use (v), we obtain z_j* = 0 and hence . This concludes the proof.    

Our next main objective is to derive an important reciprocity theorem due to Frobenius, which in its original form states that if ρ is an irreducible complex representation of a finite group and σ is an irreducible complex representation of a subgroup H, then the multiplicity of ρ in σ^G is the same as the multiplicity of σ in the representation ρ_H of H. As we shall show below, this can be proved by a simple calculation with characters. We shall first give a conceptual proof of an extension of Frobenius’ theorem. Our proof will be based on the important concept of intertwining numbers for representations and on the following proposition.

PROPOSITION 5.6.   If H is a subgroup of finite index in G, U a module for B = F[H], and V a module for A = F[G], then hom_A(U^G, V) and hom_B(U, V_H) are isomorphic as vector spaces over F.

Proof. We remark that this result can be deduced from a general result on adjoint functors applied to the restriction functor from F[G]-mod to F[H]-mod and its left adjoint. However, we shall give an independent proof. As before, let u be the map of U into U^G sending x l x. This is a B-homomorphism. Hence if ζ ∈ hom_A(U^G, V), then ζu ∈ hom_B(U, V_H). The map ζ ζu is an F-linear map of hom _A(U^G, V) into hom_B(U, V_H). The result we proved in diagram (87) is that the map ζ ζu is surjective. It is also injective. For, if ζu = 0, then ζ (lx) = 0 for all x ∈ U. Since ζ is an A-homomorphism, ζ(ax) = ζ(a(lx)) = aζ(lx) = 0 for all a ∈ A, x ∈ U. Hence ζ = 0. Thus ζ ζu is an isomorphism of hom_A(U^G, V) and hom_B(U, V_H) as vector spaces over F.    

If ρ and τ are representations of G acting on F and W respectively, then the dimensionality

is called the intertwining number l(ρ, τ) of ρ and τ(or of V and W as F[G]-modules). If V = V₁ … V_m where the V_i are ρ(G)-invariant subspaces, then we write ρ = ρ₁ … ρ_m where ρ_i = ρ|V_i. Similarly, let τ = τ₁ … τ_l where τ_j = τ|W_j and W_j is τ-invariant. We have the vector space decomposition

This implies that

If ρ and τ are completely reducible and the ρ_l and τ_j are irreducible, then this formula reduces the calculation of l(ρ, τ) to that of the l(ρ_i, τ_j) where ρ_i and τ_i are irreducible. If ρ and τ are irreducible and inequivalent, then l(ρ, τ) = 0 by Schur’s lemma. It is clear also that l(ρ, τ) is unchanged if we replace ρ or τ by an equivalent representation. Hence if ρ and τ are equivalent, then l(ρ, τ) = l(ρ, ρ) = [End_A(V, V):F]. Hence if ρ and τ are irreducible and F is algebraically closed, then

An immediate consequence of (93) and (94) is

PROPOSITION 5.7.   If F is algebraically closed, ρ is irreducible, and τ is completely reducible, then l(ρ, τ) = l(τ, ρ) = multiplicity of ρ in τ (that is, the number of irreducible components of τ isomorphic to ρ).

We can now prove the fundamental

FROBENIUS RECIPROCITY THEOREM.   Let F be an algebraically closed field, ρ an irreducible representation of G over F, σ an irreducible representation of a subgroup H of finite index in G. Assume that σ^G and ρ_H are completely reducible. Then the multiplicity of ρ in σ^G is the same as the multiplicity of a in ρ_H.

Proof.   By Proposition 5.6, l(σ^G, ρ) = l(σ, ρ_H). By Proposition 5.7, l(σ^G, ρ) is the multiplicity of ρ in σ^G and i(σ, ρ_H) is the multiplicity of σ in ρ_H. Hence these multiplicities coincide.    

The Frobenius reciprocity theorem is usually stated for finite groups and algebraically closed fields whose characteristics do not divide the group order. In this situation the hypothesis that H is of finite index and that σ^G and ρ_H are completely reducible is automatically satisfied. The proof of this result and some of the others we gave on induced representations of finite groups can be made by calculations with characters. As an illustration of this method, we shall now prove the Frobenius reciprocity theorem for G finite and F = by using characters. Let the notation be as above and let χ_ρ, χ_σ,etc. denote the characters of ρ, σ, etc. We recall that if ρ is irreducible and τ is arbitrary, then (χ_ρ|χ_τ) = (χ_τ|χ_ρ) is the multiplicity of ρ in τ. The Frobenius reciprocity theorem is equivalent to

We use the formula (85) for χ_σ^G Then we obtain

On the other hand,

since χ_ρ is a class function on G. Comparison of the two formulas gives (95).

EXERCISES

        1. Construct a character table for G = A₅. Sketch: There are five conjugacy classes in G with representatives 1, (12) (34), (123), (12345), (13524) = (12345)² with respective cardinalities 1, 15, 20, 12, 12. Let χ₁ be the character of the unit representation, v the permutation representation obtained from the natural action of G on {1, 2, 3, 4,5}. We have the following table

Calculation of (χ_v|χ_v) and (χ_v|χ₁) gives χ_v = χ₁ + χ₂ where χ₂ is irreducible. This and the above table give χ₂. To obtain the three missing irreducible characters, we look at the induced characters obtained from linear (degree 1) characters of the subgroup

We find that has index two in H, so we have the linear characters λ₁ and λ₂ on H where λ₁ ≡ 1 and . The set

is a cross section of G relative to H. Using this we can compute λ₁^G and λ₂^G as

Using (λ_i^G|λ_i^G), i = 1, 2, and (λ₁^G|χ₁), we obtain λ₁ = χ₁ + χ₃ where χ₃ is irreducible. Also we have , which implies that λ₂^G = χ₄ + χ₅ where χ₄ and χ₅ are irreducible. The values of χ_i(l), 1 ≤ i ≤ 3, and give χ₄ = 3 = χ₅(1). It follows that we have the table

Since (54321) = (12345)^{– 1} is conjugate to (12345), χ₄ and χ₅ are real (exercise 7, p. 279). Using χ₄ + χ₅ = λ₂^G, we obtain a + a' = ^{– 2}, b + b' = 0, c + c' = 1 = d + d'. Using (χ₄|χ_i) = δ_i4 for i = 1, 2, 3, 4, we obtain a system of equations for a, b, c, d whose solutions are a = – 1, b = 0, c = (1 ± )/2, d = 1 – c. Either determination can be used to complete the table.

        2. Prove Theorems 5.17 and 5.18 for complex representations of finite groups by character calculations.

        3. Let B be a subring of a ring A, M a (left) B-module. Then hom_B(A, M) becomes a left A-module if we define (af)x = f(xa) for f ∈ hom_B(A, M), a, x ∈ A (Proposition 3.4, p. 134). This is called the produced module of A determined by M. Note that the map v: f f1 of hom_B(A, M) into M is a B-homomorphism. Show that the pair (hom_B(A, M), v) constitutes a universal from the restriction of scalars functor R to M (p. 137).

        4. Let G be a group, H a subgroup of finite index, A = F[G], B = F[H], and let U be the B-module determined by a representation σ of H. Then hom_B(A, U) is finite dimensional over F and defines the produced representation ^Gσ of G. Show that ^G(σ*) is equivalent to (σ^G)*. (This shows that in the theory of finite dimensional representations, produced modules are superfluous.)

        5. Let G be finite, and let F be an algebraically closed field whose characteristic does not divide |G|. Let ρ be the permutation representation of G determined by the action of G on G/H where H is a subgroup. Use Frobenius reciprocity to show that the dimensionality of the space Inv ρ = {x|gx = x, g ∈ G} is one.

        6. (Addendum to Clifford’s theorem.) Let ρ be an irreducible representation of G acting on V/F and let . Let W be a homogeneous component of V as F[H]-module. (Recall that ρ_H is completely reducible by Clifford’s theorem.) Let T be the subgroup of G stabilizing W. Show that (i) H ⊂ T ⊂ G and [G: T] = m, the number of homogeneous components of V as F[H]-module, and (ii) if ψ is the representation of T acting on W, then ρ = ψ^G and ψ is irreducible.

        7. (Mackey.) Let H be a subgroup of finite index in G and let U be an F[H]-module. Let V be the set of maps f:G → U such that f(hg) = hf(g). Using the usual vector space structure on V and an action of G defined by (gf) (x) = f(xg), f ∈ V, g, x ∈ G. Show that V becomes an F[G]-module isomorphic to U^G.

5.11   FURTHER RESULTS ON INDUCED MODULES

One of the main problems that we shall consider in this section is that of obtaining useable criteria for an induced representation (or character) to be irreducible. To this end we shall need to study the following problem: Suppose that H and K are subgroups of a group G, H of finite index in G, and σ is a representation of H acting on the vector space U/F. What can be said about the structure of σGK = (σG)K, the restriction to K of the induced representation of σG of G? If V is an F[G]-module, we denote V regarded as an F[K]-module by VK. Thus we are interested in the structure of UGK ≡ (UG)K. We identify U with the F[H]-submodule 1U of UG. If g ∈ G, then gU is stabilized by the action of the subgroup gH = {ghg– 1|h ∈ H} since if x ∈ U, then [ghg– l)gx = ghx ∈ gU. Hence we have the submodules (gU)gH and (gU)K ∩ gH of UGgH and UGK ∩ gH respectively. We remark that since gH is of finite index in G, K ∩ gH is of finite index in K. Hence we can form the induced F[K]-module . We shall obtain a decomposition of UGK into submodules of the form . To describe this, we require some simple remarks on double cosets.

We recall that G is a disjoint union of the double cosets KgH (BAI,p. 53). Moreover, KgH is a union of left cosets kgH, k ∈ K, and since H is of finite index in G, the number of left cosets of H contained in KgH is finite. We now note that if k₁, k₂ ∈ K, then k₁gH = k₂gH if and only if k₁(K ∩ ^gH) = k₂(K ∩ ^gH). For, k₁gH = k₂gH if and only if k₂g = k₁gh, h ∈ H, and this holds if and only if k_l^{– 1} k₂ = ghg^{– l}, h ∈ H. This last condition is equivalent to k₁(K ∩ ^gH) = k₂(K ∩ ^gH). It now follows that {k_i} is a cross section of K with respect to K ∩ ^gH if and only if {k_ig} is a set of representatives of the left cosets of H contained in KgH.

We can use this result to prove the following important theorem:

MACKEY’S DECOMPOSITION THEOREM.   Let H be a subgroup of finite index in G, K an arbitrary subgroup of G, and let U be an F[H]-module. Identify U with the F[H]-submodule 1 U of U^G and let Δ be the set of double cosets D = KgH. Then

where is an F[K]-submodule of U^G_K and for any g ∈ D.

Proof. Let {s₁, …, s_r} be a cross section of G with respect to H. Then U^G = s₁U … s_rU (Proposition 5.5, p. 290). If g ∈ G, then D = KgH = s_i1H ∪ … ∪ s_ikH where the s_ij are distinct. Then DU = KgHU = s_i1U …. s_ikU. It follows that . Now DU = KgU is an F[K]-submodule of U^G_K and DU contains gU, which is an F[K ∩ ^gH]-submodule. We have seen that if {k_i|l ≤ i ≤ q] is a cross section of K with respect to K ∩ ^gH, then {k_ig} is a set of representatives of the left cosets of H contained in D = KgH. Thus we may assume that {k_ig} = {s_i1, …, s_ik}, so q = k and

By Proposition 5.5, this implies that DU as F[K]-module is isomorphic to .    

We shall derive next a theorem on the structure of the tensor product of two induced representations. This is

THEOREM 5.19 (Mackey).   Let H_i, i = 1, 2, be a subgroup of finite index in G, σ_i a representation of H_i acting on U_i/F, and let R be a set of representatives of the double cosets H₁gH₂. Put H^{(1, g)} = H₁ ∩ ^gH₂. Then the F[G] module

Proof. By Theorem 5.17.3, σ_l^Gσ₂^G is equivalent to (σ₁(σ₂^G)_H1)^G, and by Mackey’s decomposition, (U₂^G)_H1 is a direct sum of the F[H₁]-submodules isomorphic to the modules ((gU₂)_{H^{(1, g)}})^H₁, g ∈ R. Hence U₁^GU₁^G is F[G] isomorphic to

By Theorem 5.17.3,

Hence, by the transitivity of induction (Theorem 5.17.1),

Substituting this in the formula for U₁^GU₂^G proves the theorem.    

The foregoing result can be used to obtain a criterion for an induced module to be irreducible. We shall require also some general results on modules, some of which have been indicated previously in exercises (p. 251). We note first that if V_i, i = 1, 2, is a finite dimensional vector space over a field F, then we have a canonical isomorphism between V₁*_FV₂, V₁* the dual space of linear functions on V_l, and hom_F(V₁, V₂). This maps u*v for u* ∈ V₁*, v ∈ V₂, into the linear map [u*, v] such that [u*, v](x) = u*(x)v for x ∈ V₁ (p. 165). The fact that we have such an isomorphism can be seen by noting that u*v is bilinear. Hence we have a linear map η of V₁*_FV₂ into hom_F(V_l, V₂) such that η(u*v) = [u*, v]. It is readily seen that η is surjective and since the two spaces V₁*_F V₂ and hom_F(V_l, V₂) have the same dimensionality [V₁ : F] [V₂ : F], it follows that η is a linear isomorphism.

Now suppose ρ_i, i = 1, 2, is a representation of a group G acting on V_i/F. As on p. 272, we obtain a representation ρ of G acting on hom_F(V_l, V₂) if we define ρ(g)l = ρ₂(g)lρ₁(g)^{– l}, g ∈ G, l ∈ hom_F(V_l, V₂). The isomorphism η of V₁*_F V₂ with hom_F(V_l, V₂) is an equivalence of the representations ρ₁ * ρ₂ and ρ. To see this, let u* ∈ V₁*, v ∈ V₂. Then η(ρ_l *(g)u*ρ₂(g)v) is the map

Hence η(ρ_l *(g)u*ρ₂(g)v) = ρ(g)η(u*v), which implies that η is an F[G]-isomorphism of V₁*V₂ onto hom_F(V_l, V₂).

We now consider the vector space hom_F[G](V_l, V₂). Evidently, this is a subspace of hom_F(V_l, V₂) and we can identify this subspace with the set of l ∈ hom_F(V_l, V₂) such that ρ(g)l = l, g ∈ G. For this condition is equivalent to ρ₂(g)lρ₁(g)^{– 1} = l and to ρ₂(g)l = lρ₁(g), which is the condition that l ∈ hom_F[G](V_l, V₂). In general, if ρ is a representation of G acting on V, then the set of x ∈ V such that ρ(g)x = x for all g ∈ G is denoted as Inv ρ. Evidently this is a subspace. It is clear also that x ≠ 0 is in Inv ρ if and only if Fx is a onedimensional ρ(G)-invariant subspace and the restriction of ρ to Fx is the unit representation. Hence if ρ is a completely reducible representation of G, then [Inv ρ : F] is the multiplicity of the unit representation in ρ.