Contents

Representations of Finite Groups

Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5Specifics on Modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5Equivalent Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

EndZ(G) of an Abelian Group G is a Ring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6A Left R-Module Structure on an Abelian Group M is Equivalent to a Ring-Homomorphism R→ End(M) 6Associative, Unital A-Algebra E Where A Commutative Equivalent to Ring E with Ring-Homomorphism

f : A→ Z(E). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7A Commutative Ring and M an A-Module, Then EndA(M) is an A-Module & A-Algebra . . . . . . . . . . . . . . 7R Commutative Ring, A an R-Algebra, Then Left A-Module M Equivalent to R-Module M With

R-Algebra Homomorphism A→ EndR(M) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8The Group Ring or Monoid Ring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8Basic Definitions of Representation Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

Basic Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10Representations of a Ring Form a Category . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

Semisimplicity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13Matrices and Linear Maps Over Non-Commutative Rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13Semisimplicity: Three Equivalent Definitions of Semisimplicity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

Properties of Semisimple Rings/Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15Submodules and Quotients of Semisimple Modules are Semisimple (Obvious Generalization to Rings and

Algebras) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16A Ring (or Algebra) A is Semisimple iff Every (Left) A-module is Semisimple . . . . . . . . . . . . . . . . . . . . . . . . . 16

The Density Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16Jacobson’s Density Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16Burnside’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17Wedderburn’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18The Artin-Wedderburn Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

Structure Theorem For Semisimple Rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21Structure Theorem for Modules Over Semisimple Rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

Representation of Finite Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27Representation and Semisimplicity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

Equivalence of R-Algebra Representations of R[G] on an R-Module E and Representations of G on E . . . . 27Equivalence Between R[G]-Modules and (G,R)-Modules & Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27A R[G]-Linear Map of R[G]-Modules is Equivalent to a G-Homomorphism of (G,R)-Modules . . . . . . . . . . . 28G-Invariant Submodule and G-Homomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28Notation Convention . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28Representations Arising From a Given One . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

HomR(E,F ) and Dual Representations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28Tensor Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29Sums of Representations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

The Trace . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30Maschke’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

1

2 Contents

Characters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32Character of a Representation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32Effective and Irreducible Characters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

1-Dimensional Representations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35Weyl’s Trick . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37The Space of Class Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

Representations of Finite Groups

Preliminaries

Specifics on Modules

Reminder. Let G and G′ be monoids. A monoid-homomorphism f : G → G′ is a map such that f(xy) = f(x)f(y)for all x, y ∈ G which maps the identity in G to the identity in G′ (i.e., f(eG) = eG′). When G and G′ are groups, we donot need to check that f(e) = e, since this is necessarily satisfied by virtue of the existence of inverses. Indeed, if G haseven one element with an inverse, say g ∈ G, then f(e) = e since f(e) = f(gg−1) = f(g)f(g)−1 = e.

Definition (Module). Let R be a ring. A left module over R, or a left R-module, M an abelian group together withan operation of R on M such that, for all a, b ∈ R and for all m,n ∈M ,

Associativity: a(bm) = (ab)mIdentity: 1Rm = m

Distributivity: (a+ b)m = am+ bm anda(m+ n) = am+ an.

A right module over R is defined analogously (written on the right). Unless otherwise specified, we consider our modulesto be left-modules and we shall typically write 1R as simply 1. It turns out that this is equivalent to an abelian group Mand a ring-homomorphism R→ End(M), where End(M) is the ring of group-homomorphisms M →M .

Definition (Algebra). Let A be a commutative ring. An A-algebra is an A-module E together with an A-bilinear map(or, multiplication) E × E → E. This bilinear map can be viewed as a law of composition on E and it need not have anidentity or even be associative.

Remark. That A is a commutative ring is absolutely essential to the above definition. If we let E × E → E be anA-bilinear map on the A-module E, then ∀x, y ∈ E, ∀a ∈ A, (ax)y = a(xy) = x(ay) since the map is A-bilinear. Inparticular, ∀a, b ∈ A, ∀x, y ∈ E,

(ax)(by) = (ab)(xy) = (ba)(xy).

This leads to degeneracy if A is not assumed to be commutative.

Proposition. Let R be a ring and M an R-module. Then 0Rx = 0M for all x ∈M .

Proof. 0Rx+ 0Rx = (0R + 0R)x = 0Rx. Subtracting yields the desired equality. �

Corollary. Let R be a ring and M an R-module. Then (−a)x = −ax for each a ∈ R and each x ∈M .

Proof. Notice that (−1)x + x = (−1)x + (1)x = ((−1) + (1))x = (−1 + 1)x = 0x = 0 by distributivity. The general casefollows easily from this. �

Proposition. There is precisely one Z-module structure on any abelian group.

Proof. Let M be an abelian group and suppose also that M is a Z-module. Note that 1 generates Z. By distributivity,0 = (x−x) = (1x+(−1)x) for each x ∈M , 1x+1x = (1+1)x = 2x. By induction, for each n ≥ 0, x ∈M an nx =

∑nj=1 x.

Similarly, for each n ≤ 0, nx =∑nj=1(−x). This defines the module structure, obviously. �

Corollary. Every abelian group is a Z-module in precisely one way.

5

6 Preliminaries

Definition (Endomorphism). Let A be a category and A ∈ Ob(A). An endomorphism is an element of Mor(A,A).We also write End(A) in place of Mor(A,A) = Hom(A,A).

If R is a ring and we let A = RMod, the category of left R-modules, and A ∈ RMod, then we also write EndR(A)to emphasize that the maps should be R-linear, as opposed to simply being group-homomorphisms of A. Similarly forHomR(A,B). It is easy to see that EndR(A) ⊆ EndZ(A) and, moreover, that EndR(A) is the centralizer of the image ofR in End(A) defining the module structure.

Equivalent Structures

(I am indebted to George Bergmann of UC Berkeley for his “Companion to Lang’s Algebra,” from which this section takesmuch of its material.)

EndZ(G) of an Abelian Group G is a Ring

Proposition. Let M be an abelian group. Then End(M) def= EndZ(M) is a ring with addition given by the usual additionof maps and multiplication given by composition.

Proof. Let ϕ,ψ ∈ End(M). Then ϕ+ψ = ψ+ϕ since M is abelian, obviously. If ϕ ∈ End(M), the map (−ϕ) which sendsx to −ϕ(x) is a homomorphism and is the additive inverse to ϕ. The 0 map is the additive identity.

Composition of maps is naturally associative and the identity map idM is the multiplicative identity. It is clear thatϕ ◦ (ψ1 + ψ2) = ϕ ◦ ψ1 + ϕ ◦ ψ2 and that (ϕ+ ψ) ◦ γ = ϕ ◦ γ + ψ ◦ γ. �

A Left R-Module Structure on an Abelian Group M is Equivalent to aRing-Homomorphism R → End(M)

Remark. The following proposition and its corollaries is a “module-theoretic” analogue of group actions being in bijec-tion/equivalent to group homomorphism into permutation groups. It is useful to know.

Proposition. Let R be a ring and M an abelian group. A left R-module structure on M is equivalent to a ring-homomorphism R→ EndZ(M).

Proof. Suppose M is a left R-module. For each a ∈ R, define a map ϕa : M →M as follows: For each x ∈M , ϕa(x) = ax.We assert that this is an endomorphism ofM—indeed, for each x, y ∈M , ϕa(x+y) = a(x+y) = ax+ay = ϕa(x)+ϕa(y).Now define ϕ : R → End(M) by a 7→ ϕa. We assert that this is a ring-homomorphism. Notice that ϕ(ab) = ϕab and∀x ∈ M , ϕab(x) = (ab)x = a(bx) = ϕa(ϕb(x)). Hence, ϕ(ab) = ϕa ◦ ϕb. Similarly, ϕ(a + b) = ϕa+b and ∀x ∈ M ,ϕa+b = (a+ b)x = ax+ bx = ϕa(x) + ϕb(x) and hence ϕ(a+ b) = ϕa + ϕb. Thus, ϕ is a ring-homomorphism.

Conversely, suppose ϕ : R → End(M) is a ring-homomorphism. For the sake of notation convenience, denote theelement ϕ(a) by ϕa ∈ End(M). We proceed to define a left R-module structure on M . For each a ∈ R, define ax to beϕa(x). We assert that this defines a left R-module structure on M . Since ϕ is a ring-homomorphism, for all a, b ∈ R,ϕab = ϕa ◦ ϕb and ϕa+b = ϕa + ϕb. Hence, for all a, b ∈ R and for all x ∈ M , (ab)x = ϕab(x) = ϕa(ϕb(x)) = a(bx) andϕa+b(x) = (ϕa + ϕb)(x) = ϕa(x) + ϕb(x). Moreover, ϕa(x+ y) = ϕa(x) + ϕa(y) since ϕa ∈ End(M) is a homomorphism.Finally, observe that idM (x) = x for all x ∈M and that, since ϕ is a monoid-homomorphism, 1 7→ idM . This thus describesa left R-module structure on M . �

Corollary. There is a bijection between the set of left R-module structures on an abelian group M and the set of ringhomomorphisms R→ End(M).

Proof. The constructions in the above proof are reversible. That is, starting from the first construction and then using thesecond construction, we get back what we started with, and starting with the second construction and then using the first,we also get back what we started with. Hence, these yield a map with a two-sided inverse—whence the asserted bijection.

Corollary. The number of left and right R-module structures on an abelian group M is the same.

Proof. We can define right R-module structures on M by x(ϕaϕb) = ϕb(ϕa(x)). Since these are obviously in bijectionwith ring-homomorphism R→ End(M), the result follows. �

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Corollary. If k is a field and E is a non-zero k-vector space, then the ring-homomorphism k → End(E) is an embeddingand therefore k× ↪→ Aut(E).

Proof. Suppose E is a non-zero k-vector space. If αx = 0 for some α ∈ k× and x 6= 0, then x = 1 · x = (α−1α)x =α−1(αx) = 0 so x = 0. Hence, in particular, αx 6= 0 for all nonzero x so k× maps into Aut(E). Suppose now that αx = βxfor all x ∈ E. Then (α − β)x = 0 for all x ∈ E. But α − β ∈ k so if α − β 6= 0, then α − β ∈ k× so α − β defines anautomorphism, and this would be impossible from what we have seen. We are forced to conclude that α = β. Thus, thisis an embedding. �

Associative, Unital A-Algebra E Where A Commutative Equivalent to Ring E withRing-Homomorphism f : A → Z(E).

Lemma. Let A be a commutative ring. The following structures are equivalent.

(a) An associative unital A-algebra, E (i.e., an A-module E given with an associative, unital, bilinear multiplication).(b) An associative, unital ring E, together with a homomorphism f : A→ Z(E) where Z(E) denotes the center of E.

Proof. (a)⇒(b) The multiplication map, being viewed as a bilinear map of abelian groups, obviously makes E into ring,being associative and unital. Let’s call the bilinear map b : E×E → E for future reference. Let f : A→ E be the map takinga ∈ A to a1E where 1E ∈ E is the unit and where this multiplication is defined due to the A-module structure of E. Weassert that, in fact, f : A→ Z(E). Indeed, if y ∈ E, then (a1E)y = b(a1E , y) = ab(1E , y) = ab(y, 1E) = b(y, a1E) = y(a1E),since 1E is the unit element.

Conversely, given an associative unital ring E and a homomorphism f : A → Z(E), we may make E an A-moduleby defining the product of a ∈ A and x ∈ E to mean the product f(a)x in E. It is easy to verify that this makes Ean A-module such that the ring multiplication of E is A-bilinear. Thus, E becomes an A-algebra, and the conditions ofassociativity and unitality hold in this algebra because they hold in E as a ring.

Finally, it is easy to verify that if we pass from a structure of unital associative A-algebra to a structure of a ring givenwith a map of A into its center, and then back to an algebra structure in the manner we did above, we get precisely thestructure we started with; and likewise if we if we go from a structure of ring with a map of A into its center to an algebrastructure and back. Thus, the constructions we have defined are inverse to one another, and give a bijection between thetwo sorts of structure on a set E. �

Corollary. Let A be a commutative ring. The following structures are equivalent.

(a) A commutative associative unital A-algebra E.(b) A commutative associative unital ring E together with a homomorphism f : A→ E.

A Commutative Ring and M an A-Module, Then EndA(M) is an A-Module &A-Algebra

Proposition. Let A be a commutative ring and X and Y be A-modules. Then HomA(X,Y ) is an A-module.

Proof. Define af for a ∈ A and f ∈ HomA(X,Y ) to be the map such that

(af)(x) = af(x).

It is easily verified that af is A-linear and that this makes HomA(X,Y ) an A-module.If A is not commutative, then this does not make HomA(X,Y ) an A-module. Indeed, (ab)f(x) = abf(x) = af(bx) =

(af)(bx) = b(af)(x) = (ba)f(x) but ab 6= ba, necessarily. �

Corollary. Let A be a commutative ring and M an A-module. Then EndA(M) is an A-module and an A-algebra.

Proof. This is the special case thatM = X = Y , in the above proposition. It is an A-algebra under the ring-homomorphismA → EndA(M) given by a 7→ a idM . Indeed, the law of composition is A-bilinear and this is clear because each map inEndA(M) is A-linear so bilinearity is immediately seen. �

8 Preliminaries

R Commutative Ring, A an R-Algebra, Then Left A-Module M Equivalent toR-Module M With R-Algebra Homomorphism A → EndR(M)

We now turn our attention to modules over algebras.

Lemma. Let R be a commutative ring and A a possibly non-commutative unital R-algebra. The following structures areequivalent.

(a) A left A-module M—that is, an abelian group M given with a ring homomorphism A→ End(M).(b) An R-module M given with an R-algebra homomorphism A→ EndR(M).

Proof. It is clear that a structure as in (b) gives, in particular, a structure as in (a), since EndR(M) is a subring ofEnd(M), and an algebra homomorphism into the former is, in particular, a ring-homomorphism into the latter. On theother hand, given a structure as in (a), note that the map of R into Z(A), the center of A, composed with the mapA→ End(M) gives a map R→ End(M)—that is, an R-module structure on M . The action of each a ∈ A commutes withthe actions of all elements of R (why?), hence each such element in fact gives an R-module endomorphism of M ; so weget a structure as in (b).

We claim that these construction furnish two-sided inverses to each other and leave this as an exercise. �

The Group Ring or Monoid Ring

Definition (The Group Ring or Monoid Ring). Let A be a commutative ring. Let G be a monoid, written multi-plicatively. Let AG = A[G] be the set of all maps α : G→ A such that α(x) = 0 for all but finitely many x ∈ G. We defineaddition in A[G] in the obvious way for mappings into an abelian group (pointwise summation). If α, β ∈ A[G], we definetheir product αβ by the rule

(αβ)(z) =∑xy=z

α(x)β(y)

the sum taken over all pairs (x, y) with x, y ∈ G and xy = z. This sum is finite because there is only a finite number ofpairs of elements (x, y) ∈ G×G such that α(x)β(y) 6= 0. We see also that (αβ)(t) = 0 for all but finitely many t, so thatαβ ∈ A[G]. This makes A[G] a ring. Associativity follows by observing

((αβ)γ)(z) =∑xy=z

(αβ)(x)γ(y)

=∑xy=z

[∑uv=x

α(u)β(v)]γ(y)

=∑

(u,v,y)uvy=z

α(u)β(v)γ(y).

This last sum is now symmetric, and if we had computed (α(βγ))(z), we would have found this sum also. Thus, this isequal to (α(βγ))(z), proving associativity.

The unit element of A[G] is the function δ such that δ(e) = 1 and δ(x) = 0 for all x ∈ G \ {e}. It is trivial to verifythat α = δα = αδ for all α ∈ A[G].

Now let us make this clearer. Let’s adopt the convention that, if a ∈ A, x ∈ G, then a ·x (or sometimes just ax) denotesthe function whose value at x is a and 0 everywhere else. Then the unit element δ of A[G] can be written as 1 · e. Thenα ∈ A[G] =

∑x∈G α(x) · x. Then (∑

x∈Gax · x

)∑y∈G

by · y

=∑x,y

axby · xy

and ∑x∈G

ax · x+∑x∈G

bx · x =∑x∈G

(ax + bx) · x.

Then A and G embed in A[G] naturally. ϕ0 : G→ A[G] embeds G by mapping ϕ0(x) = 1 · x. f0 : A→ A[G] embeds A byf0(a) = a · e.

When G is a monoid, we call A[G] the monoid ring or monoid algebra of G over A. When G is a group, we callA[G] the group ring or group algebra when G is a group.

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Proposition. Let A be a commutative ring and let G be a group. Then the group ring A[G] is an A-algebra—whence thename group algebra.

Proof. Recall that we proved that if A is a commutative ring, then every commutative associative unital A-algebra Earises from a commutative associative unital ring E together with a ring-homomorphism f : A → E and conversely. Thenatural inclusion f0 : A ↪→ A[G] defined by sending a ∈ A to a · e—that is, f(a) = a · e—is a ring-homomorphism, clearly.This is yields the A-algebra structure we sought. �

Remark. Let us now illustrate some interesting properties of the monoid ring. In particular, we now consider the evalu-ation and reduction homomorphisms in the present context of monoids.

Theorem. Let ϕ : G→ G′ be a homomorphism of monoids and let A be a commutative ring. Then there exists a uniquehomomorphism h : A[G]→ A[G′] extending ϕ and inducing the identity on A—that is, such that h(x) = ϕ(x) for all x ∈ Gand h(a) = a for all a ∈ A.

Proof. In fact, let α =∑axx ∈ A[G] and define

h(α) =∑

axϕ(x).

Then h is immediately verified to be a homomorphism of the underlying abelian group (i.e., additive) structure of A[G]and A[G′] and that h(x) = ϕ(x) for all x ∈ G. Let’s verify that it is a monoid-homomorphism on the multiplicativestructure. Let β =

∑byy. Then

h(αβ) =∑z

(∑xy=z

axby

)ϕ(z).

Since ϕ is a group-homomorphism, ϕ(z) = ϕ(xy) = ϕ(x)ϕ(y). Hence, it is immediately seen that h(αβ) = h(α)h(β).Finally, if e ∈ G is the unit element of G, then e (i.e., 1 · e) is the unit element of A[G] and by definition, ϕ(e) = e′ wheree′ ∈ G′ (i.e., 1 · e′ ∈ A[G′]) is the unit element of G′. Hence, this is indeed the ring-homomorphism we sought. �

Theorem. Let G and G′ be monoids and let f : A → B be a ring-homomorphism of commutative rings. Then there is aunique homomorphism

h : A[G]→ A[G′]

such that

h

(∑x∈G

axx

)=∑x∈G

f(ax)x.

Proof. Since every element of A[G] has a unique expression as a sum∑axx, the formula giving h gives a well-defined

map from A[G] → B[G]. This map is obviously a homomorphism of the underlying abelian group structure of A[G] andB[G] (i.e., the additive structure). As for multiplication, let

α =∑

axx and β =∑

byy.

Then

h(αβ) =∑z∈G

f

(∑xy=z

axby

)z

=∑z∈G

∑xy=z

f(ax)f(by)z

= h(α)h(β).

We have trivially h(1) = 1 (or h(e) = e′, if we wish to write this as 1 · e and therefore associate it with the term e) so h isa ring-homomorphism, as was to be shown. �

Remark. Observe that viewing A as a subring of A[G], the restriction of h, as defined above, to A is the homomorphismf itself. In other words, if e is the unit element of G, then

h(ae) = f(a)e.

10 Preliminaries

Basic Definitions of Representation Theory

Basic Definitions

Definition (Automorphism Group). Let A be a category and let A ∈ Ob(A). We define the automorphism group,Aut(A), of A in the category A to be that collection of arrows f ∈ Mor(A,A) for which there exists g ∈ Mor(A,A)such that f ◦ g = idA and g ◦ f = idA. This is indeed a group under composition, as one can easily check.

Definition (Operation (Groups)). Let A be a category, let A ∈ Ob(A) and let G be a group. By an operation of Gon A, we shall mean a group-homomorphism of G into the group Aut(A), G→ Aut(A).

Remark. In practice, an object A is a set with elements and an automorphism in Aut(A) operates on A as a set (i.e.,induces a permutation of A). Thus, if we have a group-homomorphism

ρ : G→ Aut(A),

then for each x ∈ G, we have an automorphism ρ(x) of A which is a permutation of A.

Definition (Representation of a Group). Let A be a category. An operation of a group G on an object A ∈Ob(A)—that is, a group-homomorphism ρ : G→ Aut(A)—is also called a representation of G on A, and one then saysthat G is represented as a group of automorphisms on A.

Definition (G-Homomorphisms). Let A be a category and let A,A′ ∈ Ob(A). Suppose ρ : G → Aut(A) and ρ : G →Aut(A′) are representations. Then an element f ∈ Mor(A,A′) is said to be a G-homomorphism or a G-morphism ifis satisfies:

f(σx) = σf(x) for all x ∈ A and σ ∈ G.

Remark. It follows from this that representations of a group form a category. The construction is analogous to that ofrepresentations of ring. We shall explicate the fact that representations of a ring form a category below, in the next titledpart of this section.

Definition (Representation of a Ring). Let A and R be rings and let M be an A-module. Then M is a module overthe ring EndA(M). Let ρ : R→ EndA(M) be a ring-homomorphism. Then ρ is called a representation of R on M .

Definition (Representation of Algebras). Let k be a commutative ring and E a k-module. Then Endk(E) is a k-algebra with the obvious structure. Let R be a (associative, unital) k-algebra given by a ring-homomorphism f : k → R.By a representation of R in E, we mean a ring-homomorphism

ρ : R→ Endk(E)

making TFDC:

R Endk(E)

k

ρ

f g

In other words, ρ is a k-algebra homomorphism. Thus, a representation of R in E is simply a k-algebra homomorphismρ : R → Endk(E). In other words, a representation of R in E is a (left) R-module structure on E extending the givenk-module structure.

Observe that E may be viewed as an Endk(E)-module. Hence, E may be viewed as a (left) R-module, defining theoperation of R on E by letting

(x, v) 7→ ρ(x)v

for x ∈ R and v ∈ E. We usually write xv instead of ρ(x)v.

Representations of a Ring Form a Category

It turns out that representations of a ring form a category. (Perhaps representations of algebras do too?) We explicate thering case here.

11

Representations of a fixed ring form a category as follows. Fix a ring R. Define a morphism of a representationρ : R→ EndA(M) into a representation ρ′ : R→ EndA(M), or in other words, a homomorphism of one representationof R to another , to be an A-module homomorphism h : M →M ′ such that TFDC for all α ∈ R:

M M ′

Commutes for all α ∈ R

M M ′

h

ρ(α) ρ′(α)

h

In other words, for all α ∈ R,ρ′α ◦ h = h ◦ ρα.

In the case that h is an isomorphism, we can simplify this. We may replace the above diagram by the commutative diagram

EndA(M)

R

EndA(M ′)

[h]

ρ

ρ′

where [h] denotes conjugation by h—that is, the map defined on each f ∈ EndA(M) by [h]f = h ◦ f ◦ h−1.

Semisimplicity

Matrices and Linear Maps Over Non-Commutative Rings

Let K be a ring. We define (ϕij) with coefficients in K just as we did for commutative rings. Products of matrices aredefined by the same formula. If K is a division ring, the every non-zero K-module has a basis, and the cardinalities of twobases are equal. The argument is the same as in the case that K is a field (commutative division ring). Because of this,we call a module over a division ring a vector space.

Let R be a ring and letE = E1 ⊕ · · · ⊕ En, F = F1 ⊕ · · · ⊕ Fm

be R-modules expressed as direct sums of R-submodules. We wish to describe the most general R-homomorphism of Einto F .

Suppose first that F = F1. Letϕ : E1 ⊕ · · · ⊕ En → F

be a homomorphism. Let ϕj : Ej → F be the restriction of ϕ to the fact Ej . Every element x ∈ E has a unique expressionx = x1 + · · ·+xn, with xj ∈ Ej . We may therefore associate with x the column vector X = t(x1, . . . , xn) whose componentsare in E1, . . . , En respectively. We can associate with ϕ the row vector (ϕ1, . . . , ϕn) where ϕj ∈ HomR(Ej , F ), and theeffect of ϕ on the element x of E is described by matrix multiplication of the row vector times the column vector.

Now, more generally, consider a homomorphism

ϕ : E1 ⊕ · · · ⊕ En → F1 ⊕ · · · ⊕ Fm

and let πi : F1⊕ · · · ⊕Fm → Fi be the projection on the i-th factor. Then we can apply our previous remarks to πi ◦ϕ foreach i. In this way, we see that exist unique elements ϕij ∈ HomR(Ej , Fi), such that ϕ has a matrix representation

M(ϕ) =

ϕ11 · · · ϕ1n...

...ϕm1 · · · ϕmn

whose effect on an element x is given by matrix multiplication, namelyϕ11 · · · ϕ1n

......

ϕm1 · · · ϕmn

x1

...xn

.

Conversely, given a matrix (ϕij) with ϕij ∈ HomR(Ej , Fi), we can define an element of HomR(E,F ) by means of thismatrix. We have an additive group-isomorphism between HomR(E,F ) and this group of matrices.

In particular, let E be a fixed R-module and let K = EndR(E). Then we have a ring-isomorphism

EndR(E(n))→ Matn(K)

which to each ϕ ∈ EndR(E(n)) associates the matrix

13

14 Semisimplicityϕ11 · · · ϕ1n...

...ϕm1 · · · ϕmn

determined as before, with ϕ1j : Hom(Ej , Fj), and operating on the left on column vectors of E(n) with components in E.

Remark. Let E be a 1-dimensional vector space over a division ring D and let {v} be a basis. For each a ∈ D, thereexists a unique D-linear map

fa : E → E

such that fa(v) = av. Then we have the rulefafb = fba

Thus, when we associate a matrix with a linear map, depending on a basis, the multiplication gets twisted. Indeed, thisremark is correct. The point is that we took ϕij ∈ EndR(E), and not in D, in the special case that R = D. Thus K isnot isomorphic to D (in the non-commutative case), but anti-isomorphic. This is the only point of difference of the formalelementary theory of linear maps in the commutative or non-commutative case.

Proposition (Schur’s Lemma). Let E and F be simple R-modules. Then every non-zero homomorphism of E into Fis an isomorphism and therefore EndR(E) is a division ring.

Proof. The first part is trivial. Thus, in the case that E = F , if f 6= 0, then f has an inverse. �

Remark. The next proposition describes completely the ring of endomorphisms of a direct sum of simple modules.

Proposition. Let E = E(n1)1 ⊕ · · · ⊕ E(nr)

r be a direct sum of simple modules, the Ei being non-isomorphic, and eachEi being repeated ni times in the sum. Then the E1, . . . , Er are uniquely determined (up to permutation of indices) upto isomorphisms, and the multiplicities n1, . . . , nr are uniquely determined. The ring EndR(E) is isomorphic to a ring ofmatrices of block type

M1M2

. . .Mr

where Mi is an ni×ni matrix over EndR(Ei). (The isomorphism is the one with respect to our direct sum decomposition.)

Proof. The last statement follows from our previous considerations, taking into account the preceding proposition and theresult preceding that as well as the isomorphism types.

Suppose now that we have two R-modules with direct sum decompositions into simple submodules, and an isomorphism

E(n1)1 ⊕ · · · ⊕ E(nr)

r ≈ F (m1)1 ⊕ · · · ⊕ F (ms)

s

such that the Ei are non-isomorphic and the Fj are non-isomorphic. From the first proposition, we conclude that each Eiis isomorphic to some Fj , and conversely. It follows that r = s, and that after a permutation, Ei ≈ Fi. Furthermore, theisomorphism must induce an isomorphism

E(ni)i ≈ F (mi)

i

for each i. Since Ei ≈ Fi, we may assume WLOG that in fact Ei = Fi. Thus, we are reduced to proving: if a module isisomorphic to E(n) and to E(m), with some simple module E, then n = m. But EndR(E(n)) is isomorphic to the n × nmatrix ring over the division ring K = EndR(E). Furthermore, this isomorphism is verified at once to be an isomorphismas K-vector spaces. The dimension of Matn(K) over K is n2. This proves that the multiplicity n is uniquely determined,thereby proving our proposition. �

When E admits a finite direct sum decomposition of simple submodules, the umber of times that a simple submodule ofa given isomorphism class occurs in a decomposition will be called the multiplicity of the simple module.

Semisimplicity: Three Equivalent Definitions of Semisimplicity

Let R be a ring. Unless otherwise specified in this section all modules and homomorphisms will be R-modules and R-homomorphisms.

15

Reminder (Simple Module). Let E be an R-module. We say that E is simple if E 6= 0 and the only R-submodulesof E are 0 and E itself.

Lemma. Let E be an R-module. Suppose we can write E =∑i∈I Ei as a (not necessarily direct) sum of simple submodules.

Then there exists a subset J ⊆ I such that E =⊕

j∈J Ej.

Proof. By Zorn’s lemma, we may let J ⊆ I be a maximal subset of I such that the sum∑j∈J Ej is direct. We contend

that this sum is in fact equal to E—towards this end, it clearly suffices to prove that each Ei is contained in∑j∈J Ej .

Fix i ∈ I. Then Ei ∩ (∑j∈J Ej) ⊆ Ei; hence, Ei ∩ (

∑j∈J Ej) = 0 or Ei ∩ (

∑j∈J Ej) = Ei. If it is equal to 0, then J is

not maximal, since we can adjoin i to it, contradicting the assumption of maximality on J . Hence, Ei is contained in thesum, and our lemma is proved. �

Theorem. Let E be a (left) R-module. The following conditions on E are equivalent.

SS 1. E is the sum of a family of simple submodules.SS 2. E is the direct sum of a family of simple submodules.SS 3. Every submodule F of E is a direct summand (i.e., there exists a submodule F ′ such that E = F ⊕ F ′).

Proof. (SS 1⇒SS 2) is proved by our lemma. (SS 2⇒SS 3) Let F be a submodule. Since E =⊕

i∈I Ei is a direct sumof simple submodules, F certainly intersects sum of the Ei. If F ∩ Ei 6= {0} for all i ∈ I, then clearly F = E. Thus, byZorn’s lemma, we may let J ⊆ I be a maximal subset of I such that the sum F +

∑i∈J Ej is direct. The same reasoning

as before shows that this is equal to E.(SS 3⇒SS 1) Finally, suppose that every submodule F of E is a direct summand. To show SS 1, we shall first show

that every nonzero submodule of E contains a simple module. Let v ∈ E, v 6= 0. Then by definition, Rv is a principalsubmodule, and the kernel of the (left) R-homomorphism R→ Rv (i.e,. the annihilator of v in R) is a left-ideal of R, sayL. Hence, L is contained in a maximal left ideal M 6= R by Zorn’s lemma. Then M/L is a maximal submodule of R/Lwhich is not equal to R/L, and henceMv is a maximal submodule of Rv, which is not equal to Rv, corresponding toM/Lunder the isomorphism R/L→ Rv. Thus, we may write E = Mv ⊕M ′ for some submodule M ′ by assumption. Then

Rv = Mv ⊕ (M ′ ∩Rv)

because every element x ∈ Rv can be written uniquely as a sum x = αv + x′ with α ∈ M and x′ ∈ M ′, and x′ = x− αvlies in Rv. Since Mv is maximal in Rv, it follows that M ′ ∩Rv is simple, as desired.

Now, let E0 be the submodule of E which is the sum of all simple submodules of E. If E0 6= E, then E = E0 ⊕ F withF 6= 0, and there exists a simple submodule of F , contradicting the definition of E0. Whence SS 3 implies SS 1. �

Definition (Semisimple Module). A module satisfying our three equivalent conditions is said to be semisimple or asemisimple module.

Definition (Semisimple Ring). A ring R is called semisimple or is said to be a semisimple ring if 1 6= 0 and R issemisimple as a left module over itself.

Definition (Semisimple Algebra). An algebra A is called semisimple or is said to be a semisimple algebra if 1 6= 0and A is semisimple as a left module over itself.

Properties of Semisimple Rings/Algebras

Proposition. Every left ideal of a semisimple ring R is a subring of R and is generated by an idempotent.

Proof. Let I be a left ideal of R. First let us prove that I is principal, generated by an idempotent. Since R is semisimpleand I ⊆ R is an ideal, R = I⊕J as an R-module and we may thus write 1 = eI +eJ uniquely. Then eI must be idempotentsince eI = e2

I + eIeJ so clearly eIeJ ∈ I ∩ J = {0} so that e2I = eI . That eI generates I follows because, on the one hand,

clearly then (eI) ⊆ I and conversely if x ∈ I, then x = xeI + xeJ and therefore x = xeI so I ⊆ ReI . Now, we just sawthat eI acts as a unit element on the right. On the other hand, if x ∈ (eI), then x = 1 · x = (eI + eJ)x = eIx + eJx. Ingeneral, since IJ, JI ⊆ I ∩ J for any ideals I and J , and since I ∩ J = 0 in our case, it must be that eJx = 0. ThereforeeIx = x. Therefore I is a subring of R. �

Proposition. Let R be a semisimple ring.

(a) Whenever we write R as a direct sum of simple left ideals, the direct sum is finite.(b) R is finitely generated as an R-module.

16 Semisimplicity

(c) Every nonzero left ideal of R contains a simple left ideal of R.(d) Every nonzero left ideal of R is a sum of simple left ideals.

Proof. (a) We can write R =⊕

i∈I Li as an R-module where Li is a simple left ideal by semisimplicity. WLOG we canwrite 1 =

∑mj=1 βj with βj ∈ Ej for j = 1, . . . ,m uniquely. Then clearly R =

⊕mj=1 Rβj =

⊕mj=1 Lj . So in fact this direct

sum must have been finite.(b) This is an immediate corollary of (a).(c) Let L be a nonzero left ideal of R. WLOG suppose L is not simple. If L does not contain a simple left ideal, then

R = L⊕F where every simple left ideal lives in F . But R is the sum of simple left ideals by semisimplicity. Hence, F = R.Hence, L = 0, a contradiction.

(d) Let L be a nonzero left ideal of R which is not simple. Let (Li)i∈I be the family of all simple left ideals contained inR. Let F =

∑i∈I L. By semisimplicity, R = F ⊕ I for some left ideal I. If F 6= L, then J = L ∩ I is nontrivial. Whatever

the case, J must contain a simple left ideal, L0; but then L0 ⊆ F . This contradicts the assumption that R = F ⊕ I as adirect sum decomposition. Hence, F = L. Hence, L is a sum of simple ideals. �

Submodules and Quotients of Semisimple Modules are Semisimple (ObviousGeneralization to Rings and Algebras)

Proposition. Every submodule and quotient of a semisimple module is semisimple.

Proof. Let F be a submodule. Let F0 be the sum of all simple submodules of F . Write E = F0⊕F ′0. Every element x ∈ Fhas a unique expression x = x0 + x′0 but x′0 = x − x0 ∈ F . Hence, F is the direct sum F = F0 ⊕ (F ∩ F ′0) and we musttherefore have F0 = F , which is semisimple. As for quotients, write E = F ⊕F ′. Then F ′ is a sum of its simple submodulesand the canonical map E → E/F induces an isomorphism of F ′ onto E/F . Hence, E/F is semisimple. �

A Ring (or Algebra) A is Semisimple iff Every (Left) A-module is Semisimple

Theorem. A ring (or algebra) A is semisimple iff every (left) A-module is semisimple.

Proof. (⇒) Let E be an (left) A-module and let S be a generating set for E. Then E is a quotient of the free A-modulegenerated by S and this free A-module is obviously isomorphic to F =

⊕s∈S A. Since A is a semisimple ring, by definition

A is a semisimple module over itself. Hence, F is a direct sum of A-simple submodules, proving that F is semisimple.Since E is a quotient of F , E is semisimple by the preceding proposition. (⇐) If every A-module is semisimple, then, inparticular, A is a semisimple module over itself, which is precisely the definition of being semisimple for a ring (or algebra),as we sought to prove. �

Remark. We can really leave out the ‘(left)’ part in the statement and proof above. In the context of the foregoing result,we shall sometimes present proofs in this more general case of every A-module being semisimple.

The Density Theorem

Jacobson’s Density Theorem

Definitions (Commutant and Bicommutant). Let E be a semisimple R-module. Let R′ = R′(E) = EndR(E). ThenE is also a R’-module, the action of R′ being given by

(ϕ, x) 7→ ϕ(x)

for ϕ ∈ R′ and x ∈ E. Each α ∈ R induces a R′-homomorphism fα : E → E by the map fα(x) = αx—this clearlycommutes with R-linear maps: After all, this is what is meant by the condition

ϕ(αx) = αϕ(x).

We let R′′ = R′′(E) = EndR′(E). We call R′ = EndR(E) the commutant of R and callR′′ = EndR′(E) the bicommutantof R.

17

Remark. We can think of R′ as the subset of EndZ(E) which commutes with each element of R′. That is, it is the subsetof EndZ(E) which commutes with the maps EndR(E) (i.e., commutes with the maps which commute with R). Thus, iff ∈ R′ and g ∈ EndR(E), then f ◦ g = g ◦ f . Similarly for R′′.

Thus, we get a ring-homomorphism

R→ R′′ = EndR′(E) = R′′(E) by α 7→ fα.

It is natural to wonder how large the image can be. Jacobson’s density theorem states that it is indeed fairly large.

Lemma. Let E be semisimple over R and let F ⊆ E be any R-submodule of E. Then F is also a R′′-submodule of E.

Proof. Write E = F ⊕ F ′ as an R-module by semisimplicity. Let π : E → F be the projection map. Then π ∈ R′, andhence any element R′′ commutes with π. Therefore for any r′′ ∈ R′′, r′′F = (r′′ ◦ π)(E) = (π ◦ r′′)(E) ⊆ π[E] = N . �

Lemma. Let E be semisimple over R. Let R′ = EndR(E), f ∈ EndR′(E) as above. Let x ∈ R. There exists an elementα ∈ R such that αx = f(x).

Proof. Since E is semisimple, we can write an R-direct sum

E = Rx⊕ F.

Let π : E → Rx be the projection. Then π ∈ R′ and hence

f(x) = f(πx) = πf(x).

This shows that f(x) ∈ Rx, as desired. �

Remark. Jacobson’s density theorem generalizes this lemma by dealing with a finite number of elements of E instead ofjust one.

Theorem (Jacobson’s Density Theorem). Let E be semisimple over R, and let R′ = EndR(E). Let f ∈ EndR′(E).Let x1, . . . , xn ∈ E. Then there exists an element α ∈ R such that

αxi = f(xi) for i = 1, . . . , n.

If E is finitely generated over R′, then the natural map R→ EndR′(E) is surjective.

Proof. Suppose that E is semisimple. Then E(n) is semisimple over R, clearly. The R-action on En is the obvious diagonalaction. Let R′ = EndR(E) and R′′ = EndR′(E) act on E(n) by the diagonal action too (E(n) is a R, R′ and R′′-module).Let R′n = EndR(E(n)) and let f (n) : E(n) → E(n) be the product map:

f (n)(y1, . . . , yn) = (f(y1), . . . , f(yn)).

We know that R′n ≈ Matn(EndR(E)). Thus, since f commutes with elements of R′ in its action on E, one sees immediatelythat f (n) ∈ EndR′n(E(n)). Everything is set up so that we may now apply the lemma: By the lemma, ∃α ∈ R such that

(αx1, . . . , αxn) = (f(x1), . . . , f(xn)),

as desired. The infinite case E(I) is the same, mutatis mutandis. The notation is unwieldy, but the proof essentiallyunchanged.

We add the observation that if E is finitely generated over R′, then an element f ∈ EndR′(E) is determined by its valueon a finite number of elements of E, so the asserted surjectivity R→ EndR′(E) follows at once. (In many applications, Ewill be a FDVS over a filed k and R will be a k-algebra, so the finiteness condition is automatically satisfied.) �

Burnside’s Theorem

Reminder. If k is a field, A is a k-algebra and E is an A-module, then the action of A on E is given by a k-algebrahomomorphism A→ Endk(E).

Lemma. Let k be a field and E a nonzero k-vector space. Suppose that R ⊆ Endk(E) is a k-subspace. Then k ⊆ R′ andk ⊆ R′′.

18 Semisimplicity

Proof. Since k is a field and E a nonzero k-vector space, k certainly embeds in Endk(E) (i.e., via the k-algebra homo-morphism α 7→ α idE). As k is commutative, it follows that k is contained in the center of Endk(E) and will thereforecommute with all maps in R ⊆ Endk(E), so k ⊆ R′ = EndR(E). Therefore R′′ = EndR′(E) ⊆ Endk(E) ⊆ End(E) sincek ⊆ R′. Therefore k ⊆ R′′. �

Corollary (Burnside’s Theorem). Let k be an algebraically closed field, let E be a finite-dimensional k-vector spaceand let R be a subalgebra of Endk(E). If E is also a simple R-module, then R′ = k and therefore R = R′′(E) = Endk(E).

Proof. Since R is unital, it is a ring. Suppose E is a simple R-module. We contend that EndR(E) = k. At any rate,R′ = EndR(E) is a division ring by Schur’s lemma. We assert that R′ and R′′ contain k as a subring (meaning thatk commutes with every element of R and every element of R′). This follows by the lemma. Now, R′ = EndR(E) is ak-subspace of Endk(E) and so is finite dimensional over k because dimk E < ∞. Let α ∈ R′ and observe that k(α) is afield contained in R′ since α is either 0 or has an inverse by Schur’s lemma. Therefore k(α) is finite dimensional over kand is therefore an algebraic extension—hence, k(α) = k since k is algebraically closed and therefore α ∈ k. Since thisworks for all α ∈ R′, this proves that R′ = k; that is, that EndR(E) = k.

Now let {v1, . . . , vn} be a basis of E over k. Let A ∈ Endk(E). By the density theorem, there exists α ∈ R such that

αvi = Avi for i = 1, . . . , n.

Since the effect of A is determined by its action on a basis, we conclude that R = Endk(E). �

Remark. Notice that we did not need to assume that R was unital above (if we wished to drop that assumption fromour definition of rings).

Corollary. Let k be an algebraically closed field, let E be a finite-dimensional k-vector space and let R be a k-algebra. IfE is also a simple R-module, then the canonical k-algebra homomorphism R→ Endk(E) defining the action of R on E issurjective.

Remark. This corollary is used in the following situation (as in exercise 8 of this section in Lang). Let E be a FDVSover a field k and let G be a multiplicative submonoid of GL(E). A G-invariant subspace F of E is a subspace such thatσF ⊆ F for all σ ∈ G. We say that E is G-simple if E 6= 0 and E has no G-invariant subspace other than 0 and E itself.Let R = k[G] be the subalgebra of Endk(E) generated by G over k. Since we assumed that G is a monoid, it follows thatR consists of linear combinations ∑

aiσi

with ai ∈ k and σi ∈ G. Then we see that a subspace F of E is G-invariant iff it is k[G]-invariant. Thus, E is G-simple iffit is simple over k[G] in the sense which we have been considering. We can then restate Burnside’s theorem as he statedit:

Corollary (Burnside’s Theorem). Let E be a finite dimensional vector space over an algebraically closed field k andlet G be a multiplicative submonoid of GL(E). If E is G-simple, then k[G] = Endk(E).

Wedderburn’s Theorem

Definition (Faithful Module). Let R be a ring and E any R-module. We shall say that E is a faithful module if thefollowing condition is satisfied: If α ∈ R is such that αx = 0 for all x ∈ E, then α = 0.

Remark. In application, E is a vector space over a field k and we have a ring-homomorphism of R into Endk(E). In thisway, E is an R-module and it is faithful iff this homomorphism is injective.

Corollary (Wedderburn’s Theorem). Let R be a ring, E a simple, faithful module over R and let D = EndR(E). IfE is finite dimensional over D, then R = EndD(E) = R′′.

Proof. By Schur’s lemma, D is a division ring so that this dimensionality concept is well-defined and, moreover, it is clearthat E is a D-vector space1. Let {v1, . . . , vn} be a basis of E over D. Given A ∈ EndD(E), by the density theorem, thereexists α ∈ R such that

αvi = Avi for i = 1, . . . , n.

Hence the canonical ring-homomorphism R → EndD(E) is surjective. Our assumption that E is faithful over R impliesthat it is injective, for we would otherwise have α ∈ R such that fα = 0 ∈ EndD(E) where fα(x) = αx, which is impossiblesince E is faithful over R. �1 In the generalized sense for division rings.

19

Corollary. Let R be a finite-dimensional algebra over a field k and let E be a nonzero module over R. Let D = EndR(E).

(a) If R does not have any two-sided ideals other than 0 and R itself, then E is a faithful R-module.(b) If E is a simple R-module, then E is finite-dimensional over k and D = EndR(E) is a finite-dimensional divisionalgebra over k.(c) If E is a simple, faithful R-module, then we have an algebra representation of R as the ring of D-endomorphismof E which is an isomorphism.

Proof. (a) The kernel of the canonical k-algebra homomorphism R→ Endk(E) defining the action of R on E is a two-sidedideal which is 6= R. The first assertion is therefore clear.

(b) Suppose E is simple as an R-module. Since E is a simple R-module E 6= 0, so for some x ∈ E, the map r 7→ rx mustbe nonzero and therefore surjective, for otherwise {rx : r ∈ R} is an R-submodule of E. Suppose x ∈ E is such that r 7→ rxis onto E. Since R is finite dimensional over k, {rx : r ∈ R} is clearly finite dimensional over k—hence, dimk E <∞.

(c) Suppose E is simple as anR-module and letD = EndR(E). By Schur’s lemma,D is a division ring and is in particulara finite-dimensional k-algebra. a structure theorem for modules over algebras, E is a k-vector space with a k-algebrahomomorphism R → Endk(E). By Wedderburn’s theorem, the canonical map R → EndD(E) = EndEndR(E)(E) = R′′ isan isomorphism. Moreover, this obviously commutes with the map endowing R and EndD(E) with a k-algebra structure,as desired. �

Remark. Under the assumption that R is finite dimensional, one can find a simple module simply by taking a minimalleft ideal 6= 0. Such an deal exists merely by taking a left ideal of minimal non-zero dimension over k. An even shorterproof of Wedderburn’s theorem will be given in Rieffel’s theorem in this case.

Definition (Simple or Irreducible Representation). A simple representation or a irreducible representationis a representation (let us write this for a representation of A-algebras) ρ : R→ EndA(E) where E 6= 0 such that the onlyinvariant submodule of E is E and 0. By an invariant submodule we mean an A-submodule F of E such that RF ⊆ F .

Corollary. Let R a finite dimensional, unital algebra over an algebraically closed field k. Let V be a finite dimensionalk-vector space with an irreducible faithful representation ρ : R → Endk(V ). Then ρ is an isomorphism and thereforeR ≈ Matn(k).

Proof. The map ρ makes V a simple, faithful R-module. Since R is a ring, ρ is injective. It remains to show it surjects.Let D = EndR(V ). Obviously k ⊆ D. Note that then D is finite dimensional over k since V and R are finite dimensionalover k. Therefore V is finite dimensional over D. Thus, by Wedderburn’s theorem, R = EndD(V ). We show that D = k,completing the proof. Given α ∈ D, we note that k(α) is a commutative subfield of D, whence, a finite algebraic extensionof k, which forces k(α) = k by assumption that k is algebraically closed. As before, we conclude D = k. �

The Artin-Wedderburn Theorem

Warning. More is true. But to proceed, we shall assume knowledge of simple rings and the structure theorem. Reviewthe properties of a simple ring (i.e., in standard nomenclature, a semisimple simple ring or a (left) Artinian simple ring)in the section ahead. In particular, every module over a simple ring is faithful (otherwise we get a submodule).

Theorem (Rieffel’s Theorem). Let R be a ring with no non-trivial two sided ideals (e.g., a simple ring) and let L bea nonzero left ideal of R. Let R′(L) = EndR(L) and R′′(L) = EndR′(L)(L). Then the natural map λ : R → R′′(L) is aring-isomorphism.

Proof. If R = 0, then this is trivial, so suppose R 6= 0. Kerλ is a two-sided ideal and since Kerλ 6= 0 as R 6= 0, λ isinjective. Since LR is a two-sided ideal, LR = R and λ(L)λ(R) = λ(R). We assert that λ(L) is a left ideal of R′′. Namely,f ◦λ(x) = λ(f(x)) for any x ∈ L and f ∈ R′′, because right multiplication of L by any element y of L represents an elementof R′, and so f(xy) = f(x)y because right multiplication by y is an R-endomorphism of L, say we call it f right

y ∈ λ(L),and therefore f(xy) = f(f right

y (x)) = f righty (f(x)) = f(x)y. But this amounts to saying that λ(L) is a left ideal of R′′ since

if α ∈ L, τ ∈ R′′, then for any x ∈ L, (τ ◦ fα)(x) = τ(αx) = τ(α)x = fτ(α) ∈ λ(L). Thus

R′′ = R′′λ(R) = R′′λ(L)λ(R) = λ(L)λ(R) = λ(R),

as was to be shown. �

Lemma. If R is a simple ring and L ⊆ R be a simple left ideal of R, then L has finite dimension over the division ringD = R′(L) = EndR(L) and R ≈ EndD(L) = R′′(L).

20 Semisimplicity

Proof. The last assertion is simply Rieffel’s theorem. From what we have seen about simple rings, there is an R-moduleisomorphism ϕ :

⊕mj=1 L ≈ R for some m ∈ N. Put ϕ−1(1) = (e1, . . . , em). We assert that B = {e1, . . . , em} ⊆ L is a basis

for L overD. That B spans L overD is easy: We have a unique decomposition 1 = e1+· · ·+em. Let πi :⊕m

j=1 L→⊕m

j=1 Lbe the i-th projection (this is R-linear). Fix any x ∈ L and let fi = (ϕ ◦ πi)x for each i. Notice that fi ∈ D. Then∑fi(ei) = x, clearly. Now, we claim B is a linearly independent set. We shall write the elements of

⊕L additively

and not in component form. Since ϕ−1 is R-linear, for each i, there exists ri ∈ R such that ϕ−1(ri) = riϕ−1(1) = ei.

Hence, there exists ri such that riej = 0 if i 6= j and riei = ei (as left R-modules, we note) and therefore if fi ∈ D and∑fi(ei) = 0, then rj

∑fi(ei) =

∑fi(rjei) = fj(rjej) = fj(ej) = 0, and this holds for all j, so, in fact, fj(ej) = 0 for all

j. �

Corollary (Wedderburn’s Theorem). Let R a simple ring/algebra and let L be a simple left ideal of R. Let D = R′ =EndR(L) and R′′ = EndR′(L). Then, R ≈ R′′ = EndR′(L) = EndD(L) and therefore R ≈ Matn(EndR(L)) = Matn(D) asrings. In particular, finite dimensional matrix rings/algebras over a division ring are simple rings/algebras.

Proof. Immediate from the lemma and Rieffel’s theorem. �

Assuming we know the structure theorem for semisimple rings, we obtain what is often called the Artin-WedderburnTheorem or just sometimes just Wedderburn’s Theorem.

Corollary (Artin-Wedderburn). Let R be a semisimple ring. Then R is ring-isomorphic to a product of matrix algebrasover division rings.

Proof. By the structure theorem, we can write R as a direct product of simple rings. From here we apply the abovecorollary. �

Remark. The Artin-Wedderburn theorem holds when R is a semisimple algebra over a commutative ring A. Let f : A→Z(R) ⊆ R be the ring-homomorphism inducing the A-algebra structure on R. Observe that (for any ideal L of R),D = R′ = EndR(L) is a Z(R)-algebra (since the these elements commute). Therefore D is made an A-algebra via thehomomorphism f—namely, aϕ def= f(a)ϕ. Hence, EndR(L) is an A-algebra.

Lemma. If R is an A-algebra, then for any R-module E, then R′ = R′(E) = EndR(E) and R′′ = R′′(E) = EndR′(E)are A-algebras.

Corollary. Let A be a commutative ring and let R be a semisimple A-algebra. Then R is isomorphic as an A-algebra toa product of matrix A-algebras over division rings.

Corollary. Let k be an algebraically closed field and let R be a finite dimensional semisimple k-algebra. Then R isisomorphic as a k-algebra to a product of matrix rings over k. In particular, each simple ring Ri in the direct productdecomposition of R is a k-subalgebra of R and

Ri ≈ Endk(Li) ≈ Matdimk(Li)(k) and therefore dimk(Ri) = (dimk(Li))2.

Proof. If R = 0 this is trivial, so suppose R 6= 0. This is really a consequence of one of the corollaries after the firstWedderburn’s theorem since every non-zero representation of a simple ring is faithful (it has no proper two-sided ideals),but we prove it here anyways. The idea is the same as before. R =

∏Ri a product of a finitely many simple rings

(which are two-sided ideals of R) which are k-algebras themselves. They obviously have finite dimension over k. Thus byArtin-Wedderburn, Ri ≈ Matni

(EndRi(Li)) ≈ EndR′

i(Li) for some simple left ideal Li. Fix L = Li.

Then L has finite dimension over k since R does. But EndRi(L) ⊆ Endk(L) since the ring-homomorphism definingthe k-algebra structure on Ri is nonzero and k is a field, it must be injective, whence the inclusion. But then EndRi(L)is a finite dimensional k-subalgebra of Endk(L) and therefore since L is a simple Ri-module, by Burnside’s theorem,R′i = EndRi

(L) = Endk(L). Hence, in fact, Ri ≈ Matni(Endk(L)). By Schur’s lemma, D = Endk(L) is a division ring.

Observe that k ⊆ Endk(L) since k is commutative. Since L has finite dimension over k, Endk(L) has finite dimension overk (it is isomorphic to the obvious matrix ring). But then [D : k] <∞ is a finite extension and therefore D/k is algebraic.It must be that D = k, then, since k is algebraically closed and therefore R′′i (L) = EndR′

i(L) = Endk(L). Therefore,

in fact, Ri ≈ Matni(k). Finally, since Ri is simple, Ri the canonical map Ri → R′′i (L) must be injective (it is certainly

nonzero so it must have kernel (0)). Another application of Burnside’s theorem tells us that this map surjects and thereforeRi ≈ Endk(Li). The final isomorphism Ri ≈ Endk(Li) ≈ Matdimk(Li)(k) is now clear. �

Corollary. Let k be an algebraically closed field and let R be a finite dimensional commutative k-algebra, say dimk R = n.Then R ≈

∏ni=1 k as rings.

Proof. Matn(k) is commutative iff n = 1. Since R ≈ kn as a k-vector space, it must also be that R ≈ kn as a ring. �

21

Lemma. If R is (left/right) Noetherian (resp. Artinian), then Matn(R) is (left/right) Noetherian (resp. Artinian).

Proof. Let ϕ : R → Matn(R) by r 7→ rIn and let Eij be the matrix with 1 in the (i, j)-th component and 0 everywhereelse. Therefore Matn(R) is finitely generated by these matrices

Corollary. If R is a semisimple ring/algebra, then R is Artinian (i.e., satisfies the descending chain condition on ideals).

Proof. R is isomorphic to a finite product of matrix rings/algebras∏ni=1 Matni

(Di) for division rings Di and integersni ≥ 1 where each is a simple ring. Each Matni

(Di) is finitely generated over Di by matrices with 1 in one location and0 everywhere else. Clearly Di is Artinian and even Noetherian, being a division ring. Thus, since Matni

(Di) is finitelygenerated over Di, we claim that it is Artinian over Di. If this is true, then it must be Artinian over itself since anydescending chain of ideals is certainly a descending chain of Di-submodules. Towards this end, notice that we have asurjective map D

(ni)2

i → Matni(Di). Clearly D(ni)2

i is Artinian over Di as a finite direct sum. Therefore Matni(Di) isArtinian by the same reasoning with which one proves that a quotient of a Noetherian ring is Noetherian by lifting. �

Structure Theorem For Semisimple Rings

Remarks. Let R be a ring. A left ideal of a R is an R-module, and is thus called simple if it is simple as a module. Twoideals L and L′ are called isomorphic if they are isomorphic as modules. Thus, if L is the set of all simple left ideals ofR, then we can define an equivalence relation ∼ on L by L1 ∼ L2 iff L1 and L2 are isomorphic. We thereby may talk ofa system of distinct representatives of simple left ideals of a ring R.

Suppose now that R is a semisimple ring. We shall now decompose R as a sum of simple left ideals, and thereby get astructure theorem for R.

Let {Li}i∈I be a system of distinct representatives of simple left ideals of the semisimple ring R. We also say that thisfamily is a family of representatives for the isomorphism classes of simple left ideals. Indeed, if L is a left ideal and L ≈ Lifor some i ∈ I, then L ∈ [Li].

Lemma. Let L be a simple left ideal of a ring R and let E be a simple R-module. If L is not isomorphic to E as R-modules,then LE = 0.

Proof. We prove the contrapositive: If LE 6= 0, then L ≈ E as R-modules. First, we notice that RLE = LE so thatLE is a submodule of E. Hence, LE = E or LE = 0 since E is simple. Thus, we may suppose that LE = E since weassumed LE 6= 0. Let y ∈ E be such that Ly 6= 0. Since Ly is a submodule of E, it follows that Ly = E. Thus, the mapof f : L → E given by α 7→ αy is surjective, and hence nonzero—moreover, f is clearly R-linear. Since L is simple, thishomomorphism must have either trivial kernel or kernel equal to L—since the map itself is nontrivial, we are forced toconclude that f is in fact a R-linear isomorphism. �

Corollary. Let R be a ring and let {Li}i∈I be a system of distinct representatives of simple left ideals of the ring R. Foreach i ∈ I, let

Ri =∑L≈Li

left ideal

Ldef= additive closure of

⋃L≈Li

left ideal

L

be the sum of all the simple left ideals isomorphic to Li. Then,

(a) For each i ∈ I, Ri is a left ideal.(b) For each i, j ∈ I, if i 6= j, then RiRj = 0 if i 6= j.

Proof. (a) That Ri is a left ideal is easy since Ri is simply the left ideal generated by⋃{L ≈ Li : L a left ideal of R}.

(b) Now, suppose i, j ∈ I but i 6= j. Then RiRj is, by definition, the additive closure of the set of products{rirj : ri ∈ Ri and rj ∈ Rj} of elements of Ri and Rj . But then ri = x1 + · · · + xr and rj = y1 + · · · + yk so thatrirj is simply a sum of pairwise products xiyj . But xi ∈ L ≈ Li and yj ∈ L′ ≈ Lj and both of these are simple R-modulesso that, by the lemma, xiyj = 0 for all i, j. Hence, rirj = 0. Hence, RiRj = 0, as desired. We might also see this since

RiRj = (∑L≈Li

left ideal

L) (∑L′≈Lj

left ideal

L′) =∑L≈Li

L′≈Lj

left ideals

LL′,

which, again, by the lemma gives us what we want. �

22 Semisimplicity

Notation. Let R be a ring and let L = {Li}i∈I be a system of distinct representatives of simple left ideals of the ringR. Throughout, for each i ∈ I, we shall denote Ri to be the sum of all simple left ideals of R isomorphic to Li.

Corollary. Let R be a semisimple ring and let L = {Li}i∈I be a system of distinct representatives of simple left idealsof the ring R. Then R =

∑i∈I Ri and for each j ∈ I, Rj is a two-sided ideal of R.

Proof. Since R is semisimple, it is, by definition, semisimple as a left R-module over itself. Hence, R (as a left R-module)is the sum of a family of simple submodules (i.e., simple left ideals). Suppose R =

∑j L′j for some family {L′j}. Then each

L′j appears in some Ri, soR =

∑j

Lj ⊆∑

Ri ⊆ R

so that we are forced to conclude that∑Ri = R, as desired. Finally, note that for any j ∈ I

Rj ⊆ RjR = RjRj ⊆ Rj ,

the first inclusion because 0 6= 1 ∈ R, the equality holding because R =∑Ri so that Rj

∑Ri =

∑RjRi = RiRi by the

corollary and the last inclusion holding because Rj is a left ideal. Thus, Rj is also a right ideal and is thus a two-sidedideal. �

Corollary. Let R be a semisimple ring. Then the unit element 1 ∈ R can be expressed as a finite sum

1 =∑i∈I

ei

where ei ∈ Ri with all but finitely many ei = 0.

Proof. Since R is semisimple, R =∑Ri. Hence, 1 ∈

∑Ri. Thus, 1 is a finite sum as claimed. �

Definition (Simple or Simple Artinian Ring). A ring R is said to be a simple Artinian ring or (as Lang puts it)a simple ring if it is semisimple and if it has only one isomorphism class of simple left ideals.

Let’s prove a few properties of such rings before proceeding.

Lemma. Let R be a ring and let ψ ∈ EndR(R) be a homomorphism of R into itself viewed as a left R-module. Then∃α ∈ R such that ψ(x) = xα for all x ∈ R.

Proof. We have ψ(x) = ψ(x · 1) = xψ(1). Let α = ψ(1). �

Proposition. Let R be any simple ring.

(a) As an R-module over itself, R is a finite direct sum of its simple left ideals. In particular, for simple left ideal L,for some m ∈ N, R ≈

⊕mi=1 L.

(b) R has no two-sided ideals other than 0 and the ring itself.

Proof. (a) Let R be a simple ring. Then we can write R =⊕

i∈I Li as an R-module where Li is a simple left idealby semisimplicity. Then we can write 1 =

∑mj=1 βj (WLOG) with βj ∈ Lj for j = 1, . . . ,m uniquely. Then clearly

R =⊕m

j=1 Rβj =⊕m

j=1 Lj . So in fact this direct sum must have been finite. Now, since every simple left ideal of R isisomorphic to each other, the last assertion follows easily.

(b) Let R be a simple ring. Then, in particular, R is semisimple. Let I be a two-sided ideal of R. Suppose first that Idoes not contain a simple left ideal. Then L∩I = 0 for all simple left ideals of R and therefore by semisimplicity, R = I⊕Jas (left) R-modules and hence all simple left ideals of R live in J . But this forces J = R since R is the sum of all simpleleft ideals and hence I = 0. Suppose that I 6= 0. Then I contains a simple left ideal, say L. To prove that I = R, it sufficesto prove LR = R, and towards this end it suffices to prove that for each simple left ideal M of R, there exists α ∈ R suchthat Lα = M . Notice that LR is a left ideal. Thus, R being semisimple, LR is semisimple and so is a direct sum of simpleleft ideals, say

LR =m⊕j=1

Lj , L = L1

(m is finite due to the structure theorem we shall prove below). We also have a direct sum decomposition R = L⊕L′. Letπ : R→ L be the projection which is therefore an R-endomorphism. Let σ : L→M be an isomorphism. Then σ◦π : R→ Ris an R-endomorphism and thus by the lemma, ∃α ∈ R such that σ(π(x)) = xα for all x ∈ R. Thus, for each elementx ∈ L, σ(x) = xα. The map x 7→ xα is a R-linear map of L into M and is non-zero. Hence, it is an isomorphism. Fromthis, it follows that LR = R since R =

∑L and thus our assertion. �

23

Reminder. A subring of a ring R contains the multiplicative identity of R. However, there can exist a ring S which is asubset of R, but which is not a subring of R. This is because the multiplicative identity of S may not be equal to 1 ∈ R.We shall identify such sets as rings contained in R, but will not identify them as subrings.

Proposition. Let R be a semisimple ring and let {Li}i∈I be a system of distinct representatives for the simple left idealsof R. Write 1 = e1 + · · ·+ es where ei ∈ Ri \ {0} for i = 1, . . . , s. Ri1 , . . . , Ris . Then

(a) R = Ri1 + · · ·+Ris .(b) If ij 6= ik, then Rij ∩Rik = RijRik = 0 and therefore the sum R = Ri1 + · · ·+Ris is direct:

R =s⊕j=1

Rij as a left R-module.

(c) The presentation 1 = e1 + · · ·+ es exists and is unique.(d) For Rt with t 6= i1, . . . , is, Rt = 0. Therefore there are only finitely many non-isomorphic simple left ideals.(e) For each k = 1, . . . , s, Rik is itself a ring, contained in R, with unit element ek.(f) Each (nonzero) Ri is a simple ring.(g) R is finitely generated by the e1, . . . , en.

Proof. (a) That 1 has such a representation follows from the above corollary. WLOG suppose ik = k for each k = 1, . . . , s.Now, for every x ∈ R, x = 1 · x = e1x + · · · esx. Since ek ∈ Rk and each Rk is a two-sided ideal, ekx ∈ Rk. Thus, everyelement is a sum of elements of R1, . . . , Rs. Hence, R = R1 + · · ·+Rs.

(b) Since R =∑i∈I Ri, for any x ∈ R, we may write

x =∑i∈I

xi, xi ∈ Ri with all but finitely many xi = 0.

By distributivity and the fact that RiRj = 0 if i 6= j, for j = 1, . . . , s we have

ejx = ej∑i∈I

xj = ejxj

and by similar reasoning thatxj = 1 · xj = e1xj + · · ·+ esxj = ejxj .

Therefore ejx = ejxj = xj for each j = 1, . . . , s. Thus, given any presentation of x ∈ R as x =∑i∈I xi where xi ∈ Ri,

we have that ejx = ej∑i∈I xi = ejxj = xj for j = 1, . . . , s. Thus, given any other such presentation x =

∑i∈I yi,

ejx = yj = xj . Thus, every element x ∈ R has a unique presentation in terms of the elements xj = ejx ∈ Rj forj = 1, . . . , s. This uniqueness forces for j 6= k, j, k ∈ {1, . . . , s} that Rj ∩Rk = 0. Therefore, R =

⊕si=1 Ri as an R-module.

(c) Since R =⊕s

i=1 Ri as a left R-module, the presentation for 1 we gave (which we know exists since R is semisimple)is unique. This is basically a corollary of (b).

(d) Let y ∈ Rt. Then y = 1 · y = (e1 + · · ·+ es)y = e1y + · · ·+ esy = 0. Hence, Rt = 0. Since this holds for all such tnot belonging to our distinguished set 1, . . . , s, there are only finitely many isomorphism classes of simple left ideals of R.

(e) Clearly Rj = (ej) for j = 1, . . . , s. Now, we have seen for any xj ∈ Rj , ejxj = xj . On the other hand, similarreasoning as above shows that xj = xj · 1 = xje1 + · · ·+ xjes = xjej . Therefore ej serves as the unit element. Thereforeeach Rj is a subring of R.

(f) Since Ri is the sum of simple submodules (i.e., ideals) it is SS 1 semisimple. Clearly Ri 6= 0. Let a ⊆ Ri 6= 0 bea simple left ideal of Ri. We assert that a ≈ Li. In other words, Ri is a simple ring with the peculiar isomorphism classbeing that of Li. By the first lemma, if a is not isomorphic to Li, then aLi = 0. By contraposition, if aLi 6= 0—hence,since Li is simple, RiaLi = aLi is a submodule so must then be equal to Li—then I ≈ Li. Since a 6= 0 and since Ria = a,there exists r ∈ Ri such that ria 6= 0. By definition of the Ri, r ∈ L for some left ideal L ≈ Li. But then L is simple too,and aL 6= 0. Hence, aL = L so that a ≈ L as Ri-modules Therefore a ≈ Li. Indeed, explicitly, let ϕ : L ≈ Li be Ri-linearand observe that Li = ϕ(aL) = aϕ(L) = aLi, as desired. Therefore Ri is simple.

(g) Finally, it is clear that R is finitely generated. �

Corollary. Let R be a semisimple ring and suppose R =⊕s

i=1 Ri as an R-module. Then, in fact, R is the internal directproduct of the Ri with

R ≈s∏i=1

Ri.

Proof. We have seen that we may write 1 uniquely for a semisimple ring R in terms of the Ri. Since RiRj = 0 for i 6= j,it follows that

24 Semisimplicity

R ≈s∏i=1

Ri,

as desired. �

Remark. Indeed, what we have shown amounts to the following: Every semisimple ring R is the internal direct productof some ideals Ri. Let us summarize this here.

Theorem. Let R be a semisimple ring. Then there are only a finite number of non-isomorphic simple left ideals, sayL1, . . . , Ls. If

Ri =∑L≈Li

left ideal

L

is the sum of all simple left ideals isomorphic to Li, then Ri is a two-sided ideal which is also a ring2 (the operations beingthose induced by R) and R is the internal direct product of the Ri:

R ≈s∏i=1

Ri.

Each Ri is a simple ring and if ei is its unit element, then 1 = e1 + · · ·+ es and Ri = Rei = (ei). Furthermore, eiej = 0if i 6= j.

Remark. Notice that each Ri is simple and therefore semisimple. We have seen that semisimple rings are finite directsums and that simple rings are finite direct sums of their one isomorphism class of simple left ideal.

Structure Theorem for Modules Over Semisimple Rings

Theorem. Let R be a semisimple ring and let L1, . . . , Ls be representatives of each isomorphism class of simple left idealsof R. Let E be a nonzero R-module. As usual, denote Ri = (ei) the two-sided ideal

Ri =∑L≈Li

left ideal

L = (ei) = Rei

whose unit element is ei. For each i, let Ei denote the submodule of E consisting of the sum of all simple modules of Eisomorphic to Li.

(a) For each i = 1, . . . , s, RiE = Ei.(b) For each i = 1, . . . , s, RiE = eiE.(c) For each i = 1, . . . , s, RiE = Ei is a semisimple submodule of E.(d) We have the direct sum decomposition

E =s⊕i=1

RiE =s⊕i=1

eiE.

Proof. (a) For each i = 1, . . . , s, letEi =

∑F submodule of E

F≈Li

F.

If V is a simple submodule of E, then RV = V , and hence, clearly, for some i, LiV 6= 0 so that LiV = V . For such i,then V ≈ Li as R-modules. Therefore E1, . . . , Es capture each possible isomorphism class of simple submodules of E andeach simple submodule V of E belongs to LiE ⊆ RiE for some i. Therefore Ei ⊆ RiE for each i. On the other hand, ifx ∈ RiE, then x is a finite sum x = `1x1 + · · ·+ `nxn where xi ∈ E and the `i belong to possibly distinct simple left idealsof R isomorphic to Li. So clearly x ∈ E and therefore Ei = RiE, as desired.

(b) Since (ei) is a two sided ideal, if x ∈ E, r ∈ R, then reix = eir′x for some r′ ∈ R. Hence, RiE = eiE. (c) By

(a), each RiE is the sum of a family of simple submodules of E. Clearly RiE 6= 0. Therefore RiE is semisimple. (d)Since E is a module over semisimple ring, it is semisimple, and is therefore the sum of simple submodules. ThereforeR1E + · · ·+RsE = E. It remains to show that this sum is direct. Let x ∈ RiE ∩RjE for i 6= j. We have x = 1 · x = eix

2 Again, not necessarily a subring since this would imply Ri = R as 1 ∈ Ri.

25

since RiRj = 0. But then ei ∈ Ri and x ∈ RjE so x = rjy for some rj ∈ Rj , y ∈ E. Hence, eirjy = 0. Thus, x = 0. Thesum is therefore direct, as asserted. Finally, since (ei) is a two sided ideal, if x ∈ E, r ∈ R, then reix = eir

′x for somer′ ∈ R. Hence, RiE = eiE. �

Remark. Let k be a field and let G be a finite group. Let E be k[G]-module and suppose dimk(E) < ∞. If k[G] issemisimple (e.g., characteristic of k does not divide #(G)), then each RiE is the direct sum of finitely many isomorphiccopies of its one isomorphism class of submodule, furnishing a further decomposition.

Corollary. Let R be a semisimple ring and let L1, . . . , Ls be representatives of each isomorphism class of simple leftideals of R. . Then every simple module is isomorphic to one of the simple left ideals L1, . . . , Ls.

Corollary. A simple ring has exactly one simple module up to isomorphism.

Proposition. Let k be a field and E a finite dimensional k-vector space. Let S ⊆ Endk(E) be a subring3 and let R be thek-algebra generated by the elements of S. Then R is semisimple iff E is a semisimple R (or S)-module (with the structuregiven in the obvious way).

Proof. (⇒) If R is semisimple, then E is semisimple as we know. (⇐) Suppose E is a semisimple S-module. ThenE is semisimple as an R-module because if we F is a simple S-submodule, then its closure under k is a simple R-submodule—indeed, let F̃ =

⋃x∈F kx. Then clearly F̃ is an R-submodule and is obviously simple (complete this proof).

Therefore

E =n⊕i=1

Ei

where each Ei is simple. Then for each i, ∃vi ∈ Ei, Ei = Rvi. The map

x 7→ (xv1, . . . , xvn)

is an R-homomorphism of R into E, and is an injection since R ⊆ Endk(E). Since a submodule of a semisimple moduleis semisimple, we’re done. �

3 Lang says subset.

Representation of Finite Groups

Representation and Semisimplicity

Throughout this section, if it is not specified, we shall assume that the ring R is commutative. We will work in the categoryof modules over R, Rmod.

Equivalence of R-Algebra Representations of R[G] on an R-Module E andRepresentations of G on E

Proposition. Let R be a commutative ring, G a group and E an R-module. Then a representation of the algebra R[G]on E is equivalent to a representation of G on E (i.e., a bijection).

Proof. (⇒) In fact, every ring-homomorphism R[G]→ EndR(E) induces a group-homomorphism G→ AutR(E) and thusa representation of the ring R[G] in E. To see this, note that we have a canonical embedding ϕ0 : G→ R[G]. Hence, if theimage of g ∈ G in EndR(E) is not trivial (i.e., idR), then it has an inverse.

Now, conversely, suppose ρ : G → AutR(E) is a representation of G. We extend ρ to a representation of R[G] as anR-algebra on E (i.e., ρ : R[G]→ EndR(E) with R 3 r · e 7→ r · idE) as follows: Let α =

∑aσσ and for each x ∈ E define

ρ(α)x def=∑

aσρ(σ)x.

It is immediately verified that ρ has been extend to a ring-homomorphism of R[G] into EndR(E) since ρ(σ) is an R-linearmap. Now, in fact, our extension is an R-algebra homomorphism with the desired property, we assert. We first observethat since R is commutative, EndR(E) is an R-algebra. Now, for each x ∈ E,

ρ(rα)x = ρ(r(∑

aσσ))x =∑

raσρ(σ)x = r∑

aσρ(σ)x = rρ(∑

aσσ)x = rρ(α)x

proving that it is in fact an R-algebra homomorphism. Indeed, this means that ρ(re)x = rρ(e)x = r ·x so that r ·e 7→ r · idEand thus extends the R-algebra structure on EndR(E).

Now, given an R-algebra representation ρ : R[G]→ EndR(E), it follows that ρ(r ·e) = r · idE and that ρ(r ·g) = rρ(1 ·g)and this says that the R-algebra representation is uniquely determined by the way G operates on E. Conversely, if weextend ρ : G → AutR(E) to R[G] as above and we have another such extension ρ′ to an R[G] representation on E, thenclearly they are the same so that these two processes are obviously inverse to one another. �

Equivalence Between R[G]-Modules and (G, R)-Modules & Definitions

Definition (Faithful Representation). We say a representation ρ is faithful if it is injective.

Definition (G-Module, G-Space, (G,R)-Module). Let G be a group and R a commutative ring. An R-module Etogether with a representation ρ : G → AutR(E) will be called a G-module or G-space, or also a (G,R)-module if wewish to specify the ring R. In other words, a (G,R)-module is a pair (E, ρ : G→ AutR(E)) where E is an R-module.

Remark. Now we have the terminology to say something even more extraordinary: The R[G]-module structure isequivalent to a (G,R)-module structure.

27

28 Representation of Finite Groups

Let E be a R[G]-module. Then E is an R-module and we automatically get the G-module structure as above. Conversely,if E is a (G,R)-module, we may extend ρ : G → AutR(E) to ρ : R[G] → EndR(E) as above. But then ρ gives E a R[G]-module structure. It is clear that these processes define an equivalence.

A R[G]-Linear Map of R[G]-Modules is Equivalent to a G-Homomorphism of(G, R)-Modules

Corollary. Let R be a commutative ring, G a group and let E and F be R-modules. A G-homomorphism f : E → F as(G,R)-modules is equivalent to an R[G]-linear map f : E → F as R[G]-modules.

Proof. (⇒) E and F are (G,R)-modules, so in particular they are R[G]-modules due to our equivalence theorem. But thenf((r · σ)x) = f((r · e)(1 · σ)x) = (r · e)f((1 · σ)x) = (1 · σ)(r · e)f(x) = (r · σ)f(x), so f is R[G]-linear. (⇐) An R[G]-linearmap of R-modules behaves in the way we want to on R ⊆ R[G]. Hence, the structure on E (resp. F ) is equivalent to anR-module together with a map R[G]→ EndR(E) (resp. F ). Restricted to G, the image must be invertible, so we turn Eand F into (G,R)-modules. Finally, it is clear that f is a G-homomorphism. �

Remark. All this adds up to saying that a R[G]-module is simply a (G,R)-module and conversely.

G-Invariant Submodule and G-Homomorphisms

Definition (G-Invariant Submodule). Let E be a (G,R)-module with representation ρ : G→ AutR(E). We define theG-invariant submodule of E to be

invG(E) def= {x ∈ E : ∀σ ∈ G, σx = x} .

For, if r ∈ R, then rσx = σ(rx), so clearly this is an R-submodule of E.

Reminder. Let E and F be (G,R)-modules. Recall that a G-homomorphism f : E → F is an element f ∈ Mor(E,F )(i.e., an R-linear map) such that f(σx) = σf(x) for all x ∈ E and σ ∈ G.

Proposition. Given a G-homomorphism f : E → F , Ker f is a G-submodule of E and the cokernel F/f [E] admits anoperation of G in a unique way such that the canonical map ϕ : F → F/f [E] is a G-homomorphism.

Proof. Ker f is a G-submodule of E since we have the pair (Ker f, ρ|Ker f), we’re good. Let ρ′ : G → AutR(F ) be theoperation on F . Then, to obtain an operation of G on the cokernel, simply factor G→ AutR(F )→ AutR(F/f [E]) by wayof ρ′ and composition with the canonical projection ϕ. �

Remark. We typically make R into a G-module by making G act trivially on R (i.e., G→ AutR(R) is trivial).

Notation Convention

Given a representation ρ of G on E, we often write simply σx or ρσx instead of ρ(σ)x, whenever we deal with a fixedrepresentation throughout a discussion.

Representations Arising From a Given One

HomR(E, F ) and Dual Representations

Let E and F be (G,R)-modules. We make HomR(E,F ) a G-module under the action defined for f ∈ HomR(E,F ) by

σf is the map defined on each x ∈ E by ([σ]f)(x) def= σf(σ−1x).

That is, if ρ1 : G→ AutR(E) and ρ2 : G→ AutR(F ) are the associated representations, then

([σ]f)(x) def= ρ2(σ)(f(ρ1(σ−1)x)

).

29

The resulting homomorphism ρ : G→ AutR(HomR(E,F )) is then simply

σ 7→ (f 7→ ρ2(σ) ◦ f ◦ ρ1(σ−1))

(i.e., σ ∈ G maps to the mapping on the RHS). It is clear that this is a group-homomorphism and hence defines anoperation of G on HomR(E,F ). We could also have checked that this defines an operation of G on HomR(E,F ) by bruteforce: ([e]f)(x) = ef(e−1x) = ef(ex) = f(x) and ([σ1σ2]f)(x) = σ1σ2f(σ−1

2 σ−11 x) = ([σ1]([σ2]f))(x), giving associativity.

We are particularly concerned with the case where F = R (so with the trivial action σr = r for all r ∈ R), in whichcase HomR(E,R) = E∨ is the dual module.

Definition (Dual (or Contragredient) Representation). Let R be a commutative ring, let G be a group and let Ebe an R-module. Let ρ : G→ AutR(E) be a representation of G on E. Then we have a representation

ρ∨ : G→ AutR(E∨) given by the action σ 7→ (f 7→ f ◦ ρσ−1)

that is, the operation of σ ∈ G on E∨ is defined for f ∈ HomR(E,R) = E∨ precisely as we did above:

([σ]f)(x) = σf(σ−1x) = f(σ−1x)

with this equality holding because G acts trivially on R. The representation ρ∨ is called the dual representation or thecontragredient representation.

Reminders. Let R be a commutative ring and let E and F be a free R-module of finite rank n and m respectively.

(a) Fix a basis B = {x1, . . . , xn}. Then the dual basis B∨ of B in E∨ is the basis {x̂1, . . . , x̂n} where x̂k : E → R isthe R-linear map x̂k(xj) = δkj .(b) Let L : E → F be a R-linear homomorphism. Then the transpose (or dual or adjoint) of L is defined to be theR-linear map tL : F∨ → E∨ given by tL(f) = f ◦ L.(c) (Transposition and Inversion Commute) Suppose n = m and suppose L : E → F is a R-linear isomorphism ofE with F . Then tL : F∨ → E∨ is an isomorphism. Moreover, (tL)−1 = t(L−1). This follows because tL−1 : E∨ → F∨

has tL−1(f) = f ◦ L−1 whereas (tL)(f ◦ L−1) = f ◦ L−1 ◦ L = f , so that (tL)−1(f) = f ◦ L−1 so that tL−1 = (tL)−1.

Proposition. Let E be a free R-module of finite rank and let ρ : G → AutR(E) be a representation of G on E; letρ∨ : G→ E∨ be its dual representation. Fix a basis B of E. Then for each σ ∈ G,

MB∨

B∨ (ρ∨σ ) = t(MB

B (ρσ))−1 = tMB

B (ρσ−1). (0.1)

Tensor Products

Let E and E′ be (G,R)-modules. We can form their tensor product E⊗RE′. Then there is a unique action of G on E⊗E′such that for each σ ∈ G, we have

σ(x⊗ x′) = σx⊗ σx′.

Suppose E,F are free R-modules of finite rank. Then the R-isomorphism

E∨ ⊗ F ≈ HomR(E,F ) given by (λ, f) 7→ λ(•)f : E → F. (0.2)

is immediately verified to be a G-isomorphism.Whether or not E and F have finite rank, when E and F are free R-modules, we have an R-isomorphism (not necessarily

a G-isomorphism)invG(E∨ ⊗ F ) ≈ HomG(E,F ) (0.3)

between the elements of invG(E∨ ⊗ F ) and the hom-set of G-homomorphism E → F which is an R-module, clearly.

Sums of Representations

Definition (Sums of Representations). Let R be a commutative ring, G be a group and let E and E′ be R-modules.Let ρ : G → AutR(E) and ρ′ : G → AutR(E′) be representations of G on E and E′, respectively. Then we define theirsum ρ⊕ ρ′ to be the representation on the direct sum E ⊕ E′ with σ ∈ G acting componentwise.

30 Representation of Finite Groups

The Trace

Proposition. Let R be a commutative ring, G be a group. The G-isomorphism classes of representations have an additivemonoid structure under the sum and an associative multiplicative structure under the tensor product which distributivewith respect to the addition (direct sum).

Proof. Let ρ : G → AutRE and ρ′ : G → AutRE′ be two representations. We define their sum as above. We define theirproduct to be

ρ⊗ ρ′ : G→ AutR(E ⊗ E′) via σ 7→ ρσ ⊗ ρ′σ : E ⊗ E′ → E ⊗ E′.

We have seen that the tensor product distributes over the direct sum, so the conclusion holds. �

Definition (Trace). Let R be a commutative ring, G be a finite group and E a (G,R)-module with representationρ : G→ AutR(E). Then we can define the trace TrG : E → E which is an R-homomorphism given by

TrG(x) =∑σ∈G

ρσx =∑σ∈G

σx.

In particular, as shall prove,TrG : E → invG(E).

In fact, given (G,R)-modules E and F , we can define TrG : HomR(E,F ) → Hom(G,R)(E,F ) (i.e., TrG takes R-linearmaps to G-homomorphisms in the category of R-modules). We shall see/define these properties below.

Proposition. Let G be a finite group and E a (G,R)-module. Then TrG[E] ⊆ invG(E)—that is, ∀x ∈ E, TrG(x) ∈invG(E).

Proof. Let τ ∈ G. Then for each x ∈ E,τ TrG(x) =

∑σ∈G

τσx

and the left regular action of G on itself is a permutation. Hence, τ TrG(x) = TrG(x). �

Proposition. Let G be a finite group and let f : E → F be an R-homomorphism of G-modules. Then

TrG(f) def= TrG ◦f : E → F which acts on each x ∈ E by TrG(f)(x) =∑σ∈G

([σ]f)(x) =∑σ∈G

σf(σ−1x)

is a G-homomorphism.

Proof. Let τ ′ ∈ G and let x ∈ E. Then

TrG(f)(τx) =∑σ∈G

σf(σ−1τx) =∑σ∈G

([σ]f)(τx)

and we assert thatTrG(f)(τx) =

∑σ∈G

([σ]f)(τx) =∑σ∈G

τσf(σ−1x) = τ TrG(f)(x).

This follows becauseτσf(σ−1x) = τσf(σ−1τ−1τx) = [τσ]f(τx)

so that since τ ∈ G acts on left by simply permuting the elements of G,∑σ∈G([τσ]f)(τx) =

∑σ∈G([σ]f)(τx) as each

([α]f)(τx) appears exactly one in each sum. Thus, since ∀τ ∈ G, TrG(f)(τx) = τ TrG(f) and TrG(f) is R-linear, TrG(f)is a G-homomorphism. �

Notation. We denote this composition by TrG(f) : E → F .

Lemma. Let E1, E2 and E3 be G-modules and let

E1f1−→ E2

f2−→ E3

be R-homomorphisms. Then for each σ ∈ G,

σ(f2 ◦ f1) = (σf2) ◦ (σf1).

That is, each σ ∈ G “distributes” over composition of R-homomorphisms of G-modules. (This generalizes by induction.)

31

Proof. For each x ∈ E1,

([σ](f2 ◦ f1))(x) = σf2(f1(σ−1x)) = σf2(σ−1σf1(σ−1x)) = (([σ]f2) ◦ ([σ]f1))(x),

as desired. �

Lemma. The set of G-homomorphisms E → F of G-modules is equal to invG(HomR(E,F )).

Proof. If ϕ : E → F is a G-homomorphism, then ([σ]ϕ)(x) = σϕ(σ−1x) = ϕ(x) for each x ∈ E. Hence, clearly ϕ ∈invG(HomR(E,F )). Conversely, if ϕ ∈ invG(HomR(E,F )), then clearly σϕ(σ−1x) = ϕ(x) for all σ and x so obviously ϕis a G-homomorphism. �

Proposition. Let G be a finite group and let E′, E, F, F ′ be G-modules. Let

E′ϕ−→ E

f−→ Fψ−→ F ′

be R-homomorphisms and assume that ϕ,ψ are G-homomorphisms. Then

TrG(ψ ◦ f ◦ ϕ) = ψ ◦ TrG(f) ◦ ϕ.

Proof. We haveTrG(ψ ◦ f ◦ ϕ) =

∑σ∈G

σ(ψ ◦ f ◦ ϕ) =∑σ∈G

(σψ) ◦ (σf) ◦ (σϕ)

and since ψ and ϕ are G-homomorphisms, this is∑σ∈G

(σψ) ◦ (σf) ◦ (σϕ) = ψ ◦ (∑σ∈G

σf) ◦ ϕ = ψ ◦ TrG(f) ◦ ϕ,

as desired. �

Maschke’s Theorem

Theorem. Let G be a finite group of order n and k a field. Then k[G] is a semisimple ring iff the characteristic of k doesnot divide n (in particular, k could have characteristic 0).

Proof. (⇐) Suppose the characteristic of k does not divide n. We shall prove that every k[G]-module is semisimple byproving each submodule has a complement, thereby proving that k[G] is semisimple (we could also simply prove that everyk[G]-submodule of k[G] is semisimple, but the same idea holds in this more general case).

Let E be a k[G]-modules and F ⊆ E be a k[G]-submodule of E. Let π : E → F be any surjective k-linear map inducingthe identity on F—one can always find such a map by choosing a k-basis for F and extending it to E. Since E is ak[G]-module and k ⊆ k[G], E is a k-vector space and so there exists a k-subspace F ′ ⊆ E such that E = F ⊕ F ′ as ak-vector space. Put π : E → F the canonical projection map.

Using Weyl’s trick, we putϕ = 1

#(G) TrG(π) : E → E.

That is,ϕ = 1

n

∑σ∈G

σπ.

Since the characteristic of k does not divide the order of G, the element 1n

is well-defined and n 6= 0. Now, clearly ϕ isk-linear as π is k-linear and k commutative. We assert that ϕ is in fact k[G]-linear; towards this end it suffices to checkthat ϕ(gx) = gϕ(x) for all g ∈ G. We have

ϕ(gx) = 1n

∑σ∈G

σπ(σ−1gx) = 1n

∑σ∈G

g(g−1σ)π((σ−1g)x) = g([g−1σ]ϕ)(x) = gϕ(x),

as desired. Thus, this is a k[G]-linear map of k-vector spaces (i.e., respecting that structure) so it is a G-homomorphism,and therefore ϕ[E] is a k[G]-submodule. Moreover, ϕ remains a projection onto F . To see this, note that clearly ϕ[E] = Fand in fact for all x ∈ F , σπ(σ−1x) = x since F is a k[G]-submodule. It now suffices to show that ϕ2 = ϕ. Towards this

32 Representation of Finite Groups

end, we note that given x ∈ F and g ∈ G, gx ∈ F so π(gx) = gx. Then ϕ(x) = 1n

∑σ∈G σπ(σ−1x) = n

nx = x. Thus, if

v ∈ E, then ϕ(v) ∈ F so ϕ(ϕ(v)) = ϕ(v), proving that ϕ2 = ϕ.Similarly, the inclusion j : F → E is a G-homomorphism since F is a emphatically a G-submodule. Hence, we have a

splitting: ϕ ◦ j = idF , so that by the splitting lemma, E is the k[G]-direct sum of F and Kerϕ, thereby proving k[G] issemisimple.

(⇒) Suppose for the sake of a contradiction that k[G] is semisimple but the characteristic of k divides that of G.Then the element x =

∑σ∈G σ ∈ k[G] satisfies gx = x for all g ∈ G and x2 = #(G)x = 0. Thus, k[G]x = kx is a

submodule (hence, ideal) of k[G] containing no idempotent elements. But every ideal of a semisimple ring is generatedby an idempotent, for if R is semisimple and I ⊆ R is an ideal, then R = I ⊕ J as an R-module and we may thus write1 = eI + eJ uniquely. Then eI must be idempotent and generates I since eI = e2

I + eIeJ so clearly eIeJ ∈ I ∩ J = {0} sothat e2

I = eI and then ReI ⊆ I, clearly and conversely if x ∈ I, then x = xeI + xeJ and therefore x = xeI so I ⊆ ReI .This is clearly a contradiction, then. �

Characters

Reminder. Let R be a commutative ring and let E be a finite free R-module of rank n. Let f ∈ EndR(E) and let(aij) = MB

B (f) be the n× n matrix of f with respect to a basis B of E. By the trace of M , we mean tr(M) =∑ni=1 aii.

tr(AB) = tr(BA) and tr(B−1AB) = tr(A). Thus, the trace is independent of our choice of basis. Therefore tr : Matn(R)→R or equivalently tr : EndR(E) → R is well-defined. It is clear that, in fact, tr is R-linear. Notice that tr(AA−1) = n sothat

Theorem. The trace of an n× n matrix is the sum of the n eigenvalues of A.

(I have shamelessly stolen this beautiful proof of this fact from stackexchange.)

Proof. The characteristic polynomial cA(t) = det(A − tI) = (−1)n(tn − tr(A)tn−1 + · · · + (−1)n detA) = (−1)n(t −λ1) · · · (t− λn) where the λi are the eigenvalues of A. Comparing coefficients, tr(A) is simply the sum of the eigenvalues,as desired. �

Except in the Section on ‘Induced Representations,’ we denote by G a finite group and we denote E,Ffinite dimensional k-vector spaces where k is a field of characteristic not dividing #(G) so that k[G] issemisimple. We usually denote #(G) by n.

Character of a Representation

Definition (Degree or Dimension). Let G be a finite group of order n and k be a field of characteristic 0. Let E be ak-vector space of dimension m. Let ρ : G→ Autk(E) be a (effective) representation. Then we say that m = dimk E is thedegree or dimension of the representation ρ of G on E.

Definition (Character of a Representation). Let G be a finite group of order n and k be a field of characteristic notdividing n. Let ρ : k[G] → Endk(E) be a k-algebra representation of k[G] on E (as we have seen, this is equivalent to arepresentation G→ Autk(E)). By the character χρ of the mean representation, we shall the k-valued function

χρ : k[G]→ k such that χρ(α) = tr ρ(α) for all α ∈ k[G].

In other words, χρ = tr ◦ρ so that χρ is a k-vector space homomorphism. We sometimes write χE instead of χρ. We alsocall E the representation space of ρ.

Definition (Trivial Character). By the trivial character , we shall mean the character of the representation of Gon the k-space E = k itself such that σx = x for all x ∈ k. It is the character of the extension to k[G] of the trivialrepresentation ρ : G → Autk(k) where for all σ ∈ G, σ 7→ idk ∈ Autk(k). It is the function taking the value 1 on allelements of G.We denote it by χ0 or also by 1G if we need to specify the dependence on G.

Proposition. Characters are functions on G and the values of a character on elements of k[G] are completely determinedby its values on G.

Proof. The extension from G to k[G] is by k-linearity, as in the equivalence we proved before. �

33

Definition (Isomorphic Representations). We say that two representations ρ : G → Autk(E) and ϕ : G → Autk(F )are isomorphic if there is a G-isomorphism f : E → F . Indeed, in this case, for all σ ∈ G,

f ◦ ρσ = ϕσ ◦ f.

We then see that if ρ,ϕ are isomorphic representations, then their characters are equal. (Put another way, if E and F areG-spaces and are G-isomorphic, χE = χF .)

Remarks. In everything that follows, we are interested only in isomorphism classes of representations.If E and F are G-spaces (with ρE and ρF , respectively), then E ⊕ F and E ⊗k F are G-spaces with the operation of

G being componentwise: σ(x⊕ y) = σx⊕ σy and σ(x⊗ y) = σx⊗ σy. Indeed, the resulting maps are simply ρE ⊕ ρF andρE ⊗ ρF , respectively.

Proposition. Let E and F be G-spaces. Then

χE + χF = χE⊕F and χEχF = χE⊗F .

If χ∨ denotes the character of the dual representation on E∨, then for each σ ∈ G,

χ∨(σ) = χ(σ−1)

and, in particular, if k = C, then for each σ ∈ G, χ(σ−1) = χ(σ) and hence if k = C,

χ(σ−1) = χ(σ) = χ∨(σ).

Proof. The first relation holds because of the segmented block structure of the matrix of an element σ in the representationE⊕F . For the second assertion, if B1 = {vi} is a basis of E and B2 = {wj} is a basis of F over k, then B = {vi⊗wj} isa basis of E ⊗ F , as we know. Let (aiν) be the matrix of σ with respect to B1 and (bjµ) be the matrix of σ with respectto B2. Then

σ(vi ⊗ wj) = σvi ⊗ σwj =∑ν

aiνvν ⊗∑µ

bjµwµ =∑ν,µ

aiνbjµ(vν ⊗ wµ).

Checking where ν = i and j = µ, by definition, we find

χE⊗F (σ) =∑i

∑j

aiibjj = χE(σ)χF (σ).

Now, the statement for the character of the dual representation follows from the fact that if we put M∨ to be the matrixof ρ∨(σ) with respect to the dual basis, then M∨ = tM−1. But tr(tM−1) = tr(M−1) = χ(σ−1).

Now, suppose k = C. We shall obtain the last part as a corollary later, but we can give an easy proof right now: Since,σ ∈ G (NB: we are viewing σ ∈ G as a k-automorphism of E) has finite order, so too does its eigenvalues λ1, . . . , λm.Hence, they must have absolute value equal to 1 (this is also a consequence of the fact that the matrix σ is a unitarymatrix [i.e., a matrix equal to its conjugate transpose or hermitian conjugate]). Thus,

χ(σ) = tr(σ) =∑

λi =∑

λ−1i = tr(σ−1) = χ(σ−1),

as desired. �

Proposition. Let G be a finite group k be a field of characteristic not dividing #(G). Let ρ : G → Autk(E) be a repre-sentation of G on E of degree n. Then,

(a) χρ(e) = n.(b) For all σ, σ′ ∈ G, χρ(σσ′σ−1) = χρ(σ′).(c) For each n ∈ N, nχρ = χ⊕n

i=1ρ.

Proof. (a) We have χρ(e) = idE and tr(idE) = n since dimk E = n.(b) This is χρ

((σσ′)σ−1). Let u = σσ′ and let v = σ−1. Then vu = σ′. Now, χρ(uv) = tr(ρuv) (recall that we often

write ρα in place of ρ(α)) and since ρ : G → Autk(E) is a group-homomorphism, this is tr(ρuv) = tr(ρu ◦ ρv). But thetrace is independent of the order of the product of linear operators. Hence, invoking the group-homomorphism propertyof ρ, this is simply

χρ((σσ′)σ−1) = tr(ρu ◦ ρv) = tr(ρv ◦ ρv) = tr(ρvu) = χρ(σ′),

as desired.(c) Since nχρ =

∑ni=1 χρ, this follows from the preceding proposition. �

34 Representation of Finite Groups

Effective and Irreducible Characters

So far we have defined the notion of character associated with a representation. It is now natural to form linear combinationsof such characters with more general coefficients than positive integers. We now make a more formal distinction.Definition ((Generalized) Characters and Effective Characters). By a generalized character of G (or, moresimply, a character of G) we shall mean a function G → k which can be written as a linear combination of charactersof representations with arbitrary integer coefficients, whereas the characters associated with representations will be calledeffective characters. Everything we have defined of course depends on the field k, and we shall add over k to ourexpressions if we need to specify the field.

Proposition. Let G be a finite group and k a field whose characteristic does not divide #(G). The generalized charactersof G over k form a ring.

Proof. This is clear for if we write χ =∑ciχi and χ′ =

∑aiχ′i where χi, χ′i are effective characters of G and the ai, ci ∈ Z,

then their product simplifies to a Z-linear combination of sums of products of effective characters, and as we have seen,effective characters are closed under product and sum, so we’re good. �

Definition (Simple or Irreducible Representation). A simple representation or a irreducible representationis a representation ρ : G → EndR(E) where E 6= 0 such that the only G-invariant submodule of E is E and 0. ByG-invariant, we mean GF ⊆ F . In our case, this corresponds to simple R[G]-modules.

Definition (Simple or Irreducible Character). By a simple character or a irreducible character of G, we meanthe character of a simple representation (i.e., the character associated with a simple k[G]-module).

Lemma. Suppose dimk E = n. If E is k[G]-module, then any direct sum decomposition of E over k[G] has ≤ n summands.In particular, dimk k[G] = #(G).

Proof. As a k-vector space, E can be written as a direct sum of at most n pieces. Now, if E had a direct sum decompositionwith m > n pieces over k[G], then each piece must contain a k-linearly independent member of E. But this is impossiblesince then we have found a linearly independent set with m > n = dimk E elements. The assertion follows. It is clear thatdimk k[G] = #(G). �

Theorem. There are only a finite number of simple characters of G (over k). The characters of representations of G arethe linear combinations of the simple characters with integer coefficients ≥ 0.

Proof. By Maschke’s theorem, k[G] is semisimple. A simple character of G is a character associated with a simple k[G]-module E, say. But the simple k[G]-modules are precisely isomorphic to the (left) ideals of k[G], of which there can onlybe finitely many by the structure theorem for semisimple rings. Thus, there are finitely many irreducible characters sincethere are finitely many non-isomorphic simple (irreducible) k[G]-modules and k[G]-modules are precisely representations.

Finally, suppose we are given a k[G]-module E and a representation ρE : G → Autk(E). Since by Maschke’s theoremk[G] is semisimple, E is semisimple. Hence, E is the direct sum of simple k[G]-submodules, of which there are only finitelymany non-isomorphic classes. In fact, E is a finite direct sum of at most dimk E = n simple k[G]-submodules by thelemma. Write

E = E(n1)1 ⊕ · · · ⊕ E(nm)

m

where the ni count the multiplicity of isomorphic factors and where each Ei is a k[G]-simple-submodule of E. ThenρE = n1ρE1 + · · ·+ nmρEm , as we know. This is N-linear combination of irreducible characters, as desired. �

Remark. By the structure theorem, we may write

k[G] ≈s∏i=1

Ri

where the Ri are simple rings. We have a corresponding decomposition o the unit element of k[G], as usual,

1 = e1 + · · ·+ es

where ei is the unit element of Ri, eiej = 0 if i 6= j and also RiRj = 0 if i 6= j. We note that the integer s depends on k.

Notation. If Li is a simple left ideal of k[G], then any two representations ρ1, ρ2 : G → Autk(Li) are isomorphic by thepreceding theorem and thus have the same character. Thus, it makes sense to speak of the character associated with arepresentation on Li. Indeed, since simple modules correspond (isomorphically) precisely to simple left ideals (we provedthis before), if Li denotes a typical simple module for Ri (say one of the simple left ideals), we let χi be the character ofthe representation on Li. We let ei stand for the unit element in the decomposition k[G] ≈

∏si=1 k[G]ei =

∏si=1 Ri.

35

Remark (Orthogonality Relation). We observe that χi(α) = 0 for all α ∈ Rj if i 6= j. This is a fundamental relationof orthogonality, which is obvious, but from which all our other relations will follow.

Theorem. Suppose k has characteristic 0. Then every effective character has a unique expression as a linear combination

χ =s∑i=1

niχi, ni ∈ Z, ni ≥ 0,

where χ1, . . . , χs are the simple characters of G over k. Two representations are isomorphic iff their associated charactersare equal.

Proof. Let E be the representation space of χ. By the representation theorem its following remark,

E ≈s⊕i=1

L(ni)i

and this sum is finite by the lemma (since E is a FDVS over k). Since ei acts as a unit element on Li, we find

χi(ei) = dimk Li.

(Recall, χi is the character of the representation restricted to an isomorphism class representative.) We have already seenthat χi(ej) = 0 if i 6= j. Hence,

χ(ei) = ni dimk(Li).

Since dimk Li depends only on the structure of the group algebra, we have recovered the multiplicities n1, . . . , ns. Namely,ni is the number of times Li occurs (up to an isomorphism) in the representation space of χ, and is the value of χ(ei)divided by dimk Li (we are in characteristic 0, so this poses no issue). This proves our theorem.

Definition (Multiplicity of Irreducible Characters). As a matter of definition, we call the ni in the above theoremthe multiplicity of χi in χ.

Corollary. Suppose k has characteristic 0. As functions of G into k, the simple characters

χ1, . . . , χs

are linearly independent over k.

Proof. Suppose∑aiχi = 0 with ai ∈ k. We apply this expression to ej and get

0 = (∑

aiχi)(ej) = aj dimk Lj

and hence aj = 0. This works for all j, so all aj = 0, so these are linearly independent over k. �

Warning. Note that this does not say that the simple characters are linearly independent over k. All we’ve shown is thatif a linear combination induces the 0 function, then this function is in fact the trivial function.

Corollary. If k is a field of characteristic 0, then the function dim is a homomorphism of the monoid of effectivecharacters into Z.

Example. Let k be a field of characteristic not dividing the order of #(G) = n < ∞; let us agree to write G additivelywith the understanding that is may not be abelian. Let e1 = 1

n

∑σ∈G σ. Then for any τ ∈ G, τe1 = e1 and e2

1 = e1. If welet e′1 = 1− e1, then e′21 = e′1 and e′1e1 = e1e

′1 = 0. Thus, for any such field k,

k[G] ≈ k[G]e1 × k[G]e′1 = ke1 × k[G]e′1

a direct product decomposition. In particular, the representation of G on the group algebra k[G] itself contains a 1-dimensional representation on the component ke1 whose character is the trivial character.

1-Dimensional Representations

Definition (1-Dimensional Character). By abuse of language, even in characteristic p > 0, we say that a characteris 1-dimensional if it is a homomorphism G→ k×.

36 Representation of Finite Groups

We show here that 1-dimensional characters are precisely representations of G on 1-dimensional k-vectorspaces and that the characters of such representations live in k×.

Suppose E is a 1-dimensional vector space over k and let ρ : G → Autk(E) be a representation. Let {v} be a basis ofE over k. Then for each σ ∈ G, we have

σv = χ(σ)v

for some element χ(σ) ∈ k× with χ(σ) 6= 0 because σ induces an automorphism of E. Then for each τ ∈ G,

τσv = χ(σ)τv = χ(σ)χ(τ)v = χ(στ)v

and so we see that χ : G → k× is a homomorphism, and that our 1-dimensional character is the same type of thing thatoccurred in Artin’s theorem in Galois theory.

Conversely, let χ : G→ k× be a homomorphism. Let E be a 1-dimensional k-space with basis {v} and define σ(av) =aχ(σ)v for all a ∈ k. Then we see at once this operation of G on E gives a representation of G whose associated characteris χ. Thus, 1-dimensional characters on k[G]-modules are precisely homomorphisms χ : G→ k×.

Since G is finite, we note thatχ(σ)n = χ(σn) = χ(1) = 1

so that the values of 1-dimensional characters are nth roots of unity. The 1-dimensional characters form a group undermultiplication.

Theorem. Let G be a finite abelian group and assume that k is algebraically closed. Then

(a) k[G] ≈ k#(G) as k-algebras.(b) Every simple representation of G is 1-dimensional.(c) There are #(G) simple representations of G.(d) The simple characters of G are the homomorphisms of G into k×.

Proof. (We shall prove these all in one go.) The group ring k[G] is commutative and is also semisimple by Maschke. Thus,by the structure theorem, it is a finite direct product of simple rings which are (principal) two-sided ideals (which arerings in their own right) of k[G] which are also k-algebras. But in this case, each simple ring contains no proper idealssince a simple ring has no proper two-sided ideals and since k[G] is commutative each ideal is two-sided. Now, each simplering is a ring is isomorphic to a ring of matrices over a division k-algebra by Artin-Wedderburn and thus have dimension1 since a matrix algebra is commutative iff it is Mat1(D) = D. But then D must be a field since k[G] is commutative.Moreover, 0 < dimkD < ∞ since dimk k[G] = #(G) < ∞ but also D must be a k-vector space. Since D is a k algebra,we have a ring-homomorphism f : k → D which is not zero since D 6= 0 and therefore k ⊆ D isomorphically. Since D hasfinite dimension over k, D is an algebraic extension of k and therefore must be equal to k since k is algebraically closed.Thus, in fact, we have obtained

k[G] =m∏i=1

k[G]ei ≈m∏i=1

k

as a finite product, where ei is the idempotent generating the simple ring and k[G]ei ≈ k as k-algebras. Since dimk k[G] =#(G), clearly m = #(G). Now, k[G]ei must be simple as a k[G]-module for otherwise we would have a splitting k[G]ei =⊕Li for some simple k[G]-submodule of k[G]ei and dimensionality considerations would force us to conclude dimk Li = 0

for all but one index i0, say, which is clearly not possible unless for that index Li0 = k[G]ei.This shows that the irreducible k[G]-submodules are 1-dimensional over k and therefore by the preceding part have

1-dimensional representations. Since all irreducible representations correspond to irreducible k[G]-submodules, this takescare of (a), (b), (c) and (d). �

Remark. If we didn’t already know it, the above demonstrates that the dual G∨ of an abelian group G has the samecardinality as G.

Corollary. Let k be algebraically closed and let G be a finite group. For any character χ and σ ∈ G, the value χ(σ) isequal to a sum of roots of unity with integer coefficients (i.e., coefficients in Z or Z/pZ depending on the characteristic ofk).

Proof. Let H be the cyclic subgroup generated by σ—in particular, H is abelian. A representation of G having characterχ can be viewed as a representation for H by restriction, having the same character. Since effective characters are N-linearcombinations of simple characters and (generalized) characters Z-linear combinations, the preceding theorem tells us whatwe want. �

We can actually prove something quite powerful regarding 1-dimensional characters of finite groups.

37

Corollary. Let G be a finite group and let k be an algebraically closed field whose characteristic does not divide #(G).The number of 1-dimensional representations of G is equal to #(G/[G,G]).

Proof. Let ϕ : G → k× be a 1-dimensional character (i.e., homomorphism). Then necessarily, ϕ factors through theabelian group G/[G,G], as we know. Conversely, given a 1-dimensional character χ of G/[G,G], if π : G→ G/[G,G] is thecanonical projection, then χ′ = χ ◦ ϕ furnishes a 1-dimensional character of G. Finally, by the theorem, every irreduciblerepresentation of G/[G,G] is 1-dimensional and there are #(G/[G,G]) of these, so the result now follows immediately. �

Remark. The foregoing theorem provides an interesting necessary condition for a finite group to be abelian. In otherwords, to prove that a finite group G is not abelian, it suffices to demonstrate an irreducible representation of G onsome C-vector space, say as C is algebraically closed, whose character χ is not 1-dimensional—in other words, χ is not ahomomorphism χ : G→ C×. Below, we shall give an elementary example of how this works, and show how it relates to apermutation representation of G in the case of S3, which also links this up with our understanding of permutations.

Example. For n = 3, S3 is not abelian. To see this, fix n = 3 and put G = Sn. Let E = Cn, fix B = {e1, . . . , en} a basisof E over C and consider C-automorphisms of E determined by mapping σ ∈ Sn to the automorphism of E permutingthe elements of B accordingly. Observe that the standard inner product on Cn is G-invariant: 〈x | y〉 = 〈gx | gy〉 forall g ∈ G. Observe that the fixed points of G are precisely the set of vectors of the form (z, z, z) for z ∈ C (i.e.,(Cn)G = spanC{e1 + e2 + · · · + en}. Therefore the orthogonal complement of this space is the set of vectors whose sumis 0, having dimension n − 1 = 2, and is a subrepresentation of G. We assert it is irreducible. Any subrepresentation Vof it would be 1-dimensional. But then this would mean we would an irreducible subspace comprised of vectors of formx + y = α where α is fixed; but xe1 + ye2 + αe3 is clearly not invariant under G. Therefore this is irreducible and not1-dimensional. Therefore G is not abelian.

Indeed, there is always a manner of constructing a G-invariant hermitian form4 when considering complex representations.

Weyl’s Trick

Proposition (Weyl’s Unitary Trick). Suppose we have a representation of a finite group G on a finite dimensionalC-vector space V . Then there exists a G-invariant inner product on V .

Proof. Let 〈|〉 be the inner product (i.e., positive definite hermitian form) on a V and define 〈|〉G : V ×V → C by averaging:

〈x | y〉 = 1#(G)

∑σ∈G〈σx | σy〉 .

This is obviously sesquilinear, positive definite and conjugate linear in one component, and therefore is a positive definitehermitian form, as desired. �

Remark. This process of averaging to create things which respect the G-action we are considering is a common anduseful trick. We can do this for more than simply inner products. For example, see the proof of Maschke’s theorem. Thismore general version is typically called Weyl’s Trick.

The Space of Class Functions

Definition (Class Function). By a class function of G (over k or with values in k), we shall mean a function f : G→ ksuch that f(στσ−1) = f(τ) for all σ, τ ∈ G. It is clear that characters are class functions. Thus a class function may beviewed as a function on conjugacy classes.

Remark. We shall always extend the domain of definition of a class function to the group ring, by linearity. If

α =∑σ∈G

aσσ

and f is a class function, we definef(α) =

∑σ∈G

aσf(σ).

4 A sesquilinear map s : V ×V → C that is a linear map which is conjugate linear in only one of the arguments and satisfies s(x, y) = s(y, x).

38 Representation of Finite Groups

Definition (Conjugacy Class). Let σ0 ∈ G. Define the equivalence relation ∼ on G by σ ∼ σ′ iff σ is conjugate to σ′.An element of the group ring of type

γ =∑σ∼σ0

σ

will also be called a conjugacy class.

Proposition. An element of k[G] commutes with every element of G iff it is a linear combination of conjugacy classeswith coefficients in k.

Proof. (⇒) Let α =∑σ∈G aσσ and assume ατ = τα for all τ ∈ G. Then

τ−1ατ =∑σ∈G

aστστ−1 =

∑σ∈G

aσσ = α.

Comparing terms and observing the conjugation is an isomorphism, we conclude that if σ is conjugate to σ0, then aσ0 = aσ.Hence,

α =∑γ

aγγ

over conjugacy classes. (⇐) Conversely, let α =∑γ aγγ be a k-linear combination of conjugacy classes. Then τατ−1 =∑

γ aγτγτ−1 = α. �

Corollary. The conjugacy classes are linearly independent over k and form a basis for Z(k[G]), the center of k[G], overk.

Proof. Suppose∑γ aγγ = 0. Since conjugacy classes are distinct, it is clear that if aγ ∈ k for each conjugacy class γ, then

aγ = 0 for each γ. Thus, they are linearly independent over k. Let x ∈ Z(k[G]). We prove that x is a k-linear combinationof the conjugacy classes. This follows from the proposition. �

Assume for the rest of this section that k is algebraically closed.Under the assumption that k is algebraically closed, k[G] =

∏si=1 Ri where each Ri is a simple ring and each Ri is a

matrix algebra over k by Artin-Wedderburn. In a direct product, the center is obviously the product of the centers foreach factor.

Denote ki the image of k in Ri, in other words,ki = kei

where ei is the unit element of Ri. Notice that ∀α ∈ k×, αei 6= 0 for if this were not the case then ∃α ∈ k, αei = 0 so that∀a ∈ k, aei = aα−1αei = 0, which is impossible as Ri is a nontrivial k-algebra. Then the center of k[G] is equal to

s∏i=1

ki

which is s-dimensional over k.If Li is a typical simple left ideal of Ri, then

Ri ≈ Endk(Li).

We letdi = dimk Li.

Then

d2i = dimk Ri and

s∑i=1

d2i = #(G). (0.4)

We also have the direct sum decomposition Ri ≈ L(di)i as a (G, k)-module (i.e., as a k[G]-module). Notice that this direct

sum decomposition and the dimensionality arguments all follow from Artin-Wedderburn.The above notation will remained fixed from now on because it is so standard.We summarize some of our results as follows.

Proposition. Let k be algebraically closed. Then the number of conjugacy classes of G is equal to the number of simplecharacters of G—as both of these are equal to the number s of simple left ideals in k[G]. The conjugacy classes γ1, . . . , γsand the unit elements e1, . . . , es form bases of the center of k[G].

39

Proof. It remains to show that the conjugacy classes are equal to the number of simple characters. Since k[G] =∏si=1 Ri

and since we have found that the conjugacy classes form a basis for Z(k[G]) over k, and since Z(k[G]) = Z(∏si=1 Ri) =∏s

i=1 Z(Ri), it follows that dimk Z(k[G]) = s and therefore there are s conjugacy classes. The rest is easy. �

Definition. The number of elements in a conjugacy class γi shall be denoted by hi. The number of elements in a generalconjugacy class γ shall be denoted by hγ . We call this number the class number .

Notation. Hence forth, we shall denote the center of the group algebra by Zk(G).

Definitions (Regular Character & Regular Representation). We can view k[G] as a G-module simply by lettingk[G] act on itself on the left. The resulting character will be called the regular character and will be denoted by χreg orrG. The representation on k[G] is called the (left) regular representation. The homomorphism ρ : G → Autk[G](k[G])is simply the map g 7→ (1 · g)(•) where (1 · g)(aσσ) = aσ(gσ).

Proposition. If χreg is the character of a (left) regular representation of G over k. Then from the direct sum decompositionof k[G], we get

χreg =s∑i=1

diχi. (0.5)

Proof. s is the number of simple left ideals in k[G]. Any two representations of isomorphic simple left ideals have equalcharacters. The result is clear. �

Proposition. Let χreg be the regular character. Then

(a) χreg(σ) = 0 if σ ∈ G and σ 6= 1.(b) χreg(1) = n.

Index

(G, R)-module, 27A-algebra, 5G-homomorphism, 10, 28G-invariant, 18, 28G-module, 27G-morphism, 10G-simple, 18G-space, 27

adjoint, 29Artin-Wedderburn Theorem, 20automorphism group, 10

bicommutant, 16Burnside’s Theorem, 18

character, 32, 34character is 1-dimensional, 35class function, 37class number, 39commutant, 16conjugacy class, 38contragredient representation, 29

degree, 32dimension, 32dual, 29dual basis, 29dual representation, 29

effective characters, 34endomorphism, 6

faithful module, 18

generalized character, 34group algebra, 8group ring, 8

homomorphism of one representation of R to another, 11

invariant, 19irreducible character, 34irreducible representation, 19, 34isomorphic, 21, 33

Jacobson’s Density Theorem, 17

left module, 5

monoid algebra, 8monoid ring, 8monoid-homomorphism, 5morphism, 11multiplicity, 14, 35

operation, 10over k, 34

permutation representation, 37product, 30

regular character, 39regular representation, 39representation, 10representation space, 32represented, 10Rieffel’s Theorem, 19right module, 5

Schur’s Lemma, 14semisimple, 15semisimple algebra, 15semisimple module, 15semisimple ring, 15simple, 21simple Artinian ring, 22simple character, 34simple representation, 19, 34simple ring, 19, 22sum, 29

trace, 30, 32transpose, 29trivial character, 32

vector space, 13

Wedderburn’s Theorem, 18, 20Weyl’s Trick, 31, 37Weyl’s Unitary Trick, 37

41