Abstract Algebra — Chapter 13 Representation theory

Chapter 13
Representation theory

13.1. Semisimple modules

The following definitions will be of special interest in the case of a group ring over a field.

Definition 13.1.1. §

A module $M$ over a ring $R$ is simple, or irreducible, if it has no nonzero, proper $R$ -submodules. Otherwise, $M$ is said to be reducible.

Definition 13.1.2. §

A module $M$ over a ring $R$ is indecomposable if it is not the direct sum of two proper submodules.

Definition 13.1.3. §

A module $M$ over a ring $R$ is semisimple, or completely reducible, if it is a direct sum of irreducible submodules.

Remark 13.1.4. §

By definition, a module is simple if and only if it is both semisimple and indecomposable.

Examples 13.1.5. §

a.: Any division ring $D$ is simple as a left module over itself, as it has no nontrivial left ideals.
b.: Any vector space $V$ over a field $F$ is semisimple as an $F$ -module, in that it has a basis that allows us to express it (up to isomorphism) as a direct sum of copies of $F$ .
c.: The ring $ℤ$ is indecomposable as a $ℤ$ -module, but it is not simple, as it contains proper, nontrivial submodules $𝑛ℤ$ for $n \geq 2$ .
d.: Any simple $ℤ$ -module is a simple abelian group, so isomorphic to $ℤ ∕ 𝑝ℤ$ for some prime $p$ .
e.: The $ℤ$ -module $ℤ \oplus ℤ$ is neither semisimple nor indecomposable, as it is not a direct sum of simple $ℤ$ -modules.
f.: Let $R$ be the ring of upper-triangular matrices in $M_{2} (F)$ for $F$ a field, and consider the $R$ -module $M = F^{2}$ under left multiplication of column vectors. Then $M$ has a simple submodule $N = F \cdot e_{1}$ , so $M$ is not simple. Moreover, $M$ is not semisimple, as $M = F \cdot v$ for any $v \notin N$ , so $N$ has no complement in $M$ .

Semisimple modules have the following equivalent characterizations.

Proposition 13.1.6. §

Let $M$ be an $R$ -module. The following are equivalent:

$M$ is semisimple.
$M$ is a sum of simple submodules.
Every submodule of $M$ is a direct summand.

Proof.

That (i) implies (ii) is clear. As for (ii) implies (iii), let $N$ be a submodule of

M = ⨁_{i \in I} M_{i},

where the $M_{i}$ are simple. Then $N \cap M_{i}$ is either $0$ or $M_{i}$ for each $i$ , and $N$ is the direct sum of the $M_{i}$ for which $N \cap M_{i} = M_{i}$ .

That (iii) implies (i) is proven as follows. We first claim that any nonzero $R$ -module $M$ contains a nonzero simple submodule. To see this, choose $m \in M$ , and replace $M$ with $𝑅𝑚$ without loss of generality. Let $N$ be a maximal $R$ -submodule of $M$ not containing $n$ , which exists by Zorn’s Lemma. Then $M = N \oplus N^{'}$ for some nonzero $R$ -submodule $N^{'}$ . Now $N^{'}$ must be simple, since any $Q \subseteq N^{'}$ has $N \oplus Q$ containing $a$ and therefore equals $M$ .

Now consider the nonempty set $X$ of semisimple submodules of $M$ under inclusion. The union of any chain $C$ in $X$ is semisimple (as the reader may check), so $X$ has a maximal element $N$ by Zorn’s lemma. Let $N^{'}$ be a complement to $N$ in $M$ , so $M = N \oplus N^{'}$ . If $N^{'}$ is nonzero, then $N^{'}$ contains a simple submodule $Q$ by the claim, and $N \oplus Q$ is semisimple, contradicting the maximality of $M$ . So, $M = N$ is semisimple. □

We define semisimple rings in a manner that does not obviously relate to simple rings.

Definition 13.1.7. §

A nonzero ring is semisimple if it is semisimple as a left module over itself.

The following contains equivalent conditions for a ring to be semisimple.

Theorem 13.1.8. §

The following conditions on a nonzero ring $R$ are equivalent:

i.: $R$ is semisimple,
ii.: every $R$ -module is semisimple,
iii.: every $R$ -module is projective,
iv.: every $R$ -module is injective.

Proof.

Suppose that $R$ is semisimple, and let $M$ be an $R$ -module. Then $M$ is a sum of its cyclic submodules, so by Lemma 13.1.6, it suffices to see that quotients $Q$ of $R$ are semisimple. Again employing Lemma 13.1.6, the kernel $I$ of the quotient map $R \to Q$ is a direct summand, so $Q$ is isomorphic to a left ideal of $R$ , which is semisimple as $R$ is.

That (ii) implies (iii) is an immediate consequence of Lemma 13.1.6. Every surjection of $R$ -modules is split if and only if every injection of $R$ -modules is split by Proposition 12.4.3, so (iii) and (iv) are equivalent, noting Proposition 12.4.8 and Lemma 12.4.13. Finally, (iv) tells us that every $R$ -submodule of an $R$ -module is a direct summand, so is semisimple by Lemma 13.1.6. □

We claim that simple rings are indeed semisimple, so long as we assume that descending chains of left or right ideals terminate. This can be seen directly for matrix rings over division rings, using Morita equivalence.

Definition 13.1.9. §

A ring $R$ is left artinian (resp., right artinian if it satisfies the descending chain condition on left ideals (resp., right ideals).

Lemma 13.1.10. §

If a ring $R$ is the sum of a collection of its nonzero left ideals, then it is also a sum of a finite subcollection.

Proof.

If ${I_{x} ∣ x \in X}$ is a set of nonzero left ideals of $R$ such that $𝑅≅ \sum_{x \in X} I_{x}$ as left-modules, then we can write $1 = \sum_{j = 1}^{n} a_{j}$ for some $n \geq 1$ , where $a_{j} \in I_{x_{j}}$ for some $x_{j} \in I$ . But then the left ideals $I_{x_{j}}$ with $1 \leq j \leq n$ generate $R$ as a left $R$ -module. □

Corollary 13.1.11. §

A semisimple ring $R$ is left artinian, isomorphic as an $R$ -module to the direct sum of its finitely many minimal left ideals.

Proof.

By definition, $R$ is isomorphic to the direct sum of its minimal ideals. Since the sum is direct, no proper subcollection of the minimal ideals generates $R$ . Lemma 13.1.10 then tells us that the collection of minimal ideals must be finite. It follows that $R$ is left artinian. □

Proposition 13.1.12. §

Let $R$ be a left (or right) artinian simple ring. Then $R$ is semisimple.

Proof.

Let $R$ be left artinian and simple. We first claim that $R$ has a simple $R$ -submodule (i.e., left ideal). For this, construct a possibly finite sequence of left ideals $J_{i}$ of $R$ recursively, starting with $J_{1} = R$ , and then for $i \geq 1$ , taking $J_{i + 1}$ to be a proper simple submodule if $J_{i}$ is not simple. Since $R$ is left artinian, we must have that the process terminates, so $R$ has a simple submodule.

Now, consider the nonzero sum $M$ of all distinct simple $R$ -submodules of $R$ . Let $N$ be a simple submodule of $R$ , and let $r \in R$ . Then $𝑁𝑟$ is isomorphic to a quotient of $n$ , so is either $0$ or simple. In particular, $𝑁𝑟$ is contained in $M$ , and therefore $𝑀𝑟 \subseteq M$ . Thus, $M$ is not only a left ideal of $R$ , but a right ideal as well, and therefore $M = R$ .

By Lemma 13.1.10, the $R$ -module $R$ is then a finite sum of distinct simple left ideals: say $R$ is the sum of $N_{i}$ simple for $1 \leq i \leq k$ , where $k$ is minimal. If the intersection $N_{i}$ with the sum $M_{i}$ of the $N_{j}$ for $v \neq i$ is nonzero, then it must equal $N_{i}$ , as $N_{i}$ is simple. But then $N_{i} \subseteq M_{i}$ , so $M_{i} = R$ , which contradicts the minimality of $k$ . So, $R$ is in fact the direct sum of the $N_{i}$ , as required. □

The following is an easy but very useful fact regarding homomorphisms of simple modules.

Lemma 13.1.13 (Schur’s lemma). §

Let $R$ be a ring, and let $M$ and $N$ be simple $R$ -modules. Then any nonzero homomorphism $f : M \to N$ is an isomorphism.

Proof.

The kernel of $f$ is a proper $R$ -submodule of $M$ , hence zero, and the image of $f$ is a nonzero $R$ -submodule of $N$ , hence $N$ . Thus $f$ is bijective. □

Since every nonzero $R$ -linear endomorphism of a simple module is invertible by Schur’s lemma, we have the following corollary.

Lemma 13.1.14. §

Let $R$ be a nonzero ring. The ring ${End}_{R} (M)$ of $R$ -linear endomorphisms of a simple module $M$ is a division ring.

Recall that a ring is simple if it has no nonzero ideals, and by Remark 3.8.8, matrix rings over division algebras are simple. We have the following consequence of Schur’s lemma

Lemma 13.1.15. §

Let $M$ be a simple $R$ -module, and let $n \geq 1$ . Then ${End}_{R} (M^{n}) ≅ M_{n} (D)$ , where $D$ is the division ring ${End}_{R} (M)$ . In particular, ${End}_{R} (M^{n})$ is a simple ring.

Proof.

We define a homomorphism

Φ : M_{n} (D) \to_{}^{\sim} {End}_{R} (M^{n})

on a matrix $C = (ϕ_{𝑖𝑗}) \in M_{n} (D)$ by

Φ (C) (m_{1}, \dots, m_{n}) = (\sum_{j = 1}^{n} ϕ_{1 j} (m_{j}), \dots, \sum_{j = 1}^{n} ϕ_{𝑛𝑗} (m_{j})) .

Every endomorphism $ϕ \in {End}_{R} (M^{n})$ is determined uniquely by the collection of maps $ϕ_{𝑖𝑗} = π_{j} \circ ϕ \circ ι_{i} \in {End}_{R} (M)$ , where $π_{i}$ and $ι_{i}$ denote the $i$ th projection and inclusion maps, so this is one-to-one and onto. □

We can improve this lemma to treat a finite direct sum of arbitrary simple modules.

Lemma 13.1.16. §

Let $R$ be a nonzero ring. Let $M$ be an $R$ -module that is isomorphic to a direct sum $N_{1}^{n_{1}} \oplus N_{2}^{n_{2}} \oplus \dots \oplus N_{k}^{n_{k}}$ with the $N_{i}$ mutually nonisomorphic simple modules and $n_{i} \geq 1$ for $1 \leq i \leq k$ . Then we have an isomorphism of rings

{End}_{R} (M) ≅ \prod_{i = 1}^{k} M_{n_{i}} (D_{i}),

where $D_{i}$ is the division ring ${End}_{R} (N_{i})$ .

Proof.

Let $π_{i} : M \to N_{i}^{n_{i}}$ and $ι_{i} : N_{i}^{n_{i}} \to M$ be the projection and inclusion maps. Any $R$ -module endomorphism $f$ of $M$ determines and is determined by the homomorphisms $f_{i, j} = π_{j} \circ f \circ ι_{i} : N_{i}^{n_{i}} \to N_{j}^{n_{j}}$ for $1 \leq i, j \leq n$ . But ${Hom}_{R} (N_{i}, N_{j}) = 0$ for $i \neq j$ , so ${Hom}_{R} (N_{i}^{n_{i}}, N_{j}^{n_{j}}) = 0$ for $i \neq j$ as well. Therefore, the product of restriction maps to $N_{i}^{n_{i}}$ yields the first of the isomorphisms in

{End}_{R} (M) ≅ \prod_{i = 1}^{k} {End}_{R} (N_{i}^{n_{i}}) ≅ \prod_{i = 1}^{k} M_{n_{i}} (D_{i}),

where the second isomorphism is by Lemma 13.1.15 □

Evaluation at $1$ gives the isomorphism in the following lemma.

Lemma 13.1.17. §

We have ${End}_{R} (R) \to R^{op}$ as rings.

We now come to the Artin-Wedderburn theorem, which classifies semisimple rings.

Theorem 13.1.18 (Artin-Wedderburn theorem). §

A nonzero ring is semisimple if and only if it is isomorphic to a direct product of matrix algebras over division rings.

Proof.

For any nonzero ring $R$ , we have an isomorphism

R^{op} \to_{}^{\sim} {End}_{R} (R), r \mapsto (s \mapsto 𝑠𝑟) .

Supposing that $R$ is semisimple, we have by Corollary 13.1.11 that $𝑅≅ N_{1}^{n_{1}} \oplus N_{2}^{n_{2}} \oplus \dots \oplus N_{k}^{n_{k}}$ with the $N_{i}$ mutually nonisomorphic simple left $R$ -modules and $n_{i} \geq 1$ for $1 \leq i \leq k$ . Noting Lemma 13.1.17, we then have

R^{op} ≅ {End}_{R} (R) ≅ \prod_{i = 1}^{k} M_{n_{i}} (D_{i}),

where $D_{i} = {End}_{R} (N_{i})$ is a division ring. By taking the opposite ring of both sides, we obtain

𝑅≅ \prod_{i = 1}^{k} M_{n_{i}} {(D_{i})}^{op} = \prod_{i = 1}^{k} M_{n_{i}} (D_{i}^{op}),

and $D_{i}^{op}$ is a division ring as well.

On the other hand, suppose that $𝑅≅ \prod_{i = 1}^{k} M_{n_{i}} (E_{i})$ for some division algebras $E_{i}$ . The matrix rings $M_{n_{i}} (E_{i})$ are semisimple left modules over $M_{n_{i}} (E_{i})$ , isomorphic to a direct sum of the simple submodules of column vectors. They are also then semisimple as modules for the larger ring $R$ , since the action of $R$ on $M_{n_{i}} (E_{i})$ by left multiplication factors through $M_{n_{i}} (E_{i})$ . Thus, $R$ is a semisimple ring as a direct sum of these as a left $R$ -module. □

Here are some corollaries. The first follows directly from Proposition 13.1.12 and the Artin-Wedderburn theorem.

Corollary 13.1.19. §

A nonzero ring is left artinian and simple if and only if it is isomorphic to a matrix ring over a division ring.

Consequently, we have the following, which explains the relationship between simple and semisimple rings.

Corollary 13.1.20. §

A nonzero ring is semisimple if and only if it is isomorphic to a finite direct product of left artinian simple rings.

For algebras over a field, we obtain Wedderburn’s theorem.

Corollary 13.1.21 (Wedderburn). §

An algebra over a field $F$ is semisimple if and only if it is a product of finite-dimensional simple $F$ -algebras, and these simple algebras are isomorphic to matrix rings over finite-dimensional division algebras over $F$ .

The following greatly limits the choice of finite-dimensional division algebras over algebraically closed fields.

Proposition 13.1.22. §

Let $D$ be a finite-dimensional division algebra over an algebraically closed field $F$ . Then $D = F$ .

Proof.

Let $γ \in D$ . Note that $γ$ commutes with every element of $F$ , so $F (γ)$ is a field. Since $D$ is finite-dimensional over $F$ , the elements $γ^{i}$ for $i \geq 0$ are linearly dependent over $F$ , and therefore $γ$ is algebraic over $F$ . Thus $F (γ) = F$ , which is to say $γ \in F$ . □

Corollary 13.1.23. §

Let $A$ be a finite-dimensional, semsimple $F$ -algebra, where $F$ is an algebraically closed field. Then $A$ is isomorphic to a direct product of matrix algebras with $F$ -entries.

For commutative rings, we have this:

Corollary 13.1.24. §

A commutative semisimple ring is a finite direct product of fields. A finite-dimensional commutative semisimple algebra over a field $F$ is isomorphic to a direct product of finite field extensions of $F$ .

Definition 13.1.25. §

Let $R$ be a ring. An idempotent in $R$ is a nonzero element $e \in R$ such that $e^{2} = e$ .

Definition 13.1.26. §

Let $R$ be a ring. We say two idempotents $e, f \in R$ are orthogonal if $𝑒𝑓 = 𝑓𝑒 = 0$ .

Remark 13.1.27. §

Any finite sum of orthogonal idempotents is also an idempotent.

Definition 13.1.28. §

We say that an idempotent $e$ in a ring $R$ is primitive if $𝑒𝑅$ is a subring of $R$ that is not a product of two subrings of $R$ .

Lemma 13.1.29. §

Let $R$ be a nonzero ring and $k \geq 1$ . Then $R = R_{1} \times R_{2} \times \dots \times R_{k}$ with $R_{i}$ rings for $1 \leq i \leq k$ if and only if there exist mutually orthogonal idempotents $e_{1}, e_{2}, \dots, e_{k}$ in $Z (R)$ such that $e_{1} + e_{2} + \dots + e_{k} = 1$ . These may be chosen so that $R_{i} = (e_{i})$ in $R$ .

Proof.

If $R = \prod_{i = 1}^{k} R_{i}$ , then let $e_{i}$ be the identity in $R_{i}$ . The $e_{i}$ are then clearly mutually orthogonal, central idempotents. Set $e = \sum_{i = 1}^{k} e_{i}$ . If $1 = (r_{1}, r_{2}, \dots, r_{k}) \in R$ , then

e = e \cdot 1 = (e_{1} r_{2}, e_{2} r_{2}, \dots, e_{k} r_{k}) = r .

Conversely, given $e_{1}, e_{2}, \dots, e_{k}$ , set $R_{i} = R e_{i}$ in $R$ . For any $r \in R$ , we have $r = \sum_{i = 1}^{k} r e_{i}$ , so $R = \sum_{i = 1}^{k} R_{i}$ . If $r_{i} \in R_{i}$ for each $i$ , then set $r = \sum_{i = 1}^{k} r_{i}$ . This satisfies $r e_{j} = r_{j}$ for each $1 \leq j \leq k$ , so $r = 0$ if and only if each $r_{j} = 0$ , and thus $R = \oplus_{i = 1}^{k} R_{i}$ as left $R$ -modules. Since each $e_{i}$ is central in $R$ , each $R_{i}$ is also a right ideal and a ring with unit element $e_{i}$ (though not a subring of $R$ if $k \geq 2$ , since then $e_{i} \neq 1$ ). □

Example 13.1.30. §

If $R$ is a direct product of matrix rings, then it has a set of mutually orthogonal idempotents consisting of the identity matrices in those rings.

Lemma 13.1.31. §

Let $R = \prod_{i = 1}^{k} R_{i}$ be a direct product of rings $R_{i}$ , and let $e_{i}$ be identity element of $R_{i}$ . Let $M$ be a left $R$ -module. Then $M = \oplus_{i = 1}^{k} e_{i} M$ as an $R$ -module.

Proof.

Any $m \in M$ can be written as $m = e_{1} m + e_{2} m + \dots + e_{k} m$ , so $M = \sum_{i = 1}^{k} e_{i} M$ . If $m_{1} + m_{2} + \dots + m_{k} = m$ with $m_{i} \in e_{i} M$ for $1 \leq i \leq k$ , then

m_{i} = e_{i} (m_{1} + m_{2} + \dots + m_{k}) = e_{i} m

for each $i$ , so the representation of $m$ as an element of the sum is unique. Therefore, $M$ is the direct sum of the $e_{i} M$ . □

13.2. Representations of groups

Let $G$ be a group.

Proposition 13.2.1. §

Let $R$ be a commutative ring, let $G$ be a group, and let $M$ be an $R$ -module. There is a one-to-one correspondence between

i.: homomorphisms $ρ : G \to {Aut}_{R} (M)$ ,
ii.: $R [G]$ -module structures given by $R$ -bilinear maps $ϕ : R [G] \times M \to M$

such that $ρ$ corresponds to a unique $ϕ$ with $ρ (g) (m) = ϕ (g, m)$ for all $g \in G$ and $m \in M$ .

Proof.

If $ρ : G \to {Aut}_{R} (M)$ is a homomorphism, then we define

(\sum_{g \in G} a_{g} g) \cdot m = \sum_{g \in G} a_{g} ρ (g) (m)

(13.2.1)

for $\sum_{g \in G} a_{g} g \in R [G]$ and $m \in M$ . For a fixed $m$ , this provides the unique $R [G]$ -module homomorphism $R [G] \to M$ that sends $g$ to $ρ (g) (m)$ by the $R$ -freeness of $R [G]$ . In other words, the operation $R [G] \times M \to M$ is left distributive. Since $ρ (g) (m + m^{'}) = ρ (g) (m) + ρ (g) (m^{'})$ for $g \in G$ and $m, m^{'} \in M$ in that $ρ (g) \in {Aut}_{R} (M)$ , right distributivity follows from the definition in (13.2.1) as well.

Conversely, given an $R [G]$ -module $M$ , we define $ρ : G \to {Aut}_{R} (M)$ by $ρ (g) (m) = g \cdot m$ for $g \in G$ and $m \in M$ . Note that

ρ (g) (m + m^{'}) = g (m + m^{'}) = 𝑔𝑚 + g m^{'} = ρ (g) (m) + ρ (g) (m^{'})

and $ρ (g) (𝑟𝑚) = g (𝑟𝑚) = r (𝑔𝑚) = 𝑟𝜌 (g) (m)$ , so $ρ (g)$ is indeed an element of ${Aut}_{R} (M)$ . Moreover,

ρ (g g^{'}) (m) = (g g^{'}) m = g (g^{'} m) = ρ (g) (g^{'} m) = ρ (g) (ρ (g^{'}) (m)) = (ρ (g) \circ ρ (g^{'})) (m),

so $ρ$ is a homomorphism. □

Remark 13.2.2. §

To give an $R [G]$ -module structure on an $R$ -module $M$ , it suffices to give an operation $G \times M \to M$ such that the map $g : M \to M$ defined by left multiplication is $R$ -linear.

Remark 13.2.3. §

The trivial $R [G]$ -module $R$ on which $g \cdot r = r$ for all $g \in G$ and $r \in R$ corresponds to the homomorphism $ρ : G \to {Aut}_{R} (R) ≅ R^{\times}$ with $ρ (g) = 1$ for all $g \in G$ .

Example 13.2.4. §

Let $R$ be a commutative ring and $G$ be a group. We may view $R [G]$ as a left $R [G]$ -module under left multiplication. This corresponds to the homomorphism $ρ : G \to {Aut}_{R} (R [G])$ that takes $g$ to left multiplication by $g$ on $R [G]$ .

We now focus on the special case that $R$ is a field, which yields group representations. From now on in this section, we let $F$ denote a field.

Definition 13.2.5. §

A representation, or group representation, of a group $G$ over a field $F$ is an $F$ -vector space $V$ , together with a homomorphism $ρ : G \to {Aut}_{F} (V)$ . We also say that $V$ is an $F$ -representation of $G$ .

Remark 13.2.6. §

By Proposition 13.2.1, to make an $F$ -vector space $V$ into an $F [G]$ -module $V$ is equivalent to providing a homomorphism $ρ : G \to {Aut}_{F} (V)$ that makes it into a representation of $G$ .

Definition 13.2.7. §

We say that a representation $ρ : G \to {Aut}_{F} (V)$ is finite-dimensional if $V$ is a finite-dimensional $F$ -vector space, in which case $\dim_{F} V$ is its dimension, also known as its degree.

Representations form one of the most important tools in the study of the structure of groups.

Example 13.2.8. §

Let $G$ be a subgroup of ${GL}_{n} (F)$ . Then the inclusion $ρ : G \to {GL}_{n} (F)$ defines a representation of $G$ , and this turns $F^{n}$ into an $F [G]$ -module, where $g \in G$ acts on $v \in F^{n}$ by left multiplication of the column vector $v$ by the matrix corresponding to $g$ .

Example 13.2.9. §

The representation $ρ : ℝ \to {GL}_{2} (ℝ)$ given by

ρ (𝜃) = (\begin{matrix} \cos 𝜃 & \sin 𝜃 \\ - \sin 𝜃 & \cos 𝜃 \end{matrix}) .

is a two-dimensional real representation of the additive group $ℝ$ .

Definition 13.2.10. §

a.: The trivial representation of $G$ over $F$ is $F$ with the trivial $G$ -action.
b.: The regular representation of $G$ over $F$ is $F [G]$ with the action of $F [G]$ on itself by left multiplication.

Remark 13.2.11. §

Two $F$ -representations $V$ and $W$ of $G$ are isomorphic if $V$ and $W$ are isomorphic as $F [G]$ -modules. Phrased in terms of the corresponding homomorphisms $ρ_{V}$ and $ρ_{W}$ , this says that $ρ_{V}$ and $ρ_{W}$ are conjugate by the isomorphism $φ : V \to W$ : that is, $ρ_{W} (g) = φ \circ ρ_{V} (g) \circ φ^{- 1}$ for all $g \in G$ .

Examples 13.2.12. §

Let $V$ and $W$ be $F$ -representations of a group $G$ .

a.: The $F$ -vector space $V \otimes_{F} W$ is a representation of $G$ with respect to the diagonal $G$ -action $g \cdot (v \otimes w) = 𝑔𝑣 \otimes 𝑔𝑤$ for $g \in G$ , $v \in V$ and $w \in W$ .
b.: The $F$ -vector space ${Hom}_{F} (V, W)$ is a representation of $G$ with respect to the $G$ -action $(g \cdot φ) (v) = 𝑔𝜑 (g^{- 1} v)$ for $g \in G$ , $φ \in {Hom}_{F} (V, W)$ , and $v \in V$ .

As a special case, we have the following.

Definition 13.2.13. §

Let $V$ be an $F$ -representation of a group $G$ . The dual representation to $V$ is $V^{*} = {Hom}_{F} (V, F)$ .

The reader will easily check the following.

Lemma 13.2.14. §

Let $V$ and $W$ be $F$ -representations of a group $G$ , and suppose that $W$ is finite-dimensional. Then ${Hom}_{F} (V, W) ≅ V^{*} \otimes_{F} W$ .

Terminology 13.2.15. §

We speak of $F$ -representations of a group $G$ as being simple, indecomposable, and so forth, if the $F [G]$ -modules that define them have these properties.

Definition 13.2.16. §

An $F$ -representation $V$ of a group $G$ is called faithful if $ρ_{V} : G \to {Aut}_{F} (V)$ is injective.

Definition 13.2.17. §

A subrepresentation $W$ of an $F$ -representation $V$ of a group $G$ is an $F [G]$ -submodule of $V$ .

Remark 13.2.18. §

An irreducible (i.e., simple) representation is one that has no nonzero, proper subrepresentations.

Examples 13.2.19. §

a.: All one-dimensional representations of a group are irreducible.
b.: Let $D_{p} = ⟨ r, s ⟩$ be the dihedral group of prime order $p$ , and let $ρ : D_{p} \to {GL}_{2} (𝔽_{p})$ be the representation with $ρ (s) = (\begin{matrix} - 1 & 0 \\ 0 & 1 \end{matrix})$ and $ρ (r) = (\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix})$ . Then $ρ$ is indecomposable but not irreducible, since the $𝔽_{p}$ -submodule $W$ of $V = 𝔽_{p}^{2}$ spanned by $e_{1}$ is left stable by (i.e., is closed under) the action of $D_{p}$ , so is a subrepresentation. On the other hand, $W$ does not have a complement in $V$ (i.e., the only line in $𝔽_{p}^{2}$ that is stabilized by $D_{p}$ is $W$ ).
c.: The regular representation of a finite group is faithful, whereas the trivial representation is not faithful unless the group is trivial.

Let us rephrase Schur’s lemma in the context of representations.

Lemma 13.2.20. §

Let $V$ be an irreducible $F$ -representation of $G$ . Then ${End}_{F [G]} (V)$ is a division algebra over $F$ .

Proof.

This is an immediate consequence of Lemma 13.1.14, noting that the the endomorphisms given by multiplication by elements of $F$ are contained in the center of ${End}_{F [G]} (V)$ . □

By Proposition 13.1.22, this has the following corollary.

Corollary 13.2.21. §

Let $V$ be a finite-dimensional irreducible $F$ -representation of a group $G$ , where $F$ is algebraically closed. Then ${End}_{F [G]} (V) ≅ F$ .

Definition 13.2.22. §

Let $V$ and $W$ be representations of $G$ over a field $F$ , with $V$ semisimple and $W$ irreducible. The multiplicity of $W$ in $V$ is the largest nonnegative integer $n$ such that $W^{n}$ is isomorphic to a subrepresentation of $V$ . We say that $W$ occurs with multiplicity $n$ in $V$ .

Lemma 13.2.23. §

Let $V$ be an $F$ -representation of a finite group $G$ . Let $E ∕ F$ be a field extension. Then $E \otimes_{F} V$ is an $E [G]$ -module under the action $g \cdot (α \otimes v) = α \otimes 𝑔𝑣$ with the same character as $V$ .

Proof.

Note that $E [G] ≅𝐸 \otimes_{F} F [G]$ , and the action described is just the usual action of a tensor product of algebras on a tensor product of modules over them. □

Definition 13.2.24. §

For an $F$ -representation $V$ of a group $G$ and a field extension $E ∕ F$ , the $E$ -representation $E \otimes_{F} V$ is called the base change of $V$ from $F$ to $E$ .

13.3. Maschke’s theorem

In this section, we let $G$ be a finite group, and we let $F$ be a field of characteristic not dividing the order of $G$ .

Theorem 13.3.1 (Maschke’s theorem). §

Let $G$ be a finite group, let $F$ be a field of characteristic not dividing $| G |$ , and let $V$ be a representation of $G$ over $F$ . Then every subrepresentation of $V$ is a direct summand of $V$ as an $F [G]$ -module.

Proof.

Let $W$ be an $F [G]$ -submodule of $V$ . As $F$ -modules, we know that we can find a basis $B^{'}$ of $W$ contained in a basis $B$ of $V$ . We then have a projection map $p : V \to W$ given by $p (\sum_{v \in B} a_{v} v) = \sum_{w \in B^{'}} a_{w} w$ , where $a_{v} \in F$ equals zero for almost all $v$ . This is an $F$ -linear transformation that restricts to the identity on $W$ , but it is not necessarily a $F [G]$ -module homomorphism. So, define

π : V \to V, π (v) = \frac{1}{| G |} \sum_{g \in G} g^{- 1} p (𝑔𝑣)

for $v \in V$ . Then $π$ is clearly $F$ -linear, and moreover

π (h𝑣) = \frac{1}{| G |} \sum_{g \in G} g^{- 1} p (𝑔h𝑣) = \frac{1}{| G |} \sum_{k \in G} {(k h^{- 1})}^{- 1} p (𝑘𝑣) = h𝑝 (v),

so $π$ is an $F [G]$ -module homomorphism. Since $W$ is an $F [G]$ -submodule of $V$ , the image of $π$ is contained in $W$ , and for $w \in W$ , we have

π (w) = \frac{1}{| G |} \sum_{g \in G} g^{- 1} p (𝑔𝑤) = \frac{1}{| G |} \sum_{g \in G} g^{- 1} 𝑔𝑤 = w .

In particular, the inclusion of $W$ in $V$ splits $π$ , so $W$ is a direct summand of $V$ as an $F [G]$ -module. □

As a consequence of Maschke’s theorem and Wedderburn theory, or more specifically, Theorem 13.1.8, we have the following corollary.

Corollary 13.3.2. §

The group ring $F [G]$ is a semisimple $F$ -algebra, which is to say isomorphic to a finite direct product of matrix rings over finite-dimensional division algebras over $F$ .

This in turn yields the following corollaries. For the first, see Corollary 13.1.24.

Corollary 13.3.3. §

Let $G$ be a finite abelian group, and let $F$ be a field of characteristic not dividing $| G |$ . Then $F [G]$ is a direct product of finite field extensions of $F$ .

Example 13.3.4. §

By the Chinese remainder theorem, we have

ℚ [ℤ ∕ 𝑝ℤ] ≅ℚ [x] ∕ (x^{p} - 1) ≅ℚ [x] ∕ (x - 1) \times ℚ [x] ∕ Φ_{p} (x) ≅ℚ \times ℚ (ζ_{p}),

where $ζ_{p}$ is a primitive $p$ th root of unity in $ℂ$ . Note, however, that if we take $𝔽_{p}$ in place of $ℚ$ , then we obtain

𝔽_{p} [ℤ ∕ 𝑝ℤ] ≅ 𝔽_{p} [x] ∕ (x^{p} - 1) ≅ 𝔽_{p} [x] ∕ {(x - 1)}^{p} ≅ 𝔽_{p} [y] ∕ (y^{p})

for $y = x - 1$ , which is not a direct product of matrix rings over fields.

For the following, see Corollary 13.1.23.

Corollary 13.3.5. §

If $F$ is algebraically closed, then $F [G]$ is isomorphic to a direct product of matrix algebras over $F$ .

Proposition 13.3.6. §

Suppose that $F$ is algebraically closed, and write

F [G] ≅ \prod_{i = 1}^{k} M_{n_{i}} (F)

for some $k \geq 1$ and $n_{i} \geq 1$ for $1 \leq i \leq k$ . Then $k$ is equal to the number of conjugacy classes of $| G |$ .

Proof.

First, we remark that $Z (M_{n_{i}} (F)) = F$ , so $\dim_{F} Z (F [G]) = k$ . For any $g \in G$ , we form out of its conjugacy class $C_{g}$ the sum $N_{g} = \sum_{h \in C_{g}} h$ . If we let $G$ act on $F [G]$ by conjugation, then the $G$ -invariant module for this action is $Z (F [G])$ . Moreover, $N_{g}$ lies in this invariant group. That is, the action restricts to an action on $G$ which preserves conjugacy classes, so

k N_{g} k^{- 1} = \sum_{h \in C_{g}} 𝑘h k^{- 1} = \sum_{h \in C_{g}} h = N_{g},

The elements $N_{g}$ , where $g$ runs over a set $S$ of representatives for the conjugacy classes of $G$ , are linearly independent as they are sums over disjoint sets of group elements. And if $z = \sum_{g \in G} a_{g} g \in Z (F [G])$ and $k \in G$ , then

\sum_{g \in G} a_{g} g = \sum_{g \in G} a_{g} 𝑘𝑔 k^{- 1} = \sum_{g \in G} a_{k^{- 1} 𝑔𝑘} g,

so $a_{g} = a_{k^{- 1} 𝑔𝑘}$ for all $k$ , so $a_{h} = a_{g}$ for all $h \in C_{g}$ . Thus, $z$ is in the $F$ -span of the elements $N_{g}$ . Thus, we have $k = | S |$ , the number of conjugacy classes. □

Remark 13.3.7. §

If $F$ is algebraically closed, then we may by Corollary 13.3.5 write

F [G] ≅ \prod_{i = 1}^{k} M_{n_{i}} (F)

for some $k \geq 1$ and $n_{i} \geq 1$ for $1 \leq i \leq k$ . Then $G$ has $k$ isomorphism classes of irreducible representations of dimensions $n_{1}, n_{2}, \dots, n_{k}$ . Let $V_{i}$ be the $i$ th of these, with $\dim_{F} V_{i} = n_{i}$ . Then $V_{i}$ occurs with multiplicity $n_{i}$ in the regular representation $R [G]$ , which is to say that $R [G] ≅ V_{1}^{n_{1}} \oplus V_{2}^{n_{2}} \oplus \dots \oplus V_{k}^{n_{k}}$ . Under this isomorphism, each copy of $V_{i}$ is identified with one of the simple left ideals in $M_{n_{i}} (F)$ , isomorphic to the module $F^{n_{i}}$ of column vectors for this ring. Counting dimensions tells us that

\sum_{i = 1}^{k} n_{i}^{2} = | G | .

Example 13.3.8. §

The group $S_{3}$ has $3$ conjugacy classes, so there are $3$ isomorphism classes of irreducible representations of $G$ , and the sum of the squares of their dimensions are $6$ , so they have dimensions $1$ , $1$ , and $2$ . Thus, we have

ℂ [S_{3}] ≅ℂ \times ℂ \times M_{2} (ℂ) .

The two one-dimensional representations correspond to homomorphisms $G \to ℂ^{\times}$ , factoring through $G^{ab} ≅ℤ ∕ 2 ℤ$ . There are exactly two of these, the trivial homomorphism and the sign map $sign : S_{3} \to {\pm 1}$ . These correspond to the trivial $F [G]$ -module $F$ and the $F [G]$ -module $F$ on which $σ \in S_{3}$ acts by $σ \cdot v = sign (σ) v$ for $v \in F$ .

The irreducible two-dimensional representation $W$ of $S_{3}$ is a subrepresentation of the $3$ -dimensional permutation representation $ρ_{V} : S_{3} \to {GL}_{3} (ℂ)$ . That is, consider the standard basis ${e_{1}, e_{2}, e_{3}}$ of the corresponding $𝔽 [S_{3}]$ -module $V = 𝔽^{3}$ on which $S_{3}$ acts by permuting the indices of the basis elements. Then $W$ is spanned by $e_{1} - e_{2}$ and $e_{2} - e_{3}$ . With respect to this basis, the corresponding homomorphism $ρ_{W} : S_{3} \to {GL}_{2} (ℂ)$ satisfies

\begin{array}{rcl} ρ_{W} ((1 2)) = (\begin{matrix} - 1 & 1 \\ 0 & 1 \end{matrix}) & and & ρ_{W} ((1 2 3)) = (\begin{matrix} 0 & - 1 \\ 1 & - 1 \end{matrix}) . \end{array}

Example 13.3.9. §

Since all of the $ℂ$ -representations of $S_{3}$ take values in ${GL}_{n} (ℚ)$ for some $n$ , we have $ℚ [S_{3}] ≅ℚ \times ℚ \times M_{2} (ℚ)$ . In other words, the irreducible $ℂ$ -representations of $S_{3}$ are obtained from the irreducible $ℚ$ -representations of $S_{3}$ by base change.

13.4. Characters

Recall from Lemma 8.7.22 that the traces of similar matrices are equal.

Definition 13.4.1. §

Let $V$ be a finite-dimensional vector space over a field $F$ . The trace of $φ \in {Aut}_{F} (V)$ is the trace of the matrix representing $φ$ with respect to any choice of ordered basis of $V$ .

Definition 13.4.2. §

The character of a representation $ρ : G \to {Aut}_{F} (V)$ of a group $G$ on a finite-dimensional vector space $V$ over a field $F$ is a map $χ : G \to F$ defined by

χ (g) = tr ρ (g) .

Terminology 13.4.3. §

We say that $χ$ is a character of $G$ if it is the character of a representation of $G$ .

Notation 13.4.4. §

Given an $F [G]$ -module $V$ , we denote the corresponding representation (i.e., homomorphism) by $ρ_{V}$ and and its character by $χ_{V}$ .

Examples 13.4.5. §

a.: The character $χ : G \to F$ of a one-dimensional representation $ρ : G \to F^{\times}$ satisfies $χ (g) = ρ (g)$ for all $g \in G$ .
b.: The character of the permutation representation $ρ : S_{n} \to {GL}_{n} (F)$ satisfies $ρ (σ) = | X_{n}^{σ} |$ for every $σ \in S_{n}$ , where $X_{n} = {1, 2, \dots, n}$ .
c.: Let $W$ be as in Example 13.3.8. Then the character $χ_{W} : S_{3} \to ℂ$ satisfies $χ_{W} ((1 2)) = 0$ and $χ_{W} ((1 2 3)) = - 1$ .
d.: The character $χ$ of the regular representation $F [G]$ satisfies $χ (1) = | G |$ and $χ (g) = 0$ for all $g \in G - {1}$ .

Definition 13.4.6. §

The character of the trivial representation is called the trivial character, or principal character of $G$ .

Definition 13.4.7. §

A character $χ_{V} : G \to F$ of an $F [G]$ -module $V$ is irreducible if $V$ is irreducible.

Definition 13.4.8. §

The degree of a character $χ_{V}$ of an $F$ -representation $V$ of $G$ is $\dim_{F} (V)$ .

Definition 13.4.9. §

A class function of $G$ is a function $G \to F$ , for $F$ a field, that is constant on conjugacy classes in $G$ .

Lemma 13.4.10. §

Let $G$ be a group, let $F$ be a field, and let $V$ and $W$ be $F$ -representations of $G$ . Then

a.: $χ_{V} (e) = \dim_{F} V$ ,
b.: $χ_{V \oplus W} = χ_{V} + χ_{W}$ ,
c.: $χ_{V} = χ_{W}$ if $V$ and $W$ are isomorphic representations, and
d.: $χ_{V}$ is a class function on $G$ .

Proof.

Since $ρ_{V} (e)$ is the identity transformation, we have part $a$ . Part b follows by choosing a basis of $V \oplus W$ that is a union of bases of $V$ and $W$ and noting that the matrix representing $ρ_{V \oplus W} (g)$ for $g \in G$ with respect to that basis is block diagonal with blocks $ρ_{V} (g)$ and $ρ_{W} (g)$ . Part c holds as $ρ_{V} (g)$ and $ρ_{W} (g)$ are represented by similar matrices if $V$ and $W$ are isomorphic. Part d also holds as $ρ_{V} (g)$ and $ρ_{V} (g^{'})$ are represented by similar matrices if $g$ and $g^{'}$ are conjugate in $G$ . □

Proposition 13.4.11. §

Let $G$ be a finite group, and let $F$ be a field of characteristic zero. Let $V$ and $W$ be finite-dimensional $F$ -representations of $G$ . Then $V$ and $W$ are isomorphic if and only if $χ_{V} = χ_{W}$ .

Proof.

By Lemma 13.4.10c, we know that $𝑉≅𝑊$ implies $χ_{V} = χ_{W}$ . Write

F [G] = \prod_{i = 1}^{r} M_{n} (D_{i}) .

For $1 \leq i \leq r$ , let $e_{i}$ denote the identity of $M_{n_{i}} (D_{i})$ , let $V_{i}$ be the irreducible $F$ -representation $D_{i}^{n_{i}}$ of $G$ , and let $χ_{i}$ denote its character. Then there exist $m_{i}$ for $1 \leq i \leq r$ such that

V = ⨁_{i = 1}^{r} V_{i}^{m_{i}} .

Extend $χ_{V}$ by $F$ -linearity to a map $χ_{V} : F [G] \to F$ . Then

χ_{V} (e_{j}) = \sum_{i = 1}^{r} m_{i} χ_{i} (e_{j}) = m_{i} \dim_{F} V_{i},

so the multiplicities $m_{i}$ of the $V_{i}$ in $V$ are uniquely determined by $χ_{V}$ . That is, $χ_{V}$ determines the isomorphism class of $V$ . □

The following easy lemma, which we will use implicitly, is also quite useful for passing between groups.

Lemma 13.4.12. §

Let $V$ be an $F$ -representation of $G$ .

a.: If $H$ is a subgroup of $G$ , then $V$ may be considered as an $F$ -representation of $H$ , and its character is the restriction $χ_{V} |_{H}$ .
b.: If $N$ is a normal subgroup of $G$ and $N$ acts trivially on $V$ , then for $π : G \to G ∕ N$ the quotient map, the character of $V$ as an $F$ -representation of $G ∕ N$ is $π \circ ψ_{V}$ .

For the remainder of this section, we suppose that $G$ is finite and $F$ is algebraically closed of characteristic not dividing $| G |$ .

Proposition 13.4.13. §

Suppose that $G$ is finite and $F$ is algebraically closed. Let $V$ be a finite-dimensional $F$ -representation of $G$ . Then $ρ_{V} (g)$ is diagonalizable.

Proof.

By restricting $ρ_{V}$ to the cyclic subgroup generated by $G$ , we may suppose that $G$ is cyclic, say of order $n$ . In this case, $F [G]$ is a direct sum of $1$ -dimensional representations $V_{i}$ on which $g$ acts by multiplication by $ζ_{n}^{i}$ for $ζ_{n}$ a choice of primitive $n$ th root of unity in $F$ . As $V$ is semisimple, this tells us that $V$ is a direct sum of $1$ -dimensional representations. The automorphism $ρ_{V} (g)$ is then diagonal with respect to any basis of $V$ consisting of one basis element of each of these summands. □

Since $g$ has finite order dividing $| G |$ , the following corollary is immediate.

Corollary 13.4.14. §

Let $V$ be a finite-dimensional $F$ -representation of $G$ . Then the eigenvalues of $ρ_{V} (g)$ for $g \in G$ are all roots of unity of order dividing $| G |$ .

Lemma 13.4.15. §

Let $V$ and $W$ be finite-dimensional $F$ -representations of $G$ . Set $\bar{χ_{V}} (g) = χ_{V} (g^{- 1})$ for all $g \in G$ . Then we have

a.: $χ_{V \otimes_{F} W} = χ_{V} χ_{W}$ and
b.: $χ_{{Hom}_{F} (V, W)} = \bar{χ_{V}} χ_{W}$ .

Proof.

By the commutativity of the tensor product and direct sums and the semisimplicity of $F [G]$ , part a reduces to the case that $V$ and $W$ are irreducible. Through a simple application of Lemma 13.2.23, we may assume that $F$ is algebraically closed. By Proposition 13.4.13, we may then diagonalize the matrices $ρ_{V} (g)$ and $ρ_{W} (g)$ for $g \in G$ with respect to choices of ordered bases $(v_{1}, \dots, v_{n})$ of $V$ and $(w_{1}, \dots, w_{n})$ of $W$ . We then have that $ρ_{V \otimes_{F} W}$ is diagonal with respect to the basis of elements $v_{i} \otimes w_{j}$ with respect to the lexicographical ordering. The diagonal coordinate corresponding to $v_{i} \otimes w_{j}$ is the product of the $(i, i)$ -entry of $ρ_{V} (g)$ and the $(j, j)$ -entry of $ρ_{W} (g)$ . That is,

tr ρ_{V \otimes W} (g) = (tr ρ_{V} (g)) \cdot (tr ρ_{W} (g)),

as desired.

For part b, we recall the isomorphism

{Hom}_{F} (V, W) ≅ V^{*} \otimes_{F} W

of Lemma 13.2.14. We are then reduced by part a to the case that $W = F$ , the trivial $F [G]$ -module. Again replacing $F$ by its algebraic closure, we may diagonalize $ρ_{V} (g)$ for $g \in G$ with respect to a basis $B$ of $V$ . Let $B^{*}$ be its dual basis. For $ϕ \in B^{*}$ and $v \in B$ , we have $ϕ (g^{- 1} v) = α_{v} ϕ (v)$ , where $g^{- 1} v = α_{v} v$ . Thus, the trace of $ρ_{V^{*}} (g)$ agrees with the trace of $ρ_{V} (g^{- 1})$ , as desired. □

Remark 13.4.16. §

Let $G$ be a finite group and $F$ be a field of characteristic not dividing $| G |$ . Since $χ_{V \oplus W} = χ_{V} + χ_{W}$ and $χ_{V \otimes_{F} W} = χ_{V} \cdot χ_{W}$ , the set of $F$ -valued characters of $G$ form a ring with identity the trivial character.

Proposition 13.4.17. §

The irreducible $F$ -characters of $G$ form a basis for the $F$ -vector space of $F$ -valued class functions on $G$ .

Proof.

Let $g_{1}, \dots, g_{r}$ be representatives of the $r$ conjugacy classes of $G$ . The space of $F$ -valued class functions of $G$ has a basis consisting of the maps $𝜃_{i} : G \to F$ for $1 \leq i \leq r$ such that $𝜃_{i} (g) = 1$ if $g \in C_{g_{i}}$ and $𝜃_{i} (g) = 0$ otherwise. On the other hand, there are also $r$ irreducible $F$ -representations $V_{i}$ for $1 \leq i \leq r$ of $G$ by Proposition 13.3.6, so it suffices to see that their characters $χ_{i} = χ_{V_{i}}$ are linearly independent.

Write $F [G] ≅ \prod_{i = 1}^{r} M_{n_{i}} (F)$ in such a way that $V_{i}$ is the isomorphic to the simple module $F^{n_{i}}$ of $M_{n_{i}} (F)$ . Let $e_{i}$ denote the idempotent of $F [G]$ corresponding to the identity of $M_{n_{i}} (F)$ . We may extend $χ_{i}$ $F$ -linearly to a map $χ_{i} : F [G] \to F$ . Then $χ_{i} (x)$ for $x \in F [G]$ is the trace of the endomorphism of $V_{i}$ defined by left multiplication by $x$ . Since left multiplication by $e_{i}$ on $V_{i}$ (resp., $V_{j}$ for $j \neq i$ ) is the identity map (resp., zero map), we have $χ_{i} (e_{i}) = n_{i}$ (resp., $χ_{i} (e_{j}) = 0$ for $j \neq i$ ). Given any linear combination $ϕ = \sum_{i = 1}^{r} a_{i} χ_{i}$ with $a_{i} \in F$ , we have $ϕ (e_{j}) = a_{j} n_{j}$ , so $ϕ = 0$ if and only if $a_{i} = 0$ for all $i$ . □

We can identify the idempotents in $F [G]$ that correspond to identity matrices in terms of characters.

Proposition 13.4.18. §

Let $χ_{i}$ for $1 \leq i \leq r$ denote the irreducible $F$ -characters of $G$ , and let $n_{i}$ denote the degree of $χ_{i}$ Then the elements

e_{i} = \frac{n_{i}}{| G |} \sum_{g \in G} χ_{i} (g^{- 1}) g

are the primitive, central, orthogonal idempotents of $F [G]$ .

Proof.

Let $f_{i}$ denote the primitive central idempotent in $F [G]$ that acts on the identity on the irreducible representation $V_{i}$ with character $χ_{i}$ . Write $f_{i} = \sum_{g \in G} a_{g} g$ with $a_{g} \in F$ for $g \in G$ . For any $g \in G$ , we have

χ_{F [G]} (f_{i} g^{- 1}) = \sum_{h \in G} χ_{F [G]} (a_{h} h g^{- 1}) = a_{g} | G | .

On the other hand, we have $χ_{F [G]} = \sum_{i = 1}^{r} n_{i} χ_{i}$ , where $n_{i} = \dim_{F} V_{i}$ , so

χ_{F [G]} (f_{i} g^{- 1}) = \sum_{j = 1}^{r} n_{j} χ_{j} (f_{i} g^{- 1}) .

If $ρ_{i} : G \to {End}_{F} (V_{i})$ is the $F$ -linear map restricting to $ρ_{V_{i}}$ , then

ρ_{j} (f_{i} g^{- 1}) = ρ_{j} (f_{i}) ρ_{j} (g^{- 1}) = δ_{𝑖𝑗} ρ_{j} (g^{- 1}),

so $χ_{j} (f_{i} g^{- 1}) = δ_{𝑖𝑗} χ_{j} (g^{- 1})$ . Thus, we have

a_{g} | G | = n_{i} χ_{i} (g^{- 1}) .

It follows that

f_{i} = \frac{n_{i}}{| G |} \sum_{g \in G} χ_{i} (g^{- 1}) g .

□

13.5. Character tables

In this section, we focus on the theory of $ℂ$ -valued characters of a finite group $G$ .

Definition 13.5.1. §

A character table of a finite group $G$ is a matrix in $M_{r} (ℂ)$ , where $r$ is the number of conjugacy classes of $G$ with $(i, j)$ -entry $χ_{i} (g_{j})$ , where $χ_{i}$ for $1 \leq i \leq r$ are the distinct characters of the irreducible $ℂ$ -representations of $G$ and $g_{i}$ for $1 \leq i \leq r$ are representatives of the distinct conjugacy classes in $G$ .

Usually, a character table is written in a table format, as in the following example.

Example 13.5.2. §

Take the group $S_{3}$ . Let $χ_{1}$ denote the trivial character, $χ_{2}$ denote the sign character, and $χ_{3}$ the irreducible character of dimension $2$ . By Example 13.3.8, the character table of $S_{3}$ is then as follows:

$S_{3}$	1	$(1 2)$	$(1 2 3)$

$χ_{1}$	1	1	1
$χ_{2}$	1	$- 1$	1
$χ_{3}$	2	0	$- 1$

Recall that $\bar{α}$ denotes the complex conjugate of a complex number $α$ .

Lemma 13.5.3. §

Let $χ$ be a $ℂ$ -valued character of degree $d$ of a finite group $G$ of order $n$ . For $g \in G$ , we have $χ (g) \in ℤ [ζ_{n}]$ , $| χ (g) | \leq d$ , and $χ (g^{- 1}) = \bar{χ (g)}$ .

Proof.

Let $ρ : G \to {Aut}_{ℂ} (V)$ be the representation corresponding to $χ$ . By Corollary 13.4.14, the value $ρ (g)$ can be diagonalized to matrix with entries in $μ_{n}$ . Since the inverse of a root of unity is its complex conjugate, $ρ (g^{- 1}) = ρ {(g)}^{- 1}$ may be then be represented by the diagonal matrix $A^{- 1} = (\bar{a_{𝑖𝑗}})$ . Then $χ (g)$ is a sum of $d$ roots of unity of order dividing $n$ , which the first two statements, and we have

χ (g^{- 1}) = tr (A^{- 1}) = \bar{tr (A)} = \bar{χ (g)} .

□

Let us set $ζ_{n} = e^{2 𝜋𝑖 ∕ n} \in ℂ$ for $n \geq 1$ . The following lemma is useful for producing new characters out of old.

Lemma 13.5.4. §

Let $G$ be a group of order $n$ , and let $σ \in Gal (ℚ (ζ_{n}) ∕ ℚ)$ , where $ζ_{n} = e^{2 𝜋𝑖 ∕ n} \in ℂ$ . If $χ$ is a character of $G$ , then so is $χ^{σ} : G \to F$ defined by $χ^{σ} (g) = σ (χ (g))$ for all $g \in G$ . Moreover, if $a \in {(ℤ ∕ 𝑛ℤ)}^{\times}$ is such that $σ (ζ_{n}) = ζ_{n}^{a}$ , then $χ^{σ} (g) = χ (g^{a})$ .

Proof.

By Proposition 13.4.13, if $G$ is finite of order $n$ , then every $ℂ$ -representation of $G$ is the base change of a $ℚ (ζ_{n})$ -representation. Let $V$ be the $ℚ (μ_{n})$ -representation of $G$ with character $χ$ . As a vector space, $𝑉≅ℚ {(μ_{n})}^{d}$ for some $d \geq 0$ , and so $σ$ induces an automorphism $σ : V \to V$ as the direct sum of the automorphisms $ℚ (μ_{n}) \to ℚ (μ_{n})$ . Then $σ \circ ρ_{V}$ is again a representation, and its character is $χ^{σ}$ . Note that $σ \circ ρ_{V} (g) = ρ_{V} (g^{a})$ , since in diagonalized form, the entries of $ρ_{V} (g)$ are all elements of $μ_{n}$ upon which $σ$ acts by raising to the $a$ th power. □

We pause for a moment to discuss inner products on $ℂ$ -vector spaces.

Definition 13.5.5. §

An inner product on an $ℂ$ -vector space $V$ is a map $⟨, ⟩ : V \times V \to ℂ$ that satisifes

\begin{array}{rcl} ⟨ 𝛼𝑣 + v^{'}, w ⟩ = α ⟨ v, w ⟩ + ⟨ v^{'}, w ⟩ & and & ⟨ v, 𝛽𝑤 + w^{'} ⟩ = \bar{β} ⟨ v, w ⟩ + ⟨ v, w^{'} ⟩ \end{array}

for all $v, v^{'}, w, w^{'} \in V$ and $α, β \in ℂ$ .

Terminology 13.5.6. §

That $⟨, ⟩$ is an inner product on a $ℂ$ -vector space $V$ may be expressed as saying that it is left $ℂ$ -linear (or just linear) and right conjugate linear.

Definition 13.5.7. §

An inner product $⟨, ⟩$ on a $ℂ$ -vector space $V$ is positive definite if $⟨ v, v ⟩ \geq 0$ for all $v \in V$ , with equality only for $v = 0$ .

Definition 13.5.8. §

An inner product $⟨, ⟩$ on a $ℂ$ -vector space $V$ is Hermitian if it is positive definite and $⟨ v, w ⟩ = \bar{⟨ w, v ⟩}$ for all $v, w \in V$ .

Definition 13.5.9. §

A basis $B$ of a $ℂ$ -vector space $V$ with a Hermitian inner product $⟨, ⟩$ is orthonormal if $⟨ v, w ⟩ = δ_{v, w}$ for all $v, w \in B$ .

Definition 13.5.10. §

A complex inner product space is a pair consisting of a $ℂ$ -vector space $V$ and a Hermitian inner product on $V$ .

Example 13.5.11. §

The dot product on $ℂ^{n}$ defined by

(a_{1}, a_{2}, \dots, a_{n}) \cdot (b_{1}, b_{2}, \dots, b_{n}) = \sum_{i = 1}^{n} a_{i} \bar{b_{i}}

is a positive definite, Hermitian inner product on $ℂ^{n}$ . The standard basis of $ℂ^{n}$ is orthonormal with respect to the dot product, so $ℂ^{n}$ is an inner product space with respect to the dot product.

Definition 13.5.12. §

An inner product $⟨, ⟩$ on a $ℂ$ -representation $V$ of $G$ is said to be $G$ -invariant, or an invariant inner product, if $⟨ 𝑔𝑣, 𝑔𝑤 ⟩ = ⟨ v, w ⟩$ for all $v, w \in V$ .

The following provides a useful example.

Lemma 13.5.13. §

Let $⟨, ⟩$ be a Hermitian inner product on a $ℂ$ -representation $V$ of $G$ . Then the map $[,] : V \times V \to ℂ$ defined on $v, w \in V$ by

[v, w] = \frac{1}{| G |} \sum_{g \in G} ⟨ 𝑔𝑣, 𝑔𝑤 ⟩

is a $G$ -invariant inner product on $V$ .

Proof.

As a positive real scalar multiple of a sum of Hermitian inner products on $V$ , the pairing $[,]$ is also Hermitian. The invariance by an element of $G$ follows by reindexing the sum. □

The next lemma contains the definition of an inner product on the space of $ℂ$ -valued class functions of $G$ .

Lemma 13.5.14. §

The function which assigns to a pair $(𝜃, ψ)$ of $ℂ$ -valued class function of $G$ the value

⟨ 𝜃, ψ ⟩ = \frac{1}{| G |} \sum_{g \in G} 𝜃 (g) \bar{ψ (g)}

is a positive definite, Hermitian inner product on the space of $ℂ$ -valued class functions of $G$ .

We consider the space of class functions as a Hermitian inner product space with respect to the Hermitian inner product of Lemma 13.5.14.

Remark 13.5.15. §

If we let $h \in G$ act on the space of class functions on $G$ by $(h \cdot 𝜃) (g) = 𝜃 (h^{- 1} g)$ , then the resulting inner product is $G$ -invariant.

Remark 13.5.16. §

The inner product of the characters of any two $ℂ$ -representations $V$ and $W$ is real by Lemma 13.5.3, since

\sum_{g \in G} χ_{V} (g) \bar{χ_{W} (g)} = \sum_{g \in G} χ_{V} (g^{- 1}) \bar{χ_{W} (g^{- 1})} = \sum_{g \in G} \bar{χ_{V} (g)} χ_{W} (g) .

Lemma 13.5.17. §

Let $V$ be a $ℂ [G]$ -module of finite dimension. Then

\dim_{ℂ} V^{G} = \frac{1}{| G |} \sum_{g \in G} χ_{V} (g) .

Proof.

Let $z = \frac{1}{| G |} N_{G} \in ℂ [G]$ . Since $N_{G}^{2} = | G |$ , the element $z$ is an idempotent. The $ℂ$ -linear endomorphism $T$ of $V$ defined by left multiplication by $z$ therefore has minimal polynomial dividing $x^{2} - x = x (x - 1)$ . In particular, it is diagonalizable. The trace of $T$ is then the sum of its nontrivial eigenvalues, which is the dimension of the eigenspace $E_{1} (T)$ of $1$ . It remains then only to show that $E_{1} (T) = V^{G}$ . We check this on $v \in V$ : if $𝑧𝑣 = v$ , then $𝑔𝑣 = 𝑔𝑧𝑣 = 𝑧𝑣 = v$ for all $g \in G$ , while if $𝑔𝑣 = v$ for all $g \in G$ , then $𝑧𝑣 = \frac{1}{| G |} | G | v = v$ . □

Proposition 13.5.18. §

Let $V$ and $W$ be complex $G$ -representations. Then

⟨ χ_{V}, χ_{W} ⟩ = \dim_{ℂ} {Hom}_{ℂ [G]} (V, W) .

Proof.

Note that ${Hom}_{ℂ [G]} (V, W) = {Hom}_{ℂ} {(V, W)}^{G}$ , where $g \in G$ acts on $ϕ \in {Hom}_{ℂ} (V, W)$ by $(g \cdot ϕ) (v) = 𝑔𝜙 (g^{- 1} v)$ for every $v \in V$ . Thus,

\dim_{ℂ} {Hom}_{ℂ [G]} (V, W) = \frac{1}{| G |} \sum_{g \in G} χ_{{Hom}_{ℂ} (V, W)} (g) = \frac{1}{| G |} \sum_{g \in G} \bar{χ_{V} (g)} χ_{W} (g),

the last step by Lemma 13.4.15b. □

We may now prove the orthogonality of the basis of characters.

Theorem 13.5.19 (First orthogonality relation). §

The set of irreducible complex characters of a finite group $G$ forms an orthonormal basis of the space of $ℂ$ -valued class functions of $G$ .

Proof.

Let $V_{i}$ for $1 \leq i \leq r$ be the distinct irreducible $ℂ [G]$ -modules, and let $χ_{i}$ denote the character of $V_{i}$ . For any $1 \leq i, j \leq r$ , we have

⟨ χ_{i}, χ_{j} ⟩ = \dim_{ℂ} {Hom}_{ℂ [G]} (V_{i}, V_{j}),

by Proposition 13.5.18. The result then follows by Schur’s lemma. □

This orthogonality also gives us a sort of orthogonality of rows of the character table.

Remark 13.5.20. §

Let ${g_{1}, g_{2}, \dots, g_{r}}$ be a set of representatives of the conjugacy classes of a finite group $G$ . Let $χ_{1}, χ_{2}, \dots, χ_{r}$ be the complex irreducible character. By Theorem 13.5.19, the rows $r_{i}$ of the character table with $(i, j)$ -entry $χ_{i} (g_{j})$ are orthogonal with respect to the weighted dot product

r_{i} \cdot r_{i^{'}} = \frac{1}{| G |} \sum_{j = 1}^{r} c_{j} χ_{i} (g_{j}) \bar{χ_{i^{'}} (g_{j})} = ⟨ χ_{i}, χ_{i^{'}} ⟩,

where the weight $c_{j}$ is the order of the conjugacy class of $g_{j}$ .

We also have an orthogonality relation for columns.

Theorem 13.5.21 (Second orthogonality relation). §

Let $G$ be a finite group, and let $χ_{1}, χ_{2}, \dots, χ_{r}$ be its distinct irreducible, complex characters. For any $g, h \in G$ , we have

\sum_{i = 1}^{r} χ_{i} (g) \bar{χ_{i} (h)} = {\begin{matrix} | Z_{g} | & if g and h are conjugate, \\ 0 & otherwise, \end{matrix}

where $Z_{g}$ denotes the centralizer of $g$ in $G$ .

Proof.

Let $g_{1}, g_{2}, \dots, g_{r}$ represent the distinct conjugacy classes in $G$ , and let $A \in M_{r} (ℂ)$ be the matrix with $(i, j)$ -entry $χ_{i} (g_{j})$ . Let $C$ be the diagonal matrix with $(i, i)$ -entry $c_{i} = | C_{g_{i}} |$ . Then

{(𝐴𝐶 {\bar{A}}^{t})}_{𝑖𝑗} = \sum_{k = 1}^{r} χ_{i} (g_{k}) c_{k} \bar{χ_{j} (g_{k})} = δ_{𝑖𝑗} | G |,

the last step by Remark 13.5.20. In particular, $𝐴𝐶 {\bar{A}}^{t}$ is a scalar multiple of the identity matrix, so $𝐴𝐶 {\bar{A}}^{t} = {\bar{A}}^{t} 𝐴𝐶$ , which tells us that

δ_{𝑖𝑗} | G | = {({\bar{A}}^{t} 𝐴𝐶)}_{𝑖𝑗} = \sum_{k = 1}^{r} \bar{χ_{k} (g_{i})} χ_{k} (g_{j}) c_{j} .

Since $| Z_{g_{j}} | = | G | c_{j}^{- 1}$ by the orbit-stabilizer theorem, we are done. □

Let us study character tables in some examples.

Example 13.5.22. §

Let $n \geq 1$ , and let $ζ = e^{2 𝜋𝑖 ∕ n} \in ℂ$ . Every character of $ℤ ∕ 𝑛ℤ$ is a power of the character $χ : ℤ ∕ 𝑛ℤ \to ℂ^{\times}$ given by $χ (i) = ζ^{i}$ . Since $ℤ ∕ 𝑛ℤ$ is abelian, every element of $ℤ ∕ 𝑛ℤ$ is the lone element in its conjugacy class. The character table of $ℤ ∕ 𝑛ℤ$ is as follows:

$ℤ ∕ 𝑛ℤ$	0	1	2	$\dots$	$n - 1$

$1$	1	1	1	$\dots$	1
$χ$	1	$ζ$	$ζ^{2}$	$\dots$	$ζ^{n - 1}$
$χ^{2}$	1	$ζ^{2}$	$ζ^{4}$	$\dots$	$ζ^{n - 2}$
$⋮$	$⋮$	$⋮$	$⋮$	$⋱$	$⋮$
$χ^{n - 1}$	1	$ζ^{n - 1}$	$ζ^{n - 2}$	$\dots$	$ζ$ .

Remark 13.5.23. §

In general, the number of $1$ -dimensional complex characters of a finite group $G$ is $| G^{ab} |$ , since these are exactly the irreducible representations of $ℂ [G^{ab}]$ , which has $ℂ$ -dimension $| G^{ab} |$ .

Example 13.5.24. §

Let $G$ be any nonabelian group of order $8$ . By Theorem 7.5.2, there are two up to isomoprhism, $D_{4}$ and $Q_{8}$ . The center $Z$ of $G$ has order $2$ (it is nontrivial since $G$ is a $2$ -group and if it had order at least $4$ , it is easy to see that the group would be abelian). Furthermore, $G ∕ Z$ is abelian since all groups of order $4$ are abelian. This also means that $G ∕ Z$ is the abelianization of $G$ . It is also easy to see that this implies $G ∕ 𝑍≅ℤ ∕ 2 ℤ \oplus ℤ ∕ 2 ℤ$ . So $G$ has four characters $χ_{1}, \dots, χ_{4}$ of degree $1$ and therefore one character $χ_{5}$ of degree $2$ to make $8 = 2^{2} + 1 + 1 + 1 + 1$ . Pick representatives $g$ and $h$ in $G$ of the two summands $ℤ ∕ 2 ℤ$ . Then $g$ , $h$ , and $𝑔h$ must be representatives of distinct conjugacy classes, which are then forced to have order $2$ since $g, h, 𝑔h \notin Z$ . Finally, let $z$ generate the center. The character table is

$G$	1	$z$	$g$	$h$	$𝑔h$

$χ_{1}$	1	1	1	1	1
$χ_{2}$	1	1	1	$- 1$	$- 1$
$χ_{3}$	1	1	$- 1$	1	$- 1$
$χ_{4}$	1	1	$- 1$	$- 1$	1
$χ_{5}$	2	$- 2$	0	0	0

The last row is determined by orthogonality of the columns, since its first entry must be $2$ . Note that this implies that the isomorphism type of a group is not determined by its character.

Example 13.5.25. §

Note that $S_{4}$ has $5$ conjugacy classes, and the sums of the squares of the degrees $n_{i}$ of the $5$ irreducible characters $χ_{i}$ equals $| 24 |$ . Also $S_{4} \to ℤ ∕ 2 ℤ$ via the sign map, so there are (at least) two $1$ -dimensional characters: the trivial character $χ_{1}$ and the sign character $χ_{2}$ . Since $n_{3}^{2} + n_{4}^{2} + n_{5}^{2} = 22$ , we have $n_{3} = 2$ and $n_{4} = n_{5} = 3$ (if put in increasing order). The quotient of $S_{4}$ by the normal subgroup $⟨ (1 2) (3 4), (1 3) (2 4) ⟩$ is isomorphic to $S_{3}$ , so we obtain by composition with the irreducible two-dimensional representation $ρ : S_{3} \to {GL}_{2} (ℂ)$ a two-dimensional representation $χ_{3} : S_{4} \to {GL}_{2} (ℂ)$ , which is nonabelian and hence irreducible, being semisimple. Whatever $χ_{4}$ is, note that if we tensor its representation $V_{4}$ with the representation $V_{2}$ of the sign character $χ_{2}$ , we obtain another irreducible character $χ_{V_{4} \otimes_{ℂ} V_{2}} = χ_{4} χ_{2}$ of dimension $3$ , which we call $χ_{5}$ . (We will see that it is actually different from $χ_{4}$ .) The character table is

Color key: blue marks row χ₃ outside the identity column; red marks rows χ₄ and χ₅ outside the identity column.
$S_{4}$	$e$	$(1 2)$	$(1 2 3)$	$(1 2 3 4)$	$(1 2) (3 4)$

$χ_{1}$	1	1	1	1	1
$χ_{2}$	1	$- 1$	1	$- 1$	1
$χ_{3}$	2	0	-1	0	2
$χ_{4}$	3	1	0	$- 1$	-1
$χ_{5}$	3	$- 1$	0	$1$	-1.

The entries in blue are determined from the character table for $S_{3}$ . The entries in red are obtained by noting that $χ_{5} = χ_{4} χ_{2}$ and using orthogonality of columns. (That is, the third and fifth columns are determined using orthogonality with the first and the second and fourth, up to sign, using orthogonality with the first and each other.)

Note that we can restrict representations and characters to subgroups: for $H \leq G$ and a $G$ -representation $V$ , this amounts to considering $V$ as a module over the group ring of $H$ and restricting the function $χ_{V}$ to $H$ .

Example 13.5.26. §

The group $A_{4}$ has $4$ conjugacy classes with representatives $e$ , $(1 2 3)$ , $(1 3 2)$ , and $(1 2 3 4)$ . We have $A_{4}^{ab} ≅ℤ ∕ 3 ℤ$ , generated by the image of $(1 2 3)$ , so there are three abelian characters which are the powers of the character $χ$ such that $χ ((1 2 3)) = ω$ , where $ω = e^{2 𝜋𝑖 ∕ 3}$ . Since $| A_{4} | = 12$ , there is one more character $ψ$ , which has degree $3$ . Its values can be calculated by orthogonality of columns, yielding the character table

$A_{4}$	$e$	$(1 2 3)$	$(1 3 2)$	$(1 2) (3 4)$

$1$	1	1	1	1
$χ$	1	$ω$	$ω^{2}$	1
$χ^{2}$	1	$ω^{2}$	$ω$	1
$ψ$	3	0	0	-1.

13.6. Induced representations

Let $G$ be a finite group and $H$ a subgroup. For a commutative ring $R$ we can view a $R [G]$ -module $A$ as an $R [H]$ -module in the obvious fashion. When thinking of $A$ as an $R [H]$ -module, it is often helpful to give it a new name and symbol

Definition 13.6.1. §

An $R [G]$ -module $A$ viewed as an $R [H]$ -module is called the restriction of $A$ from $G$ to $H$ and is denoted by ${Res}_{H}^{G} (A)$ .

Together with the obvious definition on morphisms, restriction defines an exact functor

{Res}_{H}^{G} : R [G] -mod \to R [H] -mod .

The natural question arises as to whether or not ${Res}_{H}^{G}$ has an adjoint, and indeed, it has a left adjoint. We first give the construction.

Definition 13.6.2. §

Let $H$ be a subgroup of a group $G$ , and let $R$ be a commutative ring. The induced module from $H$ to $G$ of an $R [H]$ -module $B$ is the $R [G]$ -module

{Ind}_{H}^{G} (B) = {Hom}_{R [H]} (R [G], B)

where for $φ \in {Ind}_{H}^{G} (B)$ , we let $g \in G$ act by $(g \cdot φ) (x) = φ (𝑥𝑔)$ for all $x \in F [G]$ .

Remark 13.6.3. §

Since $R [G]$ is $R [H]$ -free, ${Ind}_{H}^{G}$ provides an exact functor from $R [H] -mod$ to $R [G] -mod$ , since is the functor $h_{R [G]}$ in our earlier notation.

If $H$ is of finite index in $G$ , we have an alternate description of the induced module. That is, there is another way in which to produce an $R [G]$ -module from an $R [H]$ -module $B$ using tensor products. That is, we can take the $R [G]$ -module $R [G] \otimes_{R [H]} B$ , where $g \in G$ acts on $x \otimes b$ in the tensor product by $g (x \otimes b) = 𝑔𝑥 \otimes b$ .

Proposition 13.6.4. §

Let $H$ be a finite index subgroup of $G$ , and let $R$ be a commutative ring. Given an $R [H]$ -module $B$ , there is natural isomorphism

κ : {Ind}_{H}^{G} (B) \to_{}^{\sim} R [G] \otimes_{R [H]} B

given on $φ \in {Ind}_{H}^{G} (B)$ by

κ (φ) = \sum_{\bar{g} \in H ∖ G} g^{- 1} \otimes φ (g),

where for each $\bar{g} \in H ∖ G$ , the element $g \in G$ is a choice of representative of $\bar{g}$ .

Proof.

First, we note that $χ$ is a well-defined map, as

{(h𝑔)}^{- 1} \otimes φ (h𝑔) = g^{- 1} h^{- 1} \otimes h𝜑 (g) = g \otimes φ (g)

for $φ \in R [G] \otimes_{R [H]} B$ , $h \in H$ , and $g \in G$ . Next, we see that $χ$ is an $R [G]$ -module homomorphism, as

χ (g^{'} φ) = \sum_{\bar{g} \in H ∖ G} g^{- 1} \otimes φ (g g^{'}) = g^{'} \sum_{\bar{g} \in H ∖ G} {(g g^{'})}^{- 1} \otimes φ (g g^{'}) = g^{'} χ (φ)

for $g^{'} \in G$ . As the coset representatives form a basis of $R [G]$ as a free $R [H]$ -module, we may define an inverse to $χ$ that maps

\sum_{\bar{g} \in H ∖ G} g^{- 1} \otimes b_{g} \in {Ind}_{H}^{G} (B)

to the unique $R [H]$ -linear map $φ$ that takes the value $b_{g}$ on $g$ for the chosen representative of $\bar{g} \in H ∖ G$ . □

Proposition 13.6.5. §

Let $H$ be a finite index subgroup of $G$ . Then ${Ind}_{H}^{G}$ is left adjoint to ${Res}_{H}^{G}$ .

Proof.

Using the alternate characterization of ${Ind}_{H}^{G} U$ of Proposition 13.6.4 and the adjointness of $Hom$ and $\otimes$ , we have

{Hom}_{R [G]} (R [G] \otimes_{R [H]} U, V) ≅ {Hom}_{R [H]} (U, {Hom}_{R [G]} (R [G], V)) ≅ {Hom}_{R [H]} (U, V),

the latter isomorphism being induced by evaluation at $1$ in the second variable. □

Definition 13.6.6. §

Let $H$ be a subgroup of a group $G$ , let $W$ be an $F$ -representation of $H$ .

a.: The induced representation from $H$ to $G$ of $W$ is ${Ind}_{H}^{G} (W)$ .
b.: If $H$ has finite index in $G$ and $W$ is finite-dimensional with character $ψ$ , then the induced character ${Ind}_{H}^{G} (ψ)$ of $ψ$ is the character of ${Ind}_{H}^{G} (W)$ .

Example 13.6.7. §

Let $H$ be a finite index subgroup of $G$ . For the trivial representation $F$ of $H$ , we have

{Ind}_{H}^{G} (F) ≅ F [G] \otimes_{F [H]} F ≅ F [G ∕ H],

where the latter module is the $F [G]$ -module with $F$ -basis the left $G$ -set $G ∕ H$ . That is, ${Ind}_{H}^{G} (F)$ is the permutation representation for the left action of $G$ on $G ∕ H$ . Thus, the induced character $χ = {Ind}_{H}^{G} (ψ_{1})$ of the trivial character $ψ_{1}$ on $H$ has the property that $χ (g)$ is the number of left $H$ -cosets fixed by left multiplication by $g \in G$ .

Remark 13.6.8. §

Let $H$ be a finite index subgroup of $G$ . The induced representation of the regular representation $F [H]$ of $H$ is the regular representation $F [G]$ of $G$ . In particular, all irreducible $F$ -representations of $G$ are summands of induced representations of the irreducible $F$ -representations of $H$ .

Remark 13.6.9. §

In finite group theory, one often uses the alternate tensor product characterization provided by Proposition 13.6.4 as the definition of the induced representation.

Notation 13.6.10. §

For a subgroup $H$ of a group $G$ , we denote the restriction of a character $χ$ of $G$ to $H$ by ${Res}_{H}^{G} χ$ , or more simply $χ |_{H}$ .

Notation 13.6.11. §

For the inner product of Lemma 13.5.14, we use ${⟨, ⟩}_{G}$ to indicate its dependence on the group $G$ .

Definition 13.6.12. §

For a character $χ$ of a finite-dimensional $ℂ$ -representation of a group, the multiplicity of an irreducible character $ψ$ in it is the multiplicity of the irreducible representation with character $ψ$ in the representation with character $χ$ .

We have the following corollary of Proposition 13.6.4.

Corollary 13.6.13 (Frobenius reciprocity). §

Let $G$ be a finite group and $H$ a subgroup. Let $ψ$ be a $ℂ$ -valued character of $H$ and $χ$ be a $ℂ$ -valued character of $G$ . Then

{⟨ {Ind}_{H}^{G} ψ, χ ⟩}_{G} = {⟨ ψ, {Res}_{H}^{G} χ ⟩}_{H} .

Proof.

Let $ψ = χ_{U}$ and $χ = χ_{V}$ for representations $U$ and $V$ of $H$ and $G$ , respectively. By Propositions 13.5.18 and 13.6.5, we have

{⟨ {Ind}_{H}^{G} ψ, χ ⟩}_{G} = \dim_{ℂ} {Hom}_{ℂ [G]} ({Ind}_{H}^{G} U, V) = \dim_{ℂ} {Hom}_{ℂ [H]} (U, V) = {⟨ ψ, {Res}_{H}^{G} χ ⟩}_{H} .

□

We can construct tables that contain these values of the pairings in Corollary 13.6.13.

Definition 13.6.14. §

The induction-restriction table of a subgroup $H$ of a finite group $G$ is the matrix with rows indexed by the complex irreducible characters $ψ_{i}$ of $H$ and columns by the complex irreducible characters $χ_{j}$ of $G$ with $(i, j)$ -entry ${⟨ ψ_{i}, {Res}_{H}^{G} χ_{j} ⟩}_{H}$ .

Example 13.6.15. §

Again let $χ_{i}$ for $1 \leq i \leq 5$ and $1$ , $χ$ , $χ^{2}$ , and $ψ$ be the characters of $S_{4}$ and $A_{4}$ , respectively of Examples 13.5.25 and 13.5.26. From the character tables of $G = S_{4}$ and $H = A_{4}$ , we see that $χ_{1} |_{H} = χ_{2} |_{H} = 1$ , $χ_{3} |_{H} = χ + χ^{2}$ , and $χ_{4} | H = χ_{5} |_{H} = ψ$ . The induction-restriction table is

	$χ_{1}$	$χ_{2}$	$χ_{3}$	$χ_{4}$	$χ_{5}$

$1$	1	1	0	0	0
$χ$	0	0	1	0	0
$χ^{2}$	0	0	0	1	0
$ψ$	0	0	0	1	1.

Frobenius reciprocity tells us that ${Ind}_{H}^{G} 1 = χ_{1} + χ_{2}$ , ${Ind}_{H}^{G} χ = {Ind}_{H}^{G} χ^{2} = χ_{3}$ , and ${Ind}_{H}^{G} ψ = χ_{4} + χ_{5}$ .

Proposition 13.6.16. §

Let $H$ be a finite index subfroup of a group $G$ , and let $g_{1}, \dots, g_{k}$ be a system of left coset representatives of $H$ in $G$ . For a character $ψ$ of $H$ , extend $ψ$ to a function $\tilde{ψ} : G \to ℂ$ by setting $\tilde{ψ} (g) = 0$ if $g \notin H$ . For $g \in G$ , we then have

{Ind}_{H}^{G} (ψ) (g) = \sum_{i = 1}^{k} \tilde{ψ} (g_{i}^{- 1} g g_{i}) .

Proof.

Let $W$ be an $m$ -dimensional representation of $H$ with character $ψ$ , and let $B = (w_{1}, \dots, w_{m})$ be an ordered $F$ -basis of $W$ . Recall that

{Ind}_{H}^{G} (W) ≅ F [G] \otimes_{F [H]} W

We have a basis $g_{i} \otimes w_{j}$ of $F [G] \otimes_{F [H]} W$ for $1 \leq i \leq k$ and $1 \leq j \leq m$ with the lexicographical ordering.

For a given $1 \leq i \leq k$ , any $g \in G$ satisfies

g g_{i} = g_{σ (i)} h_{i}

for some $σ \in S_{k}$ and $h_{i} \in H$ . Then

g (g_{i} \otimes w_{j}) = g_{σ (i)} \otimes h_{i} w_{j} .

With respect to the given basis, the matrix of $ρ_{W} (g)$ is a $k$ -by- $k$ matrix of blocks in $M_{m} (F)$ with one nonzero block in each row and each column, i.e., the blocks with coordinates $(i, σ (i))$ , which are those representing $ρ_{W} (h_{i})$ with respect to the basis $B$ .

Adding up the diagonal entries in the $(i, i)$ -block, we get $0$ if $i^{'} \neq i$ and $ψ (h) = ψ (g_{i}^{- 1} g g_{i})$ if $i^{'} = i$ . By definition of $\tilde{ψ}$ , this equals $\tilde{ψ} (g_{i}^{- 1} g g_{i})$ in all cases. Summing over $i$ , we obtain the result. □

Corollary 13.6.17. §

Let $G$ be a finite group and $H$ be a subgroup of $G$ . Let $ψ$ be a character of $H$ . For $g \in G$ , we then have

{Ind}_{H}^{G} (ψ) (g) = \frac{1}{| H |} \sum_{\begin{array}{c} k \in G \\ 𝑘𝑔 k^{- 1} \in H \end{array}} ψ (k^{- 1} 𝑔𝑘) .

Proof.

This follows from the formula of Proposition 13.6.16. To see that, take $\tilde{ψ}$ as in its statement, and note that for any $h \in H$ , we have

\tilde{ψ} ({(g_{i} h)}^{- 1} g (g_{i} h)) = \tilde{ψ} (h^{- 1} (g_{i}^{- 1} g g_{i}) h),

since conjugation by $h^{- 1}$ preserves $H$ and $G - H$ , and $ψ$ is a class function on $H$ . □

The following corollary is immediate.

Corollary 13.6.18. §

Let $H$ be a finite index normal subgroup of $G$ , and let $χ$ be a character of $H$ . Then ${Ind}_{H}^{G} (ψ) (g) = 0$ for all $g \notin H$ .

Example 13.6.19. §

Consider the dihedral group $G = D_{2 n}$ of order $4 n$ with $n \geq 3$ . It has abelianization the Klein $4$ -group generated by the images of $r$ and $s$ , so $G$ has four degree $1$ characters $χ_{1}, χ_{2}, χ_{3}, χ_{4}$ with $χ_{1}$ trivial, $χ_{2} (r) = χ_{3} (s) = - 1$ and $χ_{2} (s) = χ_{3} (r) = 1$ , and $χ_{4} = χ_{2} χ_{3}$ . Now, consider the cyclic subgroup $H = ⟨ r ⟩$ , which has characters $ψ^{i}$ for $0 \leq i \leq 2 n - 1$ with $ψ (r) = ζ_{2 n}$ . The induced character $𝜃_{i}$ of $ψ^{i}$ is trivial on all reflections and satisfies

𝜃_{i} (r^{j}) = ζ_{2 n}^{𝑖𝑗} + ζ_{2 n}^{- 𝑖𝑗} .

The characters $𝜃_{i} = {Ind}_{H}^{G} ψ^{i}$ with $1 \leq i \leq n - 1$ are all distinct degree $2$ characters which are clearly not sums of the $χ_{i}$ , so they are irreducible characters of $G$ . The sum of the squares of the dimensions of these characters is $4 \cdot 1^{2} + (n - 1) \cdot 2^{2} = 4 n = | G |$ , so these are all of the irreducible characters on $G$ . Setting

ξ_{k} = ζ_{2 n}^{k} + ζ_{2 n}^{- k} = 2 \cos (𝑘𝜋 ∕ n) \in ℝ,

the character table is then as follows.

$D_{2 n}$	$e$	$s$	$𝑟𝑠$	$r$	$r^{2}$	$\dots$	$r^{n - 1}$	$r^{n}$

$χ_{1}$	1	1	1	1	1	$\dots$	1	1
$χ_{2}$	1	1	$- 1$	$- 1$	1	$\dots$	${(- 1)}^{n - 1}$	${(- 1)}^{n}$
$χ_{3}$	1	$- 1$	$- 1$	1	1	$\dots$	1	1
$χ_{4}$	1	$- 1$	1	$- 1$	1	$\dots$	${(- 1)}^{n - 1}$	${(- 1)}^{n}$
$𝜃_{1}$	2	0	0	$ξ_{1}$	$ξ_{2}$	$\dots$	$ξ_{n - 1}$	$- 2$
$𝜃_{2}$	2	0	0	$ξ_{2}$	$ξ_{4}$	$\dots$	$ξ_{2 (n - 1)}$	2
$𝜃_{3}$	2	0	0	$ξ_{3}$	$ξ_{6}$	$\dots$	$ξ_{3 (n - 1)}$	$- 2$
$⋮$	$⋮$	$⋮$	$⋮$	$⋮$	$⋮$	$⋱$	$⋮$	$⋮$
$𝜃_{n - 1}$	2	0	0	$ξ_{n - 1}$	$ξ_{2 (n - 1)}$	$\dots$	$ξ_{{(n - 1)}^{2}}$	${(- 1)}^{n - 1} 2$ .

Here, one might note that $ξ_{0} = 2$ , and $ξ_{k} = ξ_{k + 2 n} = - ξ_{k + n} = - ξ_{n - k}$ for all $k$ . We remark that $𝜃_{i} |_{H} = ψ^{i} + ψ^{- i}$ for all $1 \leq i \leq n - 1$ , while $𝜃_{0} = χ_{1} + χ_{3}$ and $𝜃_{n} = χ_{2} + χ_{4}$ , consistent with Frobenius reciprocity.

We also give a formula which tells us explicitly how to determine the induced character to $G$ of an $H$ -character from the character table for $H$ and knowledge of conjugacy classes.

Proposition 13.6.20. §

Let $G$ be a finite group and $H$ be a subgroup of $G$ . Let $ψ$ be a character of $G$ , let $g \in G$ , and let $C_{g}$ be the conjugacy class of $g$ in $G$ . Write $H \cap C_{g}$ as a possibly empty disjoint union of conjugacy classes $T_{1}, \dots, T_{l}$ of $H$ . For $1 \leq i \leq l$ , let $h_{i}$ be a representative of $T_{i}$ . Then

{Ind}_{H}^{G} ψ (g) = [G : H] \sum_{i = 1}^{l} \frac{| T_{i} |}{| C_{g} |} ψ (h_{i}) .

Proof.

This is a matter of counting. That is, by Corollary 13.6.17, we must show that the number of $k \in G$ such that $k^{- 1} 𝑔𝑘$ is conjugate to $h_{i}$ in $H$ is $| Z_{g} | | T_{i} |$ , where $Z_{g}$ denotes the centralizer of $g$ in $G$ . We know that there are $| Z_{g} |$ elements of $G$ that conjugate $g$ to any particular element of $C_{g}$ . Thus, there are $| Z_{g} | | T_{i} |$ elements in $G$ that conjugate $g$ to one of the elements in $T_{i}$ , as desired. □

We may use Proposition 13.6.20 to determine the induced characters on a group from the characters on its subgroup.

Example 13.6.21. §

Take $H = S_{3}$ , which we view as a subgroup of $G = S_{4}$ . Recall that the conjugacy classes of $S_{4}$ are determined by cycle type, with conjugacy classes $C_{1}, \dots, C_{5}$ corresponding to cycle types $e$ , $(1 2)$ , $(1 2 3)$ , $(1 2 3 4)$ , and $(1 2) (3 4)$ having orders $1$ , $6$ , $8$ , $6$ , $3$ , respectively. Now, $C_{4}$ and $C_{5}$ contain no elements of $S_{3}$ , while $C_{1}$ , $C_{2}$ , and $C_{3}$ contain the conjugacy classes $T_{1}$ , $T_{2}$ , and $T_{3}$ in $S_{3}$ of $1$ , $(1 2)$ , and $(1 2 3)$ . Note that $| C_{i} | = | T_{i} |$ , $| C_{2} | = 2 | T_{2} |$ , $| C_{3} | = 4 | T_{3} |$ , and $[G : H] = 4$ . Let $ψ_{1}$ , $ψ_{2}$ , and $ψ_{3}$ be the trivial, sign, and irreducible $2$ -dimensional characters of $S_{3}$ , respectively. By Proposition 13.6.20, we obtain the following table from the character table of $S_{3}$ :

$S_{4}$	e	$(1 2)$	$(1 2 3)$	$(1 2 3 4)$	$(1 2) (3 4)$

$ϕ_{1}$	4	2	1	0	0
$ϕ_{2}$	4	-2	1	0	0
$ϕ_{3}$	8	0	-1	0	0.

We will use this to determine the characters of $S_{4}$ once again. Assume we have already found its abelian characters, the trivial character $χ_{1}$ and the alternating character $χ_{2}$ . Since

⟨ {Ind}_{H}^{G} ϕ_{i}, {Ind}_{H}^{G} ϕ_{i} ⟩ = \dim {Hom}_{ℂ [G]} (V_{i}, V_{i}),

where $V_{i}$ is the $G$ -representation induced by $ϕ_{i}$ , and these values are $2$ , $2$ , $3$ , respectively, we have that the $V_{i}$ break up into these respective numbers of irreducible representations. But note that $⟨ ϕ_{1}, χ_{1} ⟩ = 1$ and $⟨ ϕ_{1}, χ_{2} ⟩ = 0$ , so $ϕ_{1} - χ_{1}$ so is an irreducible degree $3$ character of $G$ , which we previously called $χ_{4}$ . Similarly, $ϕ_{2} - χ_{2}$ is an irreducible degree $3$ character, which we called $χ_{5}$ . We compute that $⟨ ϕ_{3}, χ_{4} ⟩ = ⟨ ϕ_{3}, χ_{5} ⟩ = 1$ , and $ϕ_{3} - χ_{4} - χ_{5}$ is an irreducible character of degree $2$ , which we called $χ_{3}$ .

13.7. Applications to group theory

Let $G$ be a group of order $n$ , and let $C_{1}, \dots, C_{r}$ be the conjugacy classes in $G$ , choose $g_{j} \in C_{j}$ and set $c_{j} = | C_{j} |$ for each $1 \leq j \leq r$ . Let $χ_{1}, \dots, χ_{r}$ be the irreducible complex characters of $G$ , and set $n_{i} = \deg χ_{i}$ for $1 \leq i \leq r$ . Let $V_{i}$ denote the irreducible representation with character $χ_{i}$ , let $ρ_{i} : ℂ [G] \to {End}_{ℂ} (V_{i})$ be the $ℂ$ -algebra homomorphism restricting to the representation $ρ_{V_{i}}$ . We also use $χ_{i}$ to denote its $ℂ$ -linear extension to map $χ_{i} : ℂ [G] \to ℂ$ .

Proposition 13.7.1. §

Set

N_{i} = {g \in G ∣ χ_{i} (g) = χ_{i} (1)}

for $1 \leq i \leq r$ . The normal subgroups of $G$ are exactly the intersections $⋂_{j \in J} N_{j}$ , where $J$ is a subset of ${1, \dots, r}$ .

Proof.

First, $N_{i}$ is a normal subgroup, since $χ_{i} (g) = χ_{i} (1)$ if and only if $g$ acts as the identity, recalling that the eigenvalues of $g$ are all roots of unity, so $N_{i} = \ker ρ_{V_{i}}$ . It follows that every intersection of the $N_{i}$ ’s is normal.

Now suppose $N$ is normal in $G$ . Let $V = ℂ [G ∕ N]$ . Let $χ_{V}$ be the character of $V$ as a $ℂ [G]$ -module. Since $\ker ρ_{V} = N$ , we have $χ_{V} (g) = χ_{V} (1)$ if and only if $g \in N$ . If $g \notin N$ , then $𝑔h𝑁 \neq h𝑁$ for any $h \in G$ , so $χ_{V} (g) = 0$ .

Now $χ_{V}$ is a sum of irreducible characters with nonnegative integer coefficients, say $χ_{V} = \sum_{i = 1}^{r} a_{i} χ_{i}$ . We claim that $N = ⋂_{i \in J} N_{i}$ , where $J$ is the set of $i$ with $a_{i} \geq 1$ . Note that for any character $ψ$ and $g \in G$ , we have $| ψ (g) | \leq ψ (1)$ since $ψ (g)$ is a sum of $ψ (1)$ roots of unity. For $g \in G$ , we have

χ_{V} (g) = | χ_{V} (g) | = | \sum_{i = 1}^{r} a_{i} χ_{i} (g) | \leq \sum_{i = 1}^{r} a_{i} χ_{i} (1) = χ_{V} (1),

with equality of the first and last term holding if and only if $g \in N$ . However, the middle inequality is an equality if and only if all $χ_{i} (g)$ for $i \in J$ are equal and have absolute value $χ_{i} (1)$ . This condition holds if and only if all $χ_{i} (g) = χ_{i} (1)$ , since one of these characters is the trivial character. □

Next, we will show how to find the center of $G$ .

Proposition 13.7.2. §

Set

Z_{i} = {g \in G ∣ | χ_{i} (g) | = χ_{i} (1)}

for $1 \leq i \leq r$ . Then $Z_{i}$ is a normal subgroup of $G$ , and the center of $G$ is equal to $⋂_{i = 1}^{r} Z_{i}$ .

Proof.

We have that $Z_{i}$ is a normal subgroup since the condition that $g \in Z_{i}$ is exactly that $g$ acts as a scalar multiple of the identity, in other words that $g$ is in the inverse image of the center of $ρ_{V_{i}} (G)$ .

We claim that $Z_{i} ∕ N_{i}$ is the center of $G ∕ N_{i}$ . Note that $ρ_{V_{i}}$ has kernel $N_{i}$ defined as in Proposition 13.7.1, and the elements of $Z_{i}$ are mapped to scalar matrices in the center of $ρ_{V_{i}} (G) ≅𝐺 ∕ N_{i}$ . So $Z_{i} ∕ N_{i}$ is contained in the center of $G ∕ N_{i}$ . Now suppose that $g N_{i}$ is in the center of $G ∕ N_{i}$ . Then $ρ_{V_{i}} (g)$ commutes with all $ρ_{V_{i}} (h)$ for $h \in G$ , which implies that left multiplication by $g$ is a $ℂ [G]$ -module isomorphism of $V_{i}$ . But $V_{i}$ is simple, so ${Hom}_{ℂ [G]} (V_{i}, V_{i}) ≅ ℂ$ . In other words, $g$ acts as scalar multiplication by some element, hence is contained in $Z_{i}$ .

Given the claim, we have that $Z (G) N_{i} ∕ N_{i} \leq Z_{i} ∕ N_{i}$ , and so $Z (G) \subseteq Z_{i}$ for all $i$ . Now suppose that $z \in Z_{i}$ for all $i$ . Let $g \in G$ . Then $𝑔𝑧 g^{- 1} z^{- 1} \in N_{i}$ by our earlier claim. But $⋂_{i = 1}^{r} N_{i}$ is trivial by Proposition 13.7.1. So, $𝑔𝑧 g^{- 1} z^{- 1} = 1$ , for all $g \in G$ , so $z \in Z (G)$ , as desired. □

Proposition 13.7.3. §

For each pair $(i, j)$ of integers with $1 \leq i, j \leq r$ and each $g \in G$ , we have

\frac{c_{j}}{n_{i}} χ_{i} (g_{j}) \in ℤ [μ_{n}] .

Proof.

Set $N_{j} = \sum_{g \in C_{j}} g \in Z (ℂ [G])$ . Multiplication by $N_{j}$ defines a $ℂ [G]$ -linear map $N_{j} : V_{i} \to V_{i}$ which by Schur’s lemma is a scalar multiple of the identity, say $N_{j} = α_{j} {id}_{V_{i}}$ for $α_{j} \in ℂ$ , which tells us that $χ_{i} (N_{j}) = α_{j} χ_{i} (1) = α_{j} n_{i}$ . But we also have

χ_{i} (N_{j}) = \sum_{g \in C_{j}} χ_{i} (g) = c_{j} χ_{i} (g_{j}),

so $α_{j} = \frac{c_{j}}{n_{i}} χ_{i} (g_{j})$ . We show that eac

Next, set

a_{𝑗𝑘𝑙} = | {(g, h) ∣ g \in C_{j}, h \in C_{k}, 𝑔h = g_{l}} | \in ℤ_{\geq 0},

and note that this number is independent of the choice of $g_{l}$ , since $𝑠𝑔 s^{- 1} \cdot 𝑠h s^{- 1} = 𝑠𝑔h s^{- 1}$ for any $s \in G$ . Then

ρ_{i} (N_{j}) ρ_{i} (N_{k}) = \sum_{g \in C_{j}} \sum_{h \in C_{k}} ρ_{i} (𝑔h) = \sum_{l = 1}^{r} a_{𝑗𝑘𝑙} \sum_{q \in C_{l}} ρ_{i} (q) = \sum_{l = 1}^{r} a_{𝑗𝑘𝑙} ρ_{i} (N_{l}) .

Since $N_{j}$ acts on $V_{i}$ by multiplication by the scalar $α_{j}$ , this implies

α_{j} α_{k} = \sum_{l = 1}^{r} a_{𝑗𝑘𝑙} α_{l} .

In particular the subring $ℤ [{α_{j} ∣ 1 \leq j \leq k}]$ of $ℂ$ has finite $ℤ$ -rank, so it is integral over $ℤ$ . In particular, each $α_{j}$ is integral over $ℤ$ . The result now follows as $α_{j} \in ℚ (μ_{n})$ for each $j$ , and $ℤ [μ_{n}]$ is the integer ring of $ℚ (μ_{n})$ . □

Corollary 13.7.4. §

The dimension of an irreducible complex representation of a finite group $G$ divides $| G |$ .

Proof.

Let $1 \leq i \leq r$ , and consider the quotient of interest

\frac{n}{n_{i}} = \frac{n}{n_{i}} ⟨ χ_{i}, χ_{i} ⟩ = \sum_{j = 1}^{r} \frac{c_{j}}{n_{i}} χ_{i} (g_{j}) \bar{χ_{i} (g_{j})},

which is a $ℤ [μ_{n}]$ -linear combination of the algebraic integers $\frac{c_{j}}{n_{i}} χ_{i} (g_{j})$ , hence an algebraic integer. Since the fraction also lies in $ℚ$ , we have that $n_{i}$ divides $n$ . □