07.02.E1 · representation-theory / character

Finite-group representation exercise pack (Serre Linear Representations supplement)

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Anchor (Master):

Formal definition of the pack Intermediate

Serre's Linear Representations of Finite Groups builds the character theory of a finite group $G$ over $C$ from Maschke's complete-reducibility theorem: every representation splits into irreducibles, the irreducible characters form an orthonormal basis of the space of class functions, the number of irreducibles equals the number of conjugacy classes, and the regular representation gives $\sum_{i} (dim V_{i})^{2} = ∣ G ∣$ . On top of this sits the theory of induction: the induced character formula, Frobenius reciprocity as the adjunction between induction and restriction, and Mackey's criterion for irreducibility of an induced representation.

This pack collects nine such exercises — three easy, four medium, two hard — each with a hint and a full solution. It is meant to be read alongside its prerequisite units rather than as a standalone development. The exercises are loosely grouped: orthogonality and the regular representation (easy), explicit character tables and dimension counts for small groups (medium), and induced representations with Frobenius reciprocity and the symmetric group (hard).

The conventions throughout are Serre's: all representations are finite-dimensional over $C$ ; the character $χ_{V} (g) = tr ρ_{V} (g)$ is a class function; the Hermitian inner product on class functions is $⟨ χ, ψ ⟩ = \frac{1}{∣ G ∣} \sum_{g \in G} χ (g) \overline{ψ (g)}$ , with respect to which the irreducible characters are orthonormal.

Key theorem with full solution Intermediate

Before the pack proper, we work one exercise in full as an exemplar of the format. The remaining eight follow the same structure (problem, hint, full answer in <details> blocks).

Lead exercise. Prove that the number of irreducible representations of a finite group $G$ over $C$ equals the number of conjugacy classes, and that $\sum_{i} n_{i}^{2} = ∣ G ∣$ with $n_{i} = dim V_{i}$ .

Solution. By Maschke's theorem $C [G]$ is semisimple, so $C [G] ≅ \prod_{i} Mat_{n_{i}} (C)$ , the product running over the distinct irreducibles $V_{i}$ with $n_{i} = dim V_{i}$ . Comparing dimensions gives $\sum_{i} n_{i}^{2} = dim C [G] = ∣ G ∣$ , which is the regular-representation identity (the regular representation contains each $V_{i}$ with multiplicity $n_{i}$ ).

For the count, consider the center $Z (C [G])$ . On one side, the center of $\prod_{i} Mat_{n_{i}} (C)$ is $\prod_{i} C$ , of dimension equal to the number of irreducibles. On the other side, an element $z = \sum_{g} a_{g} g$ is central iff $a_{g}$ is constant on conjugacy classes (since $h z h^{- 1} = z$ permutes the $a_{g}$ along conjugation orbits), so the class sums ${\sum_{g \in C} g}$ over conjugacy classes $C$ form a basis of $Z (C [G])$ . Hence $dim Z (C [G])$ equals the number of conjugacy classes.

Equating the two dimensions: the number of irreducible representations equals the number of conjugacy classes. Equivalently, the irreducible characters ${χ_{i}}$ are orthonormal class functions whose number matches the dimension of the space of class functions, so they form an orthonormal basis of that space. $□$

Exercises Intermediate

Exercise 1 (easy, numeric). Dimensions of the irreducibles of $S_{3}$ .

The group $S_{3}$ has order $6$ and three conjugacy classes. Determine the dimensions $n_{1}, n_{2}, n_{3}$ of its irreducible representations and name each.

Hint

Three irreducibles (one per conjugacy class) with $\sum n_{i}^{2} = 6$ , each $n_{i} \geq 1$ , and at least one $n_{i} = 1$ (the unit representation that sends every element to $1$ ).

Answer

There are three irreducibles (three conjugacy classes: identity, transpositions, 3-cycles), and $\sum n_{i}^{2} = ∣ S_{3} ∣ = 6$ . The only solution in positive integers with three terms is $1^{2} + 1^{2} + 2^{2} = 6$ .

So the dimensions are $1, 1, 2$ :

the unit representation $g \mapsto 1$ (dimension $1$ ),
the sign representation $sgn$ (dimension $1$ ),
the standard representation (dimension $2$ ), the orthogonal complement of the diagonal in the permutation action on $C^{3}$ .

These are all the irreducibles of $S_{3}$ .

Exercise 2 (easy, proof). Sum of squares of character values over a column.

Show that for a finite group $G$ and $g \in G$ , $$ \sum_i |\chi_i(g)|^2 = \frac{|G|}{|C(g)|}, $$ where the sum is over all irreducibles and $C (g)$ is the conjugacy class of $g$ . In particular at $g = e$ this recovers $\sum_{i} n_{i}^{2} = ∣ G ∣$ .

Hint

This is the column orthogonality relation. Derive it from row orthogonality $⟨ χ_{i}, χ_{j} ⟩ = δ_{ij}$ by viewing the character table as a (weighted) unitary matrix.

Answer

Row orthogonality says the matrix $U$ with entries $U_{i, C} = ∣ C ∣/∣ G ∣ χ_{i} (g_{C})$ (rows indexed by irreducibles, columns by conjugacy classes $C$ with representative $g_{C}$ ) satisfies $U U^{*} = I$ . Since $U$ is square (number of irreducibles equals number of classes), it is unitary, so $U^{*} U = I$ as well. The column relation $U^{*} U = I$ reads, for the diagonal entry at class $C$ ,

i \sum \frac{∣ C ∣}{∣ G ∣} ∣ χ_{i} (g_{C}) ∣^{2} = 1, i.e. i \sum ∣ χ_{i} (g) ∣^{2} = \frac{∣ G ∣}{∣ C ∣} = \frac{∣ G ∣}{∣ C ( g ) ∣} = ∣ Z_{G} (g) ∣,

the order of the centralizer (using $∣ C (g) ∣ = ∣ G ∣/∣ Z_{G} (g) ∣$ ). At $g = e$ the class is a single point, $∣ C (e) ∣ = 1$ , so $\sum_{i} χ_{i} (e)^{2} = \sum_{i} n_{i}^{2} = ∣ G ∣$ .

Exercise 3 (easy, numeric). Decompose the permutation representation of $S_{3}$ .

The permutation representation $C^{3}$ of $S_{3}$ has character $χ (g) = # {fixed points of g}$ . Decompose it into irreducibles using the inner product.

Hint

Compute $⟨ χ, χ_{triv} ⟩$ and $⟨ χ, χ_{std} ⟩$ . The fixed-point counts are $3$ (identity), $1$ (transpositions), $0$ (3-cycles).

Answer

Fixed-point character: $χ (e) = 3$ , $χ (transposition) = 1$ , $χ (3-cycle) = 0$ . Class sizes: $1, 3, 2$ .

Inner product with the unit character ( $χ_{triv} \equiv 1$ ):

⟨ χ, χ_{triv} ⟩ = \frac{1}{6} (1 \cdot 3 \cdot 1 + 3 \cdot 1 \cdot 1 + 2 \cdot 0 \cdot 1) = \frac{6}{6} = 1.

So the unit character appears once. The standard character is $χ_{std} = χ - χ_{triv}$ with values $2, 0, - 1$ ; checking $⟨ χ_{std}, χ_{std} ⟩ = \frac{1}{6} (1 \cdot 4 + 3 \cdot 0 + 2 \cdot 1) = 1$ confirms it is irreducible. Hence

C^{3} ≅ 1 \oplus V_{std},

the unit representation plus the two-dimensional standard representation. This is the general fact that any transitive permutation representation contains the unit representation exactly once.

Exercise 4 (medium, numeric). Character table of the quaternion group $Q_{8}$ .

The quaternion group $Q_{8} = {\pm 1, \pm i, \pm j, \pm k}$ has five conjugacy classes. Determine the dimensions of its irreducibles and describe the character table.

Hint

Five irreducibles with $\sum n_{i}^{2} = 8$ . The abelianization $Q_{8} / [Q_{8}, Q_{8}] = Q_{8} / {\pm 1} ≅ (Z /2)^{2}$ gives the one-dimensional characters.

Answer

The conjugacy classes are ${1}, {- 1}, {\pm i}, {\pm j}, {\pm k}$ — five classes, hence five irreducibles. With $\sum n_{i}^{2} = 8$ and five terms, the only solution is $1^{2} + 1^{2} + 1^{2} + 1^{2} + 2^{2} = 8$ : four one-dimensional and one two-dimensional.

The four linear characters factor through $Q_{8} / {\pm 1} ≅ (Z /2)^{2}$ , giving the unit character and three sign-type characters distinguishing the pairs ${\pm i}, {\pm j}, {\pm k}$ . The two-dimensional irreducible is the defining representation by Pauli-type matrices: $i\mapsto\begin{psmallmatrix}\sqrt{-1}&0\\0&-\sqrt{-1}\end{psmallmatrix}$ , $j\mapsto\begin{psmallmatrix}0&1\\-1&0\end{psmallmatrix}$ . Its character takes the value $2$ at $1$ , $- 2$ at $- 1$ , and $0$ on ${\pm i}, {\pm j}, {\pm k}$ . Notably $Q_{8}$ and the dihedral group $D_{4}$ have identical character tables but are non-isomorphic — characters do not determine the group.

Exercise 5 (medium, proof). The induced character formula.

Let $H \leq G$ with $W$ a representation of $H$ , character $ψ$ . Prove the induced character $χ = Ind_{H}^{G} ψ$ is $$ \chi(g) = \frac{1}{|H|}\sum_{\substack{s\in G\ s^{-1}gs\in H}}\psi(s^{-1}gs). $$

Hint

$Ind_{H}^{G} W = ⨁_{s \in G / H} s W$ as a vector space; $g$ permutes the cosets, and only the cosets $sH$ with $s^{- 1} g s \in H$ contribute to the trace (the block-diagonal entries).

Answer

Choose coset representatives $s_{1}, \dots, s_{m}$ for $G / H$ , so $Ind_{H}^{G} W = ⨁_{i} s_{i} \otimes W$ . The action of $g$ sends the block $s_{i} \otimes W$ to $s_{j} \otimes W$ where $g s_{i} \in s_{j} H$ . The trace of $ρ (g)$ picks up only the diagonal blocks $j = i$ , i.e. those $i$ with $s_{i}^{- 1} g s_{i} \in H$ , and on such a block $g$ acts by $ψ (s_{i}^{- 1} g s_{i})$ . Thus

χ (g) = i : s_{i}^{- 1} g s_{i} \in H \sum ψ (s_{i}^{- 1} g s_{i}) .

To symmetrise over all of $G$ , note each coset representative $s_{i}$ can be replaced by $s_{i} h$ ( $h \in H$ ), and $ψ ((s_{i} h)^{- 1} g (s_{i} h)) = ψ (h^{- 1} (s_{i}^{- 1} g s_{i}) h) = ψ (s_{i}^{- 1} g s_{i})$ since $ψ$ is a class function on $H$ . Summing over all $∣ H ∣$ elements of each coset and dividing by $∣ H ∣$ replaces the sum over representatives by a sum over all $s \in G$ with $s^{- 1} g s \in H$ :

χ (g) = \frac{1}{∣ H ∣} s \in G s^{- 1} g s \in H \sum ψ (s^{- 1} g s) . □

Exercise 6 (medium, proof). Frobenius reciprocity.

For $H \leq G$ , a representation $W$ of $H$ and $V$ of $G$ , prove $$ \langle\mathrm{Ind}_H^G\psi,\ \chi_V\rangle_G = \langle\psi,\ \mathrm{Res}_H^G\chi_V\rangle_H. $$

Hint

Plug the induced-character formula into the left inner product and swap the order of summation; the inner-class-function structure collapses the $s$ -sum to $∣ G ∣/∣ H ∣$ copies of the $H$ -inner product. (Or argue $Hom_{G} (Ind W, V) ≅ Hom_{H} (W, Res V)$ directly.)

Answer

Use the induced-character formula and expand the $G$ -inner product:

⟨ Ind_{H}^{G} ψ, χ_{V} ⟩_{G} = \frac{1}{∣ G ∣} g \in G \sum (\frac{1}{∣ H ∣} s \in G s^{- 1} g s \in H \sum ψ (s^{- 1} g s)) \overline{χ_{V} (g)} .

Substitute $h = s^{- 1} g s$ , i.e. $g = s h s^{- 1}$ ; for each $s \in G$ the inner sum runs over $h \in H$ , and $χ_{V} (g) = χ_{V} (s h s^{- 1}) = χ_{V} (h)$ because $χ_{V}$ is a class function on $G$ . The double sum becomes

\frac{1}{∣ G ∣ ∣ H ∣} s \in G \sum h \in H \sum ψ (h) \overline{χ_{V} (h)} = \frac{1}{∣ G ∣ ∣ H ∣} \cdot ∣ G ∣ h \in H \sum ψ (h) \overline{χ_{V} (h)} = \frac{1}{∣ H ∣} h \in H \sum ψ (h) \overline{Res_{H}^{G} χ_{V} (h)} .

The right-hand side is exactly $⟨ ψ, Res_{H}^{G} χ_{V} ⟩_{H}$ . Equivalently, this is the character-level shadow of the adjunction $Hom_{G} (Ind_{H}^{G} W, V) ≅ Hom_{H} (W, Res_{H}^{G} V)$ between the induction and restriction functors. $□$

Exercise 7 (medium, numeric). Induce the sign character of $S_{2}$ up to $S_{3}$ .

Let $H = ⟨(12)⟩ ≅ S_{2} \leq S_{3}$ and let $ψ$ be its sign character. Compute $Ind_{H}^{G} ψ$ and decompose it into irreducibles of $S_{3}$ .

Hint

$[G : H] = 3$ , so the induced representation is 3-dimensional. Use the induced character formula, then take inner products with the three irreducible characters of $S_{3}$ .

Answer

$∣ H ∣ = 2$ , $[G : H] = 3$ , so $dim Ind_{H}^{G} ψ = 3$ . Using the induced character formula $χ (g) = \frac{1}{2} \sum_{s : s^{- 1} g s \in H} ψ (s^{- 1} g s)$ :

$g = e$ : all six $s$ qualify, $χ (e) = \frac{1}{2} \cdot 6 \cdot ψ (e) = 3$ .
$g$ a transposition: the conjugates landing in $H$ contribute; computing, $χ (transposition) = - 1$ .
$g$ a 3-cycle: no conjugate of a 3-cycle lies in $H$ , so $χ (3-cycle) = 0$ .

So $χ = (3, - 1, 0)$ on classes (identity, transpositions, 3-cycles). Inner products against the irreducible characters $χ_{triv} = (1, 1, 1)$ , $χ_{sgn} = (1, - 1, 1)$ , $χ_{std} = (2, 0, - 1)$ :

⟨ χ, χ_{sgn} ⟩ = \frac{1}{6} (3 + 3 + 0) = 1, ⟨ χ, χ_{std} ⟩ = \frac{1}{6} (6 + 0 + 0) = 1, ⟨ χ, χ_{triv} ⟩ = \frac{1}{6} (3 - 3 + 0) = 0.

Therefore $Ind_{H}^{G} ψ ≅ sgn \oplus V_{std}$ . By Frobenius reciprocity this matches $Res$ of each irreducible to $H = S_{2}$ containing the sign with the predicted multiplicities.

Exercise 8 (hard, proof). Every irreducible of $S_{n}$ appears in the regular representation with multiplicity its dimension.

Show that the regular representation $C [S_{n}]$ decomposes as $⨁_{λ ⊢ n} (dim S^{λ}) S^{λ}$ , and deduce $\sum_{λ ⊢ n} (dim S^{λ})^{2} = n!$ . Verify for $n = 4$ .

Hint

The regular character is $χ_{reg} (g) = ∣ G ∣ δ_{g, e}$ . Compute $⟨ χ_{reg}, χ_{λ} ⟩ = dim S^{λ}$ . For $n = 4$ use the hook-length dimensions of the five partitions of $4$ .

Answer

The regular character satisfies $χ_{reg} (e) = ∣ G ∣ = n!$ and $χ_{reg} (g) = 0$ for $g \neq = e$ (the permutation matrix of left-multiplication by $g \neq = e$ has no fixed basis vectors). For any irreducible character $χ_{λ}$ ,

⟨ χ_{reg}, χ_{λ} ⟩ = \frac{1}{n !} g \sum χ_{reg} (g) \overline{χ_{λ} (g)} = \frac{1}{n !} \cdot n! \cdot \overline{χ_{λ} (e)} = dim S^{λ} .

So $C [S_{n}] ≅ ⨁_{λ ⊢ n} (dim S^{λ}) S^{λ}$ , and matching dimensions gives $\sum_{λ} (dim S^{λ})^{2} = dim C [S_{n}] = n!$ .

For $n = 4$ the five partitions and their hook-length dimensions are:

dim S^{(4)} = 1, dim S^{(3, 1)} = 3, dim S^{(2, 2)} = 2, dim S^{(2, 1, 1)} = 3, dim S^{(1^{4})} = 1.

Sum of squares: $1 + 9 + 4 + 9 + 1 = 24 = 4!$ . The decomposition and the dimension identity both check out. (Here $S^{(4)}$ is the unit representation, $S^{(1^{4})}$ is sign, $S^{(3, 1)}$ is the standard, $S^{(2, 1, 1)}$ its sign-twist, $S^{(2, 2)}$ the two-dimensional self-conjugate.)

Exercise 9 (hard, proof). Mackey's irreducibility criterion, applied to $A_{4}$ .

State Mackey's irreducibility criterion for $Ind_{H}^{G} W$ , and use it (or a direct character computation) to show that inducing a nontrivial linear character of the cyclic subgroup $C_{3} \leq A_{4}$ produces a representation that is not irreducible.

Hint

Mackey: $Ind_{H}^{G} W$ is irreducible iff $W$ is irreducible and for each $s \in G ∖ H$ the representations $W^{s}$ and $W$ have no common constituent on restriction to $H \cap sH s^{- 1}$ . Here $[A_{4} : C_{3}] = 4$ , so the induced representation is 4-dimensional, but $A_{4}$ has no 4-dimensional irreducible.

Answer

Mackey's criterion. For $H \leq G$ and an irreducible representation $W$ of $H$ , $Ind_{H}^{G} W$ is irreducible iff for every $s \in G ∖ H$ the $H_{s} := H \cap sH s^{- 1}$ -representations $Res_{H_{s}} (W)$ and $Res_{H_{s}} (W^{s})$ are disjoint (share no irreducible constituent), where $W^{s} (x) = W (s^{- 1} x s)$ .

Apply with $G = A_{4}$ (order $12$ ), $H = C_{3} = ⟨(123)⟩$ , and $W = ψ$ a nontrivial linear character. Then $dim Ind_{H}^{G} ψ = [A_{4} : C_{3}] = 4$ . The irreducible representations of $A_{4}$ have dimensions $1, 1, 1, 3$ (since $A_{4} / V_{4} ≅ C_{3}$ gives three linear characters and $1 + 1 + 1 + 9 = 12$ ). There is no $4$ -dimensional irreducible, so the $4$ -dimensional $Ind_{H}^{G} ψ$ cannot be irreducible.

Concretely, the induced character $χ$ has $χ (e) = 4$ and one computes $⟨ χ, χ ⟩ = 2 > 1$ , confirming reducibility. The decomposition (by inner products against the $A_{4}$ irreducibles) is $Ind_{C_{3}}^{A_{4}} ψ ≅ θ \oplus V_{3}$ , a linear character $θ$ plus the $3$ -dimensional irreducible $V_{3}$ . Mackey's criterion fails because the double coset $C_{3} s C_{3}$ for $s$ a double transposition has $H_{s} = {e}$ , and the disjointness condition is not met for the linear character.

Exercise pack EP1 for Chapter 07.02. Serre Linear Representations of Finite Groups supplement: characters and orthogonality, the regular representation, induced representations, Frobenius reciprocity, and representations of $S_{n}$ and small groups.

Prerequisites

07.01.03 pending
07.01.04 pending
07.01.05 pending
07.01.07 pending
07.01.08 pending
07.02.01 pending
07.05.01 pending

Tier anchors

beginner
intermediate: Serre *Linear Representations of Finite Groups* exercises (§2 characters and orthogonality, §3.1 regular representation, §5.1 induced representations, §7.2 Frobenius reciprocity, §5.8 and Part I §5 representations of $S_n$ and small groups)
master

References

Serre — Linear Representations of Finite Groups · Exercises across §2 (orthogonality of characters, column relations), §2.5 (number of irreducibles = number of conjugacy classes), §3 (the regular representation and $\sum n_i^2 = |G|$), §5 (induced representations, the induced character formula), §7 (Frobenius reciprocity and Mackey's criterion), and the worked character tables for $S_3, S_4, A_4, Q_8$
Fulton-Harris — Representation Theory: A First Course · §2-§3 (character theory of finite groups), §4 (representations of $S_d$ and the Frobenius character formula)
Isaacs — Character Theory of Finite Groups · Ch. 2-5 — orthogonality relations, induced characters, and Frobenius reciprocity

Estimated time

intermediate: 5h