University Mathematics — Year 3 · Bachelor Year 3

5Representations of Finite Groups

To understand an abstract group, make it act on a vector space and use linear algebra — eigenvalues, traces, inner products — on the action. This program, representation theory, is astonishingly effective for finite groups over $\C$ : every representation splits into irreducible ones (Maschke), the irreducible ones are pinned down by their characters (traces), and the characters satisfy orthogonality relations that make computations mechanical. The chapter builds this calculus and its first masterpieces — character tables of the small groups — and the weekend problem harvests a theorem far beyond the reach of bare group theory: Burnside’s $p^aq^b$ theorem. Throughout, $G$ is a finite group and all vector spaces are finite-dimensional over $\C$ .

5.1 Representations, Maschke, Schur

Definition 5.1

A representation of $G$ is a morphism $\rho \colon G \to GL(V)$ for a $\C$ -vector space $V$ ; $\dim V$ is its degree. A subspace $W \subseteq V$ is invariant if $\rho(g)W \subseteq W$ for all $g$ ; restriction makes $W$ a subrepresentation. $\rho$ is irreducible if $V \neq 0$ and its only invariant subspaces are $0$ and $V$ . A morphism between $(\rho, V)$ and $(\sigma, W)$ is a linear $f\colon V \to W$ with $f\rho(g) = \sigma(g)f$ for all $g$ (equivariance); bijective $f$ are isomorphisms.

Example 5.2

(a) Degree $1$ : morphisms $G \to \C^\times$ . (b) The regular representation: $V = \C^G$ with basis $(e_h)_{h \in G}$ , $\rho(g)e_h = e_{gh}$ ; degree $\abs G$ . (c) A permutation action of $G$ on a finite set $X$ gives the permutation representation on $\C^X$ : $\rho(g)e_x = e_{g\cdot x}$ . (d) $S_n$ acts on $\C^n$ by permuting coordinates; the hyperplane $\{\sum x_i = 0\}$ is invariant: the standard representation, of degree $n - 1$ .

Theorem 5.3 (Maschke)

Every invariant subspace $W$ of a representation $(V, \rho)$ admits an invariant complement. Consequently every representation is a direct sum of irreducible ones (semisimplicity).

Proof. Let $p \colon V \to V$ be any projection with image $W$ (pick any complement). Average it over the group:

\tilde p = \frac1{\abs G}\sum_{g \in G} \rho(g)\,p\,\rho(g)^{-1}.

Each term maps $V$ into $W$ ( $W$ invariant), and fixes $W$ pointwise: for $w \in W$ , $\rho(g)^{-1}w \in W$ , $p$ fixes it, and $\rho(g)$ brings it back — so $\tilde p$ is again a projection onto $W$ . It is equivariant: for $h \in G$ , $\rho(h)\tilde p\rho(h)^{-1}$ reindexes the same sum. Hence $\ker \tilde p$ is an invariant complement of $W$ . Iterating on the summands (finite dimension) decomposes $V$ into irreducibles. ∎

Theorem 5.4 (Schur’s lemma)

Let $f \colon V \to W$ be a morphism of irreducible representations. Then $f = 0$ or $f$ is an isomorphism; and if $(V,\rho) = (W,\sigma)$ , then $f = \lambda\,\mathrm{id}$ for some $\lambda \in \C$ . Hence $\dim\operatorname{Hom}_G(V, W)$ is $1$ if $V \cong W$ , $0$ otherwise.

Proof. $\ker f$ and $\operatorname{im} f$ are invariant (equivariance), so each is $0$ or everything: either $f = 0$ , or $f$ is injective with full image. If $V = W$ : $f$ has an eigenvalue $\lambda$ ( $\C$ algebraically closed); $f - \lambda\,\mathrm{id}$ is a non-injective morphism $V \to V$ , hence $0$ . For the dimension count when $V \cong W$ : fixing one isomorphism $u$ , any morphism $f$ gives the endomorphism $u^{-1}f = \lambda\,\mathrm{id}$ : $f = \lambda u$ . ∎

5.2 Characters and orthogonality

Definition 5.5

The character of $(V, \rho)$ is $\chi_\rho(g) = \operatorname{tr}\rho(g)$ . It satisfies $\chi_\rho(e) = \dim V$ , $\chi_\rho(hgh^{-1}) = \chi_\rho(g)$ (traces are conjugation-invariant): characters are class functions — elements of the space $\mathcal{CF}(G)$ of functions constant on conjugacy classes, equipped with the Hermitian inner product

\langle \varphi, \psi\rangle = \frac1{\abs G}\sum_{g \in G} \overline{\varphi(g)}\,\psi(g).

Proposition 5.6

$\rho(g)$ is diagonalizable with eigenvalues roots of unity; $\chi_\rho(g^{-1}) = \overline{\chi_\rho(g)}$ , and $\abs{\chi_\rho(g)} \leq \chi_\rho(e)$ with equality iff $\rho(g)$ is a scalar. Characters add on direct sums: $\chi_{V \oplus W} = \chi_V + \chi_W$ .

Proof. $\rho(g)^N = \mathrm{id}$ for $N = \abs G$ (Lagrange): $\rho(g)$ annihilates $X^N - 1$ , split with simple roots, so it is diagonalizable with eigenvalues $\lambda_i \in \mu_N$ . Then $\chi(g^{-1}) = \sum \lambda_i^{-1} = \sum\bar\lambda_i = \overline{\chi(g)}$ , and $\abs{\chi(g)} = \abs{\sum\lambda_i} \leq \dim V$ , with equality in the triangle inequality iff all $\lambda_i$ are equal, i.e. $\rho(g) = \lambda\,\mathrm{id}$ . Additivity on direct sums: block traces. ∎

Lemma 5.7

Let $(V, \rho)$ , $(W, \sigma)$ be representations. The averaging operator on $\operatorname{Hom}(W, V)$ ,

c(A) = \frac1{\abs G}\sum_{g}\rho(g)\,A\,\sigma(g)^{-1},

is a projection onto $\operatorname{Hom}_G(W, V)$ , and for the map $\Phi_g \colon A \mapsto \rho(g)A\sigma(g)^{-1}$ one has $\operatorname{tr}\Phi_g = \chi_\rho(g)\,\overline{\chi_\sigma(g)}$ .

Proof. $c(A)$ is equivariant (reindex the sum as in Maschke), and $c$ fixes equivariant maps (each term equals $A$ ): $c$ is a projection with image $\operatorname{Hom}_G(W, V)$ . Trace: in bases, $\Phi_g(A) = BAC$ with $B = \rho(g)$ , $C = \sigma(g)^{-1}$ ; on the basis $(E_{kl})$ of matrices, $BE_{kl}C = \sum_{m,n} b_{mk}c_{ln}E_{mn}$ , so the coefficient of $E_{kl}$ in $\Phi_g(E_{kl})$ is $b_{kk}c_{ll}$ : $\operatorname{tr}\Phi_g = \sum_{k,l}b_{kk}c_{ll} = \operatorname{tr}(B) \operatorname{tr}(C) = \chi_\rho(g)\chi_\sigma(g^{-1})$ , and $\chi_\sigma(g^{-1}) = \overline{\chi_\sigma(g)}$ . ∎

Theorem 5.8 (First orthogonality relations)

Let $\rho, \sigma$ be irreducible. Then

\langle\chi_\sigma, \chi_\rho\rangle = \begin{cases} 1 & \text{if } \rho \cong \sigma,\\ 0 & \text{otherwise:} \end{cases}

the irreducible characters form an orthonormal family in $\mathcal{CF}(G)$ .

Proof. The trace of a projection is the dimension of its image:

\dim\operatorname{Hom}_G(W, V) = \operatorname{tr} c = \frac1{\abs G}\sum_g \operatorname{tr}\Phi_g = \frac1{\abs G}\sum_g \chi_\rho(g)\overline{\chi_\sigma(g)} = \langle \chi_\sigma, \chi_\rho\rangle,

and Schur’s lemma evaluates the left side to $\delta_{\rho \cong \sigma}$ . ∎

Corollary 5.9

Decompose $V \cong \bigoplus_i V_i^{\oplus m_i}$ into distinct irreducibles ( $V_i \not\cong V_j$ ). Then $m_i = \langle \chi_{V_i}, \chi_V\rangle$ : the multiplicities — hence the representation up to isomorphism — are determined by the character. Moreover $\langle \chi_V, \chi_V\rangle = \sum_i m_i^2$ ; in particular $V$ is irreducible iff $\langle\chi_V, \chi_V\rangle = 1$ .

Proof. $\chi_V = \sum m_i\chi_{V_i}$ (Proposition 5.6); take inner products with each $\chi_{V_i}$ and use orthonormality. Two representations with equal characters have equal multiplicities, hence are isomorphic. ∎

Theorem 5.10 (The regular representation)

Let $\chi_1, \dots, \chi_r$ be the distinct irreducible characters, of degrees $n_i = \chi_i(e)$ . The regular representation decomposes with multiplicities $m_i = n_i$ ; consequently

\sum_{i=1}^{r} n_i^2 = \abs G, \qquad \sum_i n_i\chi_i(g) = 0 \quad (g \neq e).

Proof. The regular character: $\chi_{\mathrm{reg}}(g) = \#\{h : gh = h\}$ , which is $\abs G$ for $g = e$ and $0$ otherwise. So $m_i = \langle \chi_i,\chi_{\mathrm{reg}}\rangle = \frac1{\abs G}\overline{\chi_i(e)}\,\abs G = n_i$ . Evaluating $\chi_{\mathrm{reg}} = \sum n_i\chi_i$ at $e$ and at $g \neq e$ gives the two displayed identities. ∎

Theorem 5.11

The irreducible characters form an orthonormal basis of $\mathcal{CF}(G)$ : the number of irreducible representations equals the number of conjugacy classes of $G$ .

Proof. Only completeness is left: let $f \in \mathcal{CF}(G)$ be orthogonal to every $\chi_i$ ; we show $f = 0$ . For a representation $(V,\rho)$ , set $T_{f,\rho} = \frac{1}{\abs G} \sum_g \overline{f(g)}\,\rho(g)$ . It is equivariant: for $h \in G$ ,

\rho(h)T_{f,\rho}\rho(h)^{-1} = \frac1{\abs G}\sum_g \overline{f(g)}\rho(hgh^{-1}) = \frac1{\abs G}\sum_{g'}\overline{f(h^{-1}g'h)}\rho(g') = T_{f,\rho}

( $f$ is a class function). If $\rho$ is irreducible of degree $n$ , Schur gives $T_{f,\rho} = \lambda\,\mathrm{id}$ with

\lambda = \frac{\operatorname{tr}T_{f,\rho}}{n} = \frac{1}{n\abs G}\sum_g \overline{f(g)}\,\chi_\rho(g) = \frac1n\,\langle f, \chi_\rho\rangle = 0 .

So $T_{f,\rho} = 0$ on every irreducible, hence (direct sums) on every representation — in particular on the regular one. Apply it to the basis vector $e_e$ : $0 = T_{f,\mathrm{reg}}e_e = \frac1{\abs G}\sum_g\overline{f(g)}e_g$ , forcing every $\overline{f(g)} = 0$ . Thus the orthonormal family $(\chi_i)$ spans $\mathcal{CF}(G)$ , whose dimension is the number of conjugacy classes. ∎

Corollary 5.12 (Column orthogonality)

For $g, h \in G$ :

\sum_{i=1}^{r}\overline{\chi_i(g)}\,\chi_i(h) = \begin{cases} \abs{Z_G(g)} & \text{if } g, h \text{ are conjugate},\\ 0 & \text{otherwise.} \end{cases}

Proof. Let $g_1, \dots, g_r$ represent the classes, $c_j$ the class sizes. The $r \times r$ matrix $U_{ij} = \sqrt{c_j/\abs G}\;\chi_i(g_j)$ has orthonormal rows (Theorem 5.8 written classwise: $\sum_j \frac{c_j}{\abs G}\chi_i(g_j)\overline{\chi_{i'}(g_j)} = \delta_{ii'}$ ), i.e. $UU^* = I$ ; a square matrix with $UU^* = I$ also has $U^*U = I$ : columns are orthonormal, which unpacks to the displayed identity ( $\abs G/c_j = \abs{Z_G(g_j)}$ , the orbit–stabilizer relation for conjugation). ∎

Proposition 5.13 (One-dimensional characters; lifting)

(a) $G$ is abelian iff all its irreducible representations have degree $1$ ; the number of degree- $1$ characters of any $G$ is $[G : D(G)]$ (they are the characters of the abelianization). (b) If $N \trianglelefteq G$ , the irreducible representations of $G/N$ lift (compose with $G \to G/N$ ) to exactly those irreducible representations of $G$ whose kernel contains $N$ .

Proof. (a) If $G$ is abelian, each class is a singleton: $r = \abs G$ , and $\sum n_i^2 = \abs G$ forces all $n_i = 1$ ; conversely if all $n_i = 1$ , the regular representation is a sum of one-dimensional ones, so $\rho_{\mathrm{reg}}(G)$ is simultaneously diagonalizable, hence commutative, and $\rho_{\mathrm{reg}}$ is faithful: $G$ abelian. Degree- $1$ representations are morphisms $G \to \C^\times$ with abelian target: they factor through $G^{\mathrm{ab}} = G/D(G)$ (Exercise 1.9), and distinct characters of the abelian $G^{\mathrm{ab}}$ number $\abs{G^{\mathrm{ab}}}$ (it has that many classes, all degrees $1$ ). (b) Composition with the projection preserves irreducibility (invariant subspaces correspond), and a representation trivial on $N$ factors through the quotient (Theorem 1.3). ∎

5.3 Character tables

Definition 5.14

The character table of $G$ is the $r \times r$ matrix $\bigl(\chi_i(g_j)\bigr)$ : rows indexed by irreducible characters, columns by conjugacy classes (with their sizes displayed). Rows are orthonormal for the weighted product, columns orthogonal (Corollary 5.12): the table is severely overdetermined, which is what makes it computable.

Example 5.15 (The table of $S_3$ )

Classes: $e$ (size 1), transpositions (3), $3$ -cycles (2); so $r = 3$ irreducibles, of degrees $n_i$ with $\sum n_i^2 = 6$ : $1, 1, 2$ . Degree $1$ : trivial $\mathbf 1$ and signature $\varepsilon$ . The last row follows from column orthogonality (or from $\chi_{\mathrm{std}} = \chi_{\mathrm{perm}} - \mathbf 1$ ):

\begin{array}{c|ccc} S_3 & e & (1\,2)\ [3] & (1\,2\,3)\ [2]\\ \hline \mathbf 1 & 1 & 1 & 1\\ \varepsilon & 1 & -1 & 1\\ \chi_{\mathrm{std}} & 2 & 0 & -1 \end{array}

Check: $\langle\chi_{\mathrm{std}},\chi_{\mathrm{std}}\rangle = \frac{1}{6}(4 + 0 + 2) = 1$ : irreducible.

Example 5.16 (The table of $S_4$ )

Classes: $e$ [1], transpositions [6], double transpositions [3], $3$ -cycles [8], $4$ -cycles [6]: five irreducibles, $\sum n_i^2 = 24$ with two degree- $1$ ’s ( $\mathbf 1, \varepsilon$ ; $[S_4 : D(S_4)] = [S_4 : A_4] = 2$ ): degrees $1, 1, 2, 3, 3$ . The degree- $2$ lifts from $S_4/V \cong S_3$ (Proposition 5.13(b), $V$ the Klein group); degree $3$ : the standard representation and its twist by $\varepsilon$ :

\begin{array}{c|ccccc} S_4 & e\,[1] & (1\,2)\,[6] & (1\,2)(3\,4)\,[3] & (1\,2\,3)\,[8] & (1\,2\,3\,4)\,[6]\\ \hline \mathbf 1 & 1 & 1 & 1 & 1 & 1\\ \varepsilon & 1 & -1 & 1 & 1 & -1\\ \chi_2 & 2 & 0 & 2 & -1 & 0\\ \chi_{\mathrm{std}} & 3 & 1 & -1 & 0 & -1\\ \varepsilon\chi_{\mathrm{std}} & 3 & -1 & -1 & 0 & 1 \end{array}

( $\chi_{\mathrm{std}}(g) = \operatorname{fix}(g) - 1$ ; $\chi_2$ evaluates the $S_3$ -table on the image of each class mod $V$ .) All row and column orthogonality checks pass — running two of them is Exercise 5.3’s warmup.

Method 5.17

To build a character table: (1) list conjugacy classes and sizes; (2) count degree- $1$ characters via $G/D(G)$ and write them; (3) find the remaining degrees from $\sum n_i^2 = \abs G$ (small integer combinatorics); (4) obtain cheap irreducibles: lift from quotients, subtract $\mathbf 1$ from permutation characters (check $\langle\chi,\chi\rangle = 1$ ), multiply known characters by degree- $1$ ones; (5) finish unknown rows by column orthogonality — each column is orthogonal to the columns already complete, and column $e$ carries the degrees. Verify everything with a full orthogonality sweep.

The regular representation of S_3, block-diagonalized: ℂ[S_3] ℂ × ℂ × M_2(ℂ), dimensions 1 + 1 + 4 = 6 = |S_3|. In general ℂ[G] _i M_n_i(ℂ): the identity n_i2 = G is a statement about matrix blocks. — The regular representation of $S_3$ , block-diagonalized: $\C[S_3] \cong \C \times \C \times M_2(\C)$ , dimensions $1 + 1 + 4 = 6 = \abs{S_3}$ . In general $\C[G] \cong \prod_i M_{n_i}(\C)$ : the identity $\sum n_i^2 = \abs G$ is a statement about matrix blocks.

5.4 Exercises

Exercise 5.1 ★

(a) Show that the irreducible characters of $\Z/n\Z$ are the $\chi_k(\bar m) = \eu^{2\iu\pi km/n}$ , $k = 0, \dots, n-1$ , and write the character table of $\Z/4\Z$ . (b) Verify both orthogonality relations on it — and recognize the matrix: where has this book seen it before?

Solution

Solution of Exercise 5.1.

(a) $\Z/n\Z$ is abelian: all irreducibles have degree $1$ (Proposition 5.13), i.e. are morphisms $\chi \colon \Z/n\Z \to \C^\times$ , determined by $\chi(\bar 1) = \omega$ with $\omega^n = 1$ : the $n$ characters $\chi_k(\bar m) = \eu^{2\iu\pi km/n}$ . For $n = 4$ (classes $=$ elements $\bar0,\bar1,\bar2,\bar3$ ):

\begin{array}{c|cccc} & \bar0 & \bar1 & \bar2 & \bar3\\ \hline \chi_0 & 1 & 1 & 1 & 1\\ \chi_1 & 1 & \iu & -1 & -\iu\\ \chi_2 & 1 & -1 & 1 & -1\\ \chi_3 & 1 & -\iu & -1 & \iu \end{array}

(b) Rows: $\langle\chi_k,\chi_l\rangle = \frac14\sum_m \eu^{2\iu\pi(l-k)m/4} = \delta_{kl}$ (geometric sum); columns likewise. The matrix $(\eu^{2\iu\pi km/n})_{k,m}$ is the discrete Fourier transform matrix — the same roots-of-unity filter used in the Year 2 volume’s generating-functions chapter; orthogonality of characters generalizes the inversion formula of the DFT.

Exercise 5.2 ★

Let $G$ act on a finite set $X$ and $\chi$ be the character of the permutation representation $\C^X$ . (a) Show $\chi(g) = \abs{\operatorname{Fix}_X(g)}$ and $\langle \mathbf 1, \chi\rangle = \#\{\text{orbits}\}$ — Burnside’s counting lemma (Exercise 1.5) is a character computation. (b) Suppose the action is transitive, so $\chi = \mathbf 1 + \psi$ . Show that $\psi$ is irreducible iff the action is $2$ -transitive (transitive on ordered pairs of distinct points). (Compute $\langle\chi,\chi\rangle$ as the number of orbits on $X \times X$ .) (c) Conclude that the standard representation of $S_n$ ( $n \geq 2$ ) is irreducible.

Solution

Solution of Exercise 5.2.

(a) The matrix of $\rho(g)$ in the basis $(e_x)$ is a permutation matrix, of trace the number of $x$ with $g\cdot x = x$ . Then

\langle\mathbf 1, \chi\rangle = \frac1{\abs G}\sum_g \abs{\operatorname{Fix}(g)} = \#\{\text{orbits}\}

by Burnside’s counting lemma (Exercise 1.5) — equivalently, this computes the multiplicity of the trivial representation, whose isotypic space is the space of $G$ -invariant vectors, of dimension the number of orbits (one indicator per orbit).

(b) Since $\operatorname{Fix}_{X\times X}(g) = \operatorname{Fix}_X(g)^2$ , part (a) applied to $X \times X$ gives $\langle\chi,\chi\rangle = \frac1{\abs G}\sum\abs{\operatorname{Fix}(g)}^2 = \#\{\text{orbits on } X\times X\}$ ( $\chi$ is real). Writing $\chi = \mathbf 1 + \psi$ : $\langle\mathbf1,\chi\rangle = 1$ (transitivity), so $\langle\psi,\psi\rangle = \langle\chi,\chi\rangle - 1$ . The action on $X\times X$ has the diagonal as one orbit; there is exactly one other orbit iff $G$ is transitive on distinct pairs: $\langle\psi,\psi\rangle = 1$ iff $2$ -transitive (Corollary 5.9).

(c) $S_n$ is $2$ -transitive on $\{1,\dots,n\}$ (send any distinct pair anywhere): $\psi = \chi_{\mathrm{std}}$ is irreducible.

Exercise 5.3 ★

Rebuild the table of $S_3$ from scratch following Method 5.17, then verify two row and two column orthogonality relations in the $S_4$ table of Example 5.16. Decompose the permutation character of $S_4$ acting on $\{1,2,3,4\}$ and the character $\chi_{\mathrm{std}}^2$ (pointwise square) into irreducibles.

Solution

Solution of Exercise 5.3.

$S_3$ : three classes, $\sum n_i^2 = 6 = 1 + 1 + 4$ ; the two degree- $1$ characters are $\mathbf 1, \varepsilon$ ( $[S_3 : A_3] = 2$ ); the third row $(2, a, b)$ follows from column orthogonality with the $e$ -column: $1 - 1 + 2a = 0$ and $1 + 1 + 2b = 0$ : $a = 0$ , $b = -1$ — the table of Example 5.15.

$S_4$ checks (rows): $\langle\chi_{\mathrm{std}}, \varepsilon\chi_{\mathrm{std}}\rangle = \frac1{24}(9 - 6 + 3 + 0 - 6) = 0$ ; $\langle\chi_2,\chi_2\rangle = \frac1{24}(4 + 0 + 12 + 8 + 0) = 1$ . Columns: $e$ against $(1\,2)$ : $1 - 1 + 0 + 3 - 3 = 0$ ; $(1\,2)$ against itself: $1 + 1 + 0 + 1 + 1 = 4 = \abs{Z_{S_4}((1\,2))} = 24/6$ .

Permutation character on $4$ points: $(4, 2, 0, 1, 0) = \mathbf 1 + \chi_{\mathrm{std}}$ (fixed-point counts; subtract the top row). For $\chi_{\mathrm{std}}^2 = (9, 1, 1, 0, 1)$ :

\langle\mathbf1,\cdot\rangle = \tfrac{9 + 6 + 3 + 0 + 6}{24} = 1,\quad \langle\varepsilon,\cdot\rangle = 0,\quad \langle\chi_2,\cdot\rangle = \tfrac{18 + 6}{24} = 1,\quad \langle\chi_{\mathrm{std}},\cdot\rangle = 1,\quad \langle\varepsilon\chi_{\mathrm{std}},\cdot\rangle = 1:

$\chi_{\mathrm{std}}^2 = \mathbf 1 + \chi_2 + \chi_{\mathrm{std}} + \varepsilon\chi_{\mathrm{std}}$ (dimensions: $9 = 1 + 2 + 3 + 3$ ).

Exercise 5.4 ★★

Compute the character tables of $D_4$ and of $Q_8$ . Conclude that two non-isomorphic groups can have identical character tables — what group-theoretic data does the table nevertheless capture in this pair (orders of centers, abelianizations, number of involutions)? Which of these does it fail to capture?

Solution

Solution of Exercise 5.4.

Both groups have five classes and degree pattern $(1,1,1,1,2)$ (four degree- $1$ ’s from the abelianization $\cong (\Z/2\Z)^2$ , then $\sum n_i^2 = 8$ ). Ordering the classes $e$ , $z$ (the central involution: $r^2$ , resp. $-1$ ), and the three two-element classes:

\begin{array}{c|ccccc} & e & z & C_1 & C_2 & C_3\\ \hline \chi^{(1)} & 1 & 1 & 1 & 1 & 1\\ \chi^{(2)} & 1 & 1 & 1 & -1 & -1\\ \chi^{(3)} & 1 & 1 & -1 & 1 & -1\\ \chi^{(4)} & 1 & 1 & -1 & -1 & 1\\ \chi^{(5)} & 2 & -2 & 0 & 0 & 0 \end{array}

(the last row from column orthogonality). Identical tables for $D_4$ and $Q_8$ , which are not isomorphic (Problem 1.1). The table does capture: $\abs G$ , class sizes, the center ( $\{g : \abs{\chi_i(g)} = n_i\ \forall i\}$ : order $2$ in both), the abelianization, the whole lattice of normal subgroups (kernels and intersections, Exercise 5.6). It fails to capture element orders: $D_4$ has five involutions, $Q_8$ has one — so the isomorphism type is genuinely finer than the character table.

Exercise 5.5 ★★

Character table of $A_4$ : classes $e$ [1], double transpositions [3], and two classes of $3$ -cycles [4], [4]. (a) Explain the splitting of the $3$ -cycles (compare centralizers in $S_4$ and $A_4$ , as in Exercise 1.11). (b) Find the three degree- $1$ characters (via $A_4/V \cong \Z/3\Z$ ) and the degree- $3$ character (restrict $\chi_{\mathrm{std}}$ from $S_4$ ), and assemble the table. (c) Read off the normal subgroups of $A_4$ from the table (kernels $\{g : \chi(g) = \chi(e)\}$ and their intersections).

Solution

Solution of Exercise 5.5.

(a) In $S_4$ , the centralizer of $(1\,2\,3)$ has order $24/8 = 3$ : it is $\langle(1\,2\,3)\rangle \subseteq A_4$ . So $Z_{A_4}((1\,2\,3))$ has order $3$ and the $A_4$ -class has $12/3 = 4$ elements: the eight $3$ -cycles split into two $A_4$ -classes (represented by $(1\,2\,3)$ and its inverse).

(b) $A_4/V \cong \Z/3\Z$ gives three degree- $1$ characters ( $\omega = \eu^{2\iu\pi/3}$ ; the $3$ -cycle classes map to $\bar1, \bar2$ ); the restriction of $\chi_{\mathrm{std}}$ stays irreducible ( $\langle\chi,\chi\rangle = \frac1{12}(9 + 3 \cdot 1 + 0 + 0) = 1$ ):

\begin{array}{c|cccc} A_4 & e\,[1] & (1\,2)(3\,4)\,[3] & (1\,2\,3)\,[4] & (1\,3\,2)\,[4]\\ \hline \mathbf 1 & 1 & 1 & 1 & 1\\ \chi_\omega & 1 & 1 & \omega & \omega^2\\ \chi_{\bar\omega} & 1 & 1 & \omega^2 & \omega\\ \chi_3 & 3 & -1 & 0 & 0 \end{array}

(c) Kernels: $\ker\chi_\omega = \ker\chi_{\bar\omega} = V$ ; $\ker\chi_3 = \{e\}$ (no other entry has modulus $3$ ). The normal subgroups are the intersections of kernels (Exercise 5.6): $\{e\}$ , $V$ , $A_4$ — in particular $A_4$ has no normal subgroup of order $2$ or index $2$ .

Exercise 5.6 ★★

(a) Show that $\ker\chi = \{g : \chi(g) = \chi(e)\}$ is the kernel of the underlying representation (Proposition 5.6, equality case). (b) Show that every normal subgroup of $G$ is an intersection of kernels of irreducible characters. (Represent $G/N$ faithfully: its regular representation.) (c) Deduce: $G$ is simple iff $\ker\chi_i = \{e\}$ for every nontrivial irreducible $\chi_i$ — simplicity is readable from the character table.

Solution

Solution of Exercise 5.6.

(a) If $\chi(g) = \chi(e) = n$ : equality in $\abs{\chi(g)} \leq n$ forces $\rho(g) = \lambda\,\mathrm{id}$ (Proposition 5.6) with $n\lambda = n$ : $\rho(g) = \mathrm{id}$ . The converse is clear.

(b) Let $N \trianglelefteq G$ . The regular representation of $G/N$ is faithful; decompose it into irreducibles of $G/N$ and lift them to $G$ (Proposition 5.13(b)): irreducible characters $\chi_{i_1}, \dots$ of $G$ whose kernels contain $N$ and whose common kernel is exactly the preimage of $\{e\}$ , i.e. $N$ (faithfulness on the quotient). So $N = \bigcap_j \ker\chi_{i_j}$ .

(c) If $G$ is simple: for a nontrivial irreducible $\chi$ , $\ker\chi \trianglelefteq G$ is not $G$ (an irreducible representation trivial on all of $G$ is the trivial character), so $\ker\chi = \{e\}$ . Conversely, suppose all nontrivial kernels are trivial, and let $N \trianglelefteq G$ with $N \neq G$ . In (b)’s expression of $N$ as an intersection of kernels, some character involved is nontrivial (if all were trivial, the intersection would be $G$ ), and its kernel is $\{e\}$ : $N = \{e\}$ . So the only normal subgroups are $\{e\}$ and $G$ .

Exercise 5.7 ★★

Degrees of $\abs G = 8$ : show that a nonabelian group of order $8$ has degree pattern $(1,1,1,1,2)$ , and that its degree- $2$ representation is faithful. More generally show that a nonabelian group of order $p^3$ has pattern $(1^{\,p^2}, p, \dots, p)$ with $p^2$ ones and $p - 1$ characters of degree $p$ . (Use $[G : D(G)]$ and $\sum n_i^2 = \abs G$ ; here $D(G) = Z(G)$ has order $p$ .)

Solution

Solution of Exercise 5.7.

Order $8$ nonabelian: the number of degree- $1$ characters is $[G : D(G)]$ , a proper divisor of $8$ (nonabelian: $D(G) \neq \{e\}$ ), and $\sum n_i^2 = 8$ . With $k$ ones and the remaining degrees $\geq 2$ : $8 - k \equiv 0$ with squares $\geq 4$ , and $k \mid 8$ , $k < 8$ . $k = 4$ : one degree $2$ — consistent. $k = 2$ : $6$ left, not a sum of squares $\geq 4$ . $k = 1$ : impossible, since $k = [G : D(G)] \geq 2$ — $G/D(G)$ is a nontrivial abelian $2$ -group, as $G$ is a $2$ -group with $D(G) \neq G$ by solvability of $p$ -groups (Example 1.30). So the pattern is $(1,1,1,1,2)$ . Faithfulness of $\chi_5$ : the four degree- $1$ characters all contain $D(G)$ in their kernels; if $\ker\chi_5 \supseteq N \neq \{e\}$ for some minimal normal $N$ ( $N \subseteq D(G)$ or not — take $N \subseteq \ker\chi_5$ nontrivial), then $N$ would lie in all five kernels, whose intersection is trivial (the regular representation is faithful): contradiction. Order $p^3$ nonabelian: $Z(G)$ has order $p$ (order $p^2$ would make $G/Z$ cyclic, $G$ abelian), $G/Z(G)$ of order $p^2$ is abelian, so $D(G) \subseteq Z(G)$ , and $D(G) \ne \{e\}$ : $D(G) = Z(G)$ , giving $p^2$ characters of degree $1$ . The remaining degrees satisfy $\sum n_i^2 = p^3 - p^2$ with each $n_i > 1$ dividing $\abs G$ (Problem 5.1, question 8) hence $n_i \in \{p\}$ ( $n_i = p^2$ would exceed: $p^4 > p^3 - p^2$ ): exactly $p - 1$ characters of degree $p$ .

Exercise 5.8 ★★★

For finite groups $G, H$ : show that the class functions $\chi(g)\psi(h)$ on $G \times H$ , for $\chi, \psi$ irreducible characters of $G, H$ , are exactly the irreducible characters of $G \times H$ . (Orthonormality is a direct computation; completeness by counting classes.) Deduce the character table of $\Z/2\Z \times \Z/2\Z$ and re-derive Proposition 5.13(a) for finite abelian groups via the structure theorem.

Solution

Solution of Exercise 5.8.

Define, for representations $\rho, \sigma$ of $G, H$ in $V, W$ , the representation $\rho \boxtimes \sigma$ of $G \times H$ on $V \otimes W$ — concretely, on matrices: $(\rho\boxtimes \sigma)(g,h)$ is the Kronecker product $\rho(g)\otimes \sigma(h)$ , whose trace is $\operatorname{tr}\rho(g)\operatorname{tr}\sigma(h) = \chi(g)\psi(h)$ (the Kronecker product of matrices $A \otimes B$ has trace $\operatorname{tr}A\operatorname{tr}B$ : its diagonal is $a_{kk}b_{ll}$ ). So $\chi\psi$ is a character, and

\langle\chi\psi, \chi'\psi'\rangle_{G\times H} = \frac{1}{\abs G\abs H}\sum_{g,h} \overline{\chi(g)\psi(h)}\,\chi'(g)\psi'(h) = \langle\chi,\chi'\rangle_G\,\langle\psi,\psi'\rangle_H = \delta_{\chi\chi'}\delta_{\psi\psi'}.

In particular $\langle\chi\psi,\chi\psi\rangle = 1$ : each $\chi\psi$ is irreducible (Corollary 5.9). These are $r_Gr_H$ distinct irreducible characters; the classes of $G\times H$ are the products of classes ( $(g,h) \sim (g',h')$ componentwise), so there are $r_Gr_H$ of them: the list is complete (Theorem 5.11). For $(\Z/2\Z)^2$ : the four sign characters $(\pm1)\otimes(\pm1)$ — the table of Exercise 5.4’s top-left block. A finite abelian group is a product of cyclic groups (Corollary 3.13); its irreducible characters are products of the cyclic ones (Exercise 5.1): all of degree $1$ .

Exercise 5.9 ★★★

The character table of $A_5$ (classes of sizes $1, 15, 20, 12, 12$ from Exercise 1.11): (a) Show the degrees are $1, 3, 3, 4, 5$ (the only solution of $\sum n_i^2 = 60$ with $n_1 = 1$ and, using Exercise 5.6(c) with simplicity, no other $n_i = 1$ ). (b) Construct the degree- $4$ character (permutation action on $5$ points) and the degree- $5$ character (action on the six $5$ -Sylow subgroups gives degree $6 = 1 + 5$ ; check irreducibility), and complete the two degree- $3$ rows by column orthogonality: golden-ratio entries $\frac{1\pm\sqrt5}2$ appear on the $5$ -cycle classes. (c) Verify from the finished table that $A_5$ is simple (Exercise 5.6(c)).

Solution

Solution of Exercise 5.9.

(a) $D(A_5) = A_5$ (simple nonabelian), so the only degree- $1$ character is $\mathbf 1$ (Proposition 5.13). We need $n_2^2 + n_3^2 + n_4^2 + n_5^2 = 59$ with each $n_i \geq 2$ ; testing the squares $4, 9, 16, 25, 36, 49$ : the only multiset that works is $\{9, 9, 16, 25\}$ : with largest square $49$ , the remainder $10$ is not a sum of three squares $\geq 4$ ; with $36$ , the remainder $23$ is not either ( $16 + 4 + 4 = 24$ , $9 + 9 + 4 = 22$ ); with largest $25$ , one checks $25 + 16 + 9 + 9 = 59$ works and $25 + 25$ , $25 + 16 + 16$ , $25 + 16 + 4$ variants fail; with largest $16$ : $16\cdot3 = 48 < 59 - 4$ . Degrees: $1, 3, 3, 4, 5$ .

(b) Permutation on $5$ points: fixed points $(5, 1, 2, 0, 0)$ , so $\chi_4 = (4, 0, 1, -1, -1)$ with $\langle\chi_4,\chi_4\rangle = \frac{16 + 0 + 20 + 12 + 12}{60} = 1$ : irreducible. Action on the six Sylow $5$ -subgroups: an involution fixes exactly $2$ (the normalizers are dihedral of order $10$ , each containing $5$ involutions: $30$ incidences for $15$ involutions), a $3$ -element fixes $0$ (no order $3$ in $D_5$ ), a $5$ -element fixes exactly $1$ (it lies in a unique Sylow): permutation character $(6, 2, 0, 1, 1)$ , and $\chi_5 = (5, 1, -1, 0, 0)$ with norm $\frac{25 + 15 + 20 + 0 + 0}{60} = 1$ : irreducible. Two rows $(3, a, b, c, d)$ , $(3, a', b', c', d')$ remain. Column norms (Corollary 5.12): on the class of $(1\,2)(3\,4)$ , $\abs{Z} = 4$ : $1 + 0 + 1 + a^2 + a'^2 = 4$ ; column against $e$ : $1\cdot1 + 4\cdot 0 + 5\cdot1 + 3(a + a') = 0$ : $a + a' = -2$ , $a^2 + a'^2 = 2$ : $a = a' = -1$ . On the $3$ -cycles, $\abs Z = 3$ : $1 + 1 + 1 + b^2 + b'^2 = 3$ : $b = b' = 0$ . On each $5$ -class, $\abs Z = 5$ : the column against $e$ reads $1\cdot 1 + 4\cdot(-1) + 5\cdot 0 + 3(c + c') = 0$ , so $c + c' = 1$ ; and $c^2 + c'^2 = 3$ : $\{c, c'\} = \bigl\{\frac{1+\sqrt5}2, \frac{1-\sqrt5}2\bigr\}$ — the golden ratio and its conjugate; the second $5$ -class carries the swapped values (the two rows must be orthogonal).

(c) In the finished table, no entry of a nontrivial row equals its degree outside the first column: every kernel $\{g : \chi_i(g) = n_i\}$ is trivial. By Exercise 5.6(c), $A_5$ is simple.

Exercise 5.10 ★★

Let $\rho$ be an irreducible representation of degree $n$ and $z \in Z(G)$ . Show $\rho(z) = \lambda_z\,\mathrm{id}$ with $\lambda \colon Z(G) \to \C^\times$ a morphism (the central character), and deduce $\abs{\chi(z)} = n$ for central $z$ . Application: if $G$ has a faithful irreducible representation, then $Z(G)$ is cyclic.

Solution

Solution of Exercise 5.10.

$\rho(z)$ commutes with every $\rho(g)$ ( $z$ is central), i.e. $\rho(z) \in \operatorname{End}_G(V) = \C\,\mathrm{id}$ (Schur): $\rho(z) = \lambda_z\,\mathrm{id}$ , and $z \mapsto \lambda_z$ is multiplicative: a morphism $Z(G) \to \C^\times$ . Then $\chi(z) = n\lambda_z$ with $\abs{\lambda_z} = 1$ (root of unity): $\abs{\chi(z)} = n$ . If $\rho$ is faithful, $\lambda$ is injective on $Z(G)$ ( $\rho(z) = \mathrm{id} \iff \lambda_z = 1$ ), so $Z(G)$ embeds in $\C^\times$ ; a finite subgroup of the multiplicative group of a field is cyclic (Theorem 4.12).

Exercise 5.11 ★★

(Isotypic projections) Let $(V, \rho)$ be a representation of $G$ and $\chi_i$ an irreducible character of degree $n_i$ . Define

p_i = \frac{n_i}{\abs G}\sum_{g\in G} \overline{\chi_i(g)}\,\rho(g) \;\in\; \mathcal L(V).

(a) Show that $p_i$ is $G$ -equivariant, and compute its restriction to an irreducible subrepresentation $W \subseteq V$ of character $\chi_j$ : it is $\delta_{ij}\, \mathrm{id}_W$ (Schur; take traces to identify the scalar). (b) Deduce that $p_i$ is a projection onto the sum $V_i$ of all irreducible subrepresentations of character $\chi_i$ (the isotypic component), that $\sum_ip_i = \mathrm{id}_V$ , and that the decomposition $V = \bigoplus_iV_i$ is canonical — unlike the finer splitting of each $V_i$ into irreducibles. (c) For the regular representation of $S_3$ and the sign character $\varepsilon$ , write out $p_\varepsilon$ explicitly as an element of the group algebra and check $p_\varepsilon^2 = p_\varepsilon$ by hand.

Solution

Solution of Exercise 5.11.

(a) Equivariance: $\rho(h)p_i\rho(h)^{-1}$ reindexes the sum ( $g \mapsto hgh^{-1}$ , and $\chi_i$ is a class function): $p_i$ commutes with the action. On an irreducible $W$ of character $\chi_j$ , Schur makes the restriction a scalar $\lambda\,\mathrm{id}_W$ ; taking traces,

\lambda\,n_j = \frac{n_i}{\abs G}\sum_g \overline{\chi_i(g)}\,\chi_j(g) = n_i\,\langle\chi_i, \chi_j\rangle = n_i\,\delta_{ij}

(first orthogonality): $\lambda = \delta_{ij}$ .

(b) Decompose $V$ into irreducibles (Maschke): $p_i$ acts as identity on the summands of character $\chi_i$ and as $0$ on all others, so $p_i$ is the projection onto their sum $V_i$ along the sum of the rest; the image $V_i$ does not depend on the chosen decomposition (it is the set of vectors fixed by $p_i$ , defined without choices). $\sum_ip_i$ acts as identity on every irreducible summand: it is $\mathrm{id}_V$ . The finer splitting of $V_i \cong W_i^{\oplus m_i}$ involves choosing a basis of $\operatorname{Hom}_G(W_i, V)$ : canonical it is not.

(c) For $\varepsilon$ (degree $1$ ): $p_\varepsilon = \frac1{6}\sum_{g}\varepsilon(g)\,\rho(g)$ , i.e. in the group algebra

p_\varepsilon = \tfrac16\bigl(e - (1\,2) - (1\,3) - (2\,3) + (1\,2\,3) + (1\,3\,2)\bigr) .

Squaring: the coefficient of $g$ in $p_\varepsilon^2$ is $\frac1{36}\sum_{h}\varepsilon(h)\varepsilon(h^{-1}g) = \frac1{36}\,\varepsilon(g)\sum_h\varepsilon(h)^2 = \frac{6}{36}\varepsilon(g)$ : $p_\varepsilon^2 = p_\varepsilon$ . (Its image in the regular representation is the line spanned by $\sum_g\varepsilon(g)e_g$ : the sign representation appears with multiplicity $1$ , as the general theory demands.)

Exercise 5.12 ★★

(Reading a table) The character table of a certain group $G$ of order $24$ is partially known: it has $5$ classes, of sizes $1, 6, 8, 6, 3$ , and degrees $1, 1, 2, 3, 3$ . (a) Recover the full table: the two linear characters (one trivial; the other takes value $-1$ exactly on the classes of sizes $6$ and $6$ ), then the degree- $2$ character via column orthogonality with the identity column, then the two degree- $3$ characters likewise (one is $\chi_2\chi_4$ ). (b) Identify $G$ ( $\cong S_4$ : compare classes with cycle types), and extract from the table the normal subgroups via Exercise 5.6: kernels of $\chi_2$ (index $2$ : $A_4$ ) and of $\chi_3$ (the Klein $V_4$ ), and nothing else but $\{e\}, G$ . (c) Explain how the table shows $G/V_4 \cong S_3$ (which characters factor through the quotient?).

Solution

Solution of Exercise 5.12.

(a) Order the classes $e$ [1], transpositions [6], $3$ -cycles [8], $4$ -cycles [6], double transpositions [3]. The second linear character is $\chi_2 = \varepsilon$ with values $1, -1, 1, -1, 1$ . For the degree- $2$ character $\chi_3$ , column orthogonality of each column with the identity column ( $\sum_in_i\chi_i(g) = 0$ for $g \neq e$ ) gives, on transpositions: $1 - 1 + 2\chi_3 + 3(\chi_4 + \chi_5) = 0$ ; the sign trick $\chi_5 = \varepsilon\chi_4$ (a degree- $3$ character times a linear one is again irreducible — same norm) makes $\chi_4 + \chi_5$ vanish on odd classes: $\chi_3 = 0$ there. On $3$ -cycles: $1 + 1 + 2\chi_3(c_3) + 3(\chi_4 + \chi_5)(c_3) = 0$ with $\chi_5 = \chi_4$ on even classes; the column of $c_3$ with itself gives $\abs{\chi_3}^2$ data; solving the small system (use also row orthogonality of $\chi_3$ with $\mathbf 1$ and $\varepsilon$ ): $\chi_3 = (2, 0, -1, 0, 2)$ , then $\chi_4 = (3, 1, 0, -1, -1)$ and $\chi_5 = \varepsilon\chi_4 = (3, -1, 0, 1, -1)$ . The full table:

	$e$	$6\,t$	$8\,c_3$	$6\,c_4$	$3\,v$
$\chi_1$	$1$	$1$	$1$	$1$	$1$
$\chi_2$	$1$	$-1$	$1$	$-1$	$1$
$\chi_3$	$2$	$0$	$-1$	$0$	$2$
$\chi_4$	$3$	$1$	$0$	$-1$	$-1$
$\chi_5$	$3$	$-1$	$0$	$1$	$-1$

(All rows have norm $1$ ; all columns are orthogonal: checks pass.)

(b) The class data $1, 6, 8, 6, 3$ with these degrees is that of $S_4$ (cycle types $e$ , $2$ , $3$ , $4$ , $2{+}2$ ). Kernels: $\ker\chi_2 = \{g : \varepsilon(g) = 1\} = A_4$ (classes $e, c_3, v$ : $1 + 8 + 3 = 12$ , index $2$ ); $\ker\chi_3 = \{g : \chi_3(g) = 2\}$ = classes $e, v$ : the Klein group $V_4$ , of order $4$ , normal. $\chi_4, \chi_5$ are faithful ( $\chi_i(g) = n_i$ only at $e$ ). Intersections of kernels: $\{e\}$ , $V_4$ , $A_4$ , $G$ — by Exercise 5.6(b) these are all the normal subgroups of $S_4$ .

(c) The characters with $V_4 \subseteq \ker$ are $\chi_1, \chi_2, \chi_3$ : they factor through $G/V_4$ , a group of order $6$ possessing irreducible degrees $1, 1, 2$ — the table of $S_3$ . Since the quotient’s table is a complete invariant among the two groups of order $6$ ( $\Z/6\Z$ would have six linear characters), $G/V_4 \cong S_3$ : the quotient is visible inside the table as the block of rows containing $V_4$ in their kernel.

5.5 Problem: Burnside’s $p^aq^b$ theorem

Problem 5.1

Weekend problem — solvability of groups of order $p^aq^b$

Burnside proved in 1904 that every group whose order has at most two prime factors is solvable — a statement about abstract groups whose only known proofs for half a century went through character theory. This problem builds the proof in full, assembling Chapter 1 (solvability), Chapter 3 (finitely generated $\Z$ -modules) and this chapter. Throughout, $\chi_1, \dots, \chi_r$ are the irreducible characters of $G$ , $n_i = \chi_i(e)$ .

Part I — Algebraic integers. An algebraic integer is a root of a monic polynomial of $\Z[X]$ .

Show that $\alpha$ is an algebraic integer iff the ring $\Z[\alpha]$ is a finitely generated $\Z$ -module.
Deduce that the algebraic integers form a subring of $\C$ . (If $\Z[\alpha], \Z[\beta]$ are finitely generated, so is $\Z[\alpha, \beta]$ , and submodules of finitely generated $\Z$ -modules are finitely generated, by Theorem 3.5 and a presentation argument — or directly: a submodule of $\Z^n$ is free of rank $\leq n$ .)
Show that a rational algebraic integer is an integer. (Rational root theorem.)
Show that every character value $\chi(g)$ is an algebraic integer.

Part II — The class-sum relations. Fix an irreducible $(V, \rho)$ of degree $n$ and character $\chi$ . For a conjugacy class $C$ , let $S_C = \sum_{g \in C}\rho(g) \in \mathcal L(V)$ .

Show that $S_C$ is equivariant, hence $S_C = \omega_C\,\mathrm{id}$ with

$\omega_C = \frac{\abs C\,\chi(g_C)}{n} \qquad (g_C \in C \text{ any representative}).$
Show that $S_CS_{C'} = \sum_{C''} a_{CC'C''}\,S_{C''}$ where $a_{CC'C''} \in \N$ counts, for a fixed $z \in C''$ , the pairs $(x, y) \in C \times C'$ with $xy = z$ . Deduce that the $\omega_C$ satisfy $\omega_C\,\omega_{C'} = \sum_{C''} a_{CC'C''}\,\omega_{C''}$ .
Conclude that each $\omega_C$ is an algebraic integer. (The $\Z$ -module generated by $1$ and the $\omega_C$ is a finitely generated ring; apply question 1’s criterion — more precisely show $M = \Z\text{-span} (1, (\omega_C)_C)$ satisfies $\omega_{C_0} M \subseteq M$ and use a determinant/Cayley–Hamilton trick, or question 2’s subring argument.)
Deduce Frobenius divisibility: $n_i$ divides $\abs G$ for every irreducible degree. (Compute $\frac{\abs G}{n} = \frac{\abs G}{n} \langle\chi,\chi\rangle = \sum_C \omega_C\, \overline{\chi(g_C)}$ : an algebraic integer that is rational.)

Part III — Burnside’s simplicity criterion.

Let $\chi$ be irreducible of degree $n$ and $C$ a class with $\gcd(\abs C, n) = 1$ . Using Bézout and questions 4–7, show that $\frac{\chi(g_C)}{n}$ is an algebraic integer.
Suppose moreover $0 < \abs{\chi(g_C)} < n$ . Show this is impossible: the algebraic integer $\alpha = \chi(g_C)/n$ has all its conjugates — the numbers $\alpha_\sigma = \frac1n\sigma(\chi(g_C))$ for $\sigma \in \operatorname{Gal}(\Q(\zeta_{\abs G})/\Q)$ , each an average of $n$ roots of unity — of modulus $\leq 1$ , so the product $N = \prod_\sigma\alpha_\sigma$ is a rational algebraic integer with $0 < \abs N < 1$ — justify each assertion, quoting Theorem 4.23 for the Galois group and the fact that $\sigma$ permutes roots of unity. Conclude: either $\chi(g_C) = 0$ or $\rho(g_C)$ is scalar (Proposition 5.6).
(Burnside’s criterion) Let $C \neq \{e\}$ be a conjugacy class of prime power size $p^k > 1$ , and suppose $G$ is simple nonabelian. Column orthogonality on the column of $C$ against the column of $e$ gives $1 + \sum_{i \geq 2} n_i\chi_i(g_C) = 0$ . Show that some nontrivial $\chi_i$ with $p \nmid n_i$ has $\chi_i(g_C) \neq 0$ (otherwise $\frac1p$ would be an algebraic integer); by question 10, $\rho_i(g_C)$ is scalar; derive a contradiction with simplicity (the set of $g$ with $\rho_i(g)$ scalar is a normal subgroup; use faithfulness from Exercise 5.6). Conclude: no simple nonabelian group has a conjugacy class of prime power size $> 1$ .

Part IV — The theorem.

Let $\abs G = p^aq^b$ with $a + b \geq 1$ . If $G$ is simple, show it is abelian: pick $z \neq e$ in the center of a Sylow $q$ -subgroup (Theorem 1.12) and consider the size of its conjugacy class $[G : Z_G(z)]$ , a power of $p$ (why?); apply question 11.
Conclude by induction on $\abs G$ : every group of order $p^aq^b$ is solvable (Burnside). Why does the argument break for three primes — and must it, given $\abs{A_5} = 2^2\cdot3\cdot5$ ?

Part V — The character table of $A_5$ . The smallest group Burnside’s theorem cannot touch deserves its full portrait; everything below uses only this chapter plus Exercise 1.11.

Recall from Exercise 1.11 the five conjugacy classes of $A_5$ : $\{e\}$ , the $15$ double transpositions, the $20$ three-cycles, and two classes of $12$ five-cycles each, represented by $c = (1\,2\,3\,4\,5)$ and $c^2$ . Explain why the five-cycles split into two $A_5$ -classes although they form a single $S_5$ -class.
Show that the irreducible degrees of $A_5$ are exactly $1, 3, 3, 4, 5$ : use $\sum_in_i^2 = 60$ with $r = 5$ classes, and the fact that $A_5$ is perfect ( $D(A_5) = A_5$ ), so the trivial character is its only linear one; then eliminate every other multiset (write $59$ as a sum of four squares of integers $\geq 2$ : check it happens in only one way with all summands plausible degrees).
Let $\pi$ be the permutation character of $A_5$ on $\{1, \dots, 5\}$ : $\pi(g) = \#\operatorname{Fix}(g)$ , with values $5, 1, 2, 0, 0$ on the five classes. Compute $\langle\pi, \mathbf 1\rangle$ and $\langle\pi, \pi\rangle$ , and deduce that $\chi_4 = \pi - \mathbf 1$ is irreducible of degree $4$ , with values $4, 0, 1, -1, -1$ .
Same game on the $10$ unordered pairs $\{i, j\}$ : the fixed-point counts are $10, 2, 1, 0, 0$ . Compute $\langle\pi_{10}, \pi_{10}\rangle$ , $\langle\pi_{10}, \mathbf 1\rangle$ and $\langle\pi_{10}, \chi_4\rangle$ , deduce the decomposition $\pi_{10} = \mathbf 1 + \chi_4 + \chi_5$ , and obtain the irreducible $\chi_5$ of degree $5$ with values $5, 1, -1, 0, 0$ .
The two remaining irreducibles $\chi_2, \chi_3$ have degree $3$ . Column orthogonality (each nonidentity column against the identity column, and each column with itself) determines their values off the five-cycles: show $\chi_2(g) = \chi_3(g) = -1$ on double transpositions and $0$ on three-cycles.
On the five-cycle classes, set $x = \chi_2(c)$ and $y = \chi_2(c^2)$ ; symmetry lets one take $\chi_3(c) = y$ , $\chi_3(c^2) = x$ . From the column of $c$ paired with the identity column and with the column of $c^2$ , derive $x + y = 1$ and $xy = -1$ (and check the value $x^2 + y^2 = 3$ given by the column of $c$ with itself), hence

$\{x, y\} = \Bigl\{\frac{1 + \sqrt5}2,\ \frac{1 - \sqrt5}2\Bigr\} :$

the golden ratio and its conjugate. Assemble the complete character table of $A_5$ .
Run the checks: the row norm of $\chi_2$ is $1$ (use $\varphi^2 + \bar\varphi^2 = 3$ ), $\langle\chi_2, \chi_3\rangle = 0$ , and Frobenius divisibility (question 8) for all five degrees. Where in the table do you see a difference with $S_5$ , all of whose character values are rational integers?
Deduce from the table alone that $A_5$ is simple: a normal subgroup is a union of conjugacy classes containing $e$ whose cardinality divides $60$ — check that no proper sub-sum of $1 + 15 + 20 + 12 + 12$ containing the term $1$ divides $60$ . Cross-check with question 11’s criterion: verify that no class of $A_5$ has prime-power size $> 1$ .
(Icosahedral coda) $A_5$ is the rotation group of the icosahedron, and the degree- $3$ representations are the two geometric actions on $\R^3$ . Verify the trace identity: a rotation by angle $\theta$ has trace $1 + 2\cos\theta$ , and $1 + 2\cos\frac{2\pi}5 = \frac{1+\sqrt5}2 = \varphi$ . Explain without any computation why the other degree- $3$ character must carry the conjugate value: the Galois group of $\Q(\sqrt5)/\Q$ acts on the whole character table (entrywise), permuting the irreducible characters.

Part VI — Complements: a central bound and the tensor square.

(Sharper than a divisibility) Let $\chi$ be irreducible of degree $n$ . Show that $\abs{\chi(z)} = n$ for every $z \in Z(G)$ (Schur’s lemma: $\rho(z)$ is a scalar, of finite order), and deduce from $\langle\chi, \chi\rangle = 1$ the bound

$n^2 \leq [G : Z(G)] .$

Show that the nonabelian groups of order $8$ have irreducible degrees $1, 1, 1, 1, 2$ (five conjugacy classes; write $8$ as a sum of five squares) and attain equality $4 = [G : Z(G)]$ ; check the bound on $A_5$ , whose center is trivial.
(Tensor square of $\chi_2$ ) For $g$ of finite order, $\rho(g)$ is diagonalizable with root-of-unity eigenvalues; deduce the character formulas

$\chi_{\operatorname{Sym}^2 V}(g) = \frac{\chi(g)^2 + \chi(g^2)}2, \qquad \chi_{\Lambda^2 V}(g) = \frac{\chi(g)^2 - \chi(g^2)}2 .$

Apply them to $\chi_2$ of $A_5$ (note $g^2$ runs through the class of $c^2$ when $g$ runs through that of $c$ , and conversely): show $\Lambda^2\chi_2 = \chi_2$ and $\operatorname{Sym}^2\chi_2 = \mathbf 1 + \chi_5$ , hence

$\chi_2\otimes\chi_2 = \mathbf 1 + \chi_2 + \chi_5 .$

Interpret $\Lambda^2\chi_2 = \chi_2$ geometrically via the cross product on $\R^3$ .
(Final audit of the table) Verify numerically: column orthogonality between the two five-cycle columns ( $\varphi\bar\varphi = -1$ ), the value $5 = \abs{Z_{A_5}(c)}$ for the column of $c$ against itself, and the vanishing of the regular character $\sum_i n_i\chi_i(g) = 0$ on each of the four nonidentity columns of the table.

Solution

Solution of Problem 5.1.

1. If $\alpha^n + c_{n-1}\alpha^{n-1} + \dots + c_0 = 0$ ( $c_i \in \Z$ ), then $\alpha^n \in \Z\text{-span}(1, \dots, \alpha^{n-1})$ , and inductively every power is: $\Z[\alpha]$ is generated by $1, \alpha, \dots, \alpha^{n-1}$ . Conversely let $\Z[\alpha] = \Z g_1 + \dots + \Z g_m$ . Write $\alpha g_i = \sum_j m_{ij}g_j$ with $M = (m_{ij}) \in M_m(\Z)$ : the vector $g = (g_i)$ satisfies $(\alpha I - M)g = 0$ ; multiplying by the adjugate matrix, $\det(\alpha I - M)\,g_i = 0$ for every $i$ , and since $1 \in \Z[\alpha]$ is a $\Z$ -combination of the $g_i$ , $\det(\alpha I - M) = 0$ : $\alpha$ is a root of the monic $\det(XI - M) \in \Z[X]$ .

2. If $\Z[\alpha]$ is spanned by $\alpha$ ’s powers up to $n-1$ and $\Z[\beta]$ by $\beta$ ’s up to $m - 1$ , then $\Z[\alpha,\beta]$ is spanned by the $nm$ products $\alpha^i\beta^j$ (reduce any monomial). The subrings $\Z[\alpha + \beta]$ and $\Z[\alpha\beta]$ are $\Z$ -submodules of the finitely generated $\Z$ -module $\Z[\alpha, \beta]$ , hence finitely generated ( $\Z[\alpha,\beta]$ , generated by $nm$ elements, is an image of $\Z^{nm}$ ; a submodule pulls back to a submodule of $\Z^{nm}$ , free of rank $\leq nm$ by Theorem 3.5, and its image generates). By question 1, $\alpha + \beta$ and $\alpha\beta$ are algebraic integers.

3. If $\frac pq$ (lowest terms) is a root of a monic integer polynomial of degree $n$ , the rational root theorem (clear denominators: $p^n = -q(\cdots)$ ) gives $q \mid p^n$ , so $q = \pm1$ .

4. $\chi(g)$ is a sum of roots of unity (Proposition 5.6), each an algebraic integer (root of $X^N - 1$ ); conclude by question 2.

5. For $h \in G$ : $\rho(h)S_C\rho(h)^{-1} = \sum_{g\in C}\rho(hgh^{-1}) = S_C$ ( $C$ is a class). By Schur, $S_C = \omega_C\,\mathrm{id}$ ; taking traces, $\abs C\,\chi(g_C) = \omega_C\, n$ .

6. $S_CS_{C'} = \sum_{x \in C, y \in C'}\rho(xy) = \sum_{z \in G} a(z)\rho(z)$ with $a(z) = \#\{(x,y) \in C\times C' : xy = z\}$ . Conjugation by $h$ bijects the solutions for $z$ with those for $hzh^{-1}$ : $a$ is a class function with values in $\N$ , so $S_CS_{C'} = \sum_{C''}a_{CC'C''}S_{C''}$ . Substituting $S_C = \omega_C\,\mathrm{id}$ throughout and identifying the scalars: $\omega_C\omega_{C'} = \sum_{C''}a_{CC'C''}\,\omega_{C''}$ .

7. Let $M$ be the $\Z$ -module spanned by $1$ and all products $\omega_{C_1}\cdots\omega_{C_k}$ ; by question 6 every such product reduces to a $\Z$ -combination of $1$ and the $\omega_C$ : $M$ is finitely generated, and $\omega_C M \subseteq M$ for each $C$ . In particular $\Z[\omega_C] \subseteq M$ is finitely generated (submodule, as in question 2), and question 1 makes $\omega_C$ an algebraic integer.

8. For an irreducible $\chi$ of degree $n$ :

\frac{\abs G}{n} = \frac{\abs G}{n}\langle\chi,\chi\rangle = \frac1n \sum_{g}\chi(g)\overline{\chi(g)} = \sum_{C}\frac{\abs C\,\chi(g_C)}{n}\,\overline{\chi(g_C)} = \sum_C \omega_C\,\overline{\chi(g_C)} .

Each $\overline{\chi(g_C)} = \chi(g_C^{-1})$ is an algebraic integer (question 4), so the right side is one (questions 2, 7); it is rational, hence an integer (question 3): $n \mid \abs G$ .

9. Bézout: $u\abs C + vn = 1$ with $u, v \in \Z$ . Then

\frac{\chi(g_C)}{n} = u\,\frac{\abs C\,\chi(g_C)}{n} + v\,\chi(g_C) = u\,\omega_C + v\,\chi(g_C),

an algebraic integer.

10. Suppose $0 < \abs{\chi(g_C)} < n$ and let $\alpha = \chi(g_C)/n$ . All values $\chi(g)$ lie in $\Q(\zeta_N)$ , $N = \abs G$ (sums of $N$ -th roots of unity). For $\sigma \in \operatorname{Gal}(\Q(\zeta_N)/\Q)$ : $\sigma$ maps roots of unity to roots of unity ( $\sigma(\zeta^k) = \zeta^{ak}$ , Theorem 4.23), so $\sigma(\chi(g_C))$ is again a sum of $n$ roots of unity: $\abs{\sigma(\alpha)} \leq 1$ ; also $\sigma(\alpha)$ is an algebraic integer (it has the same minimal polynomial as $\alpha$ ). The product $P = \prod_\sigma \sigma(\alpha)$ is fixed by the whole Galois group, hence rational (Theorem 4.21(1)), and it is an algebraic integer with

0 < \abs P \leq \abs\alpha < 1

(no factor vanishes: $\sigma(\alpha) = 0$ would force $\alpha = 0$ ). This contradicts question 3. Hence $\chi(g_C) = 0$ or $\abs{\chi(g_C)} = n$ , and in the latter case $\rho(g_C)$ is scalar (Proposition 5.6).

11. Column orthogonality ( $C \neq \{e\}$ ): $\sum_i \chi_i(e)\overline{\chi_i(g_C)} = 0$ , i.e. $1 + \sum_{i \geq 2} n_i\overline{\chi_i(g_C)} = 0$ . If every nontrivial $\chi_i$ with $p \nmid n_i$ vanished at $g_C$ , then grouping the rest by their factor $p$ :

-\frac1p = \sum_{i \geq 2,\ p \mid n_i} \frac{n_i}{p}\,\overline{\chi_i(g_C)},

an algebraic integer — contradicting question 3. So some nontrivial $\chi_i$ has $p \nmid n_i$ and $\chi_i(g_C) \neq 0$ ; since $\abs C = p^k$ , $\gcd(\abs C, n_i) = 1$ , and question 10 makes $\rho_i(g_C)$ a scalar. Now $G$ simple nonabelian: $\chi_i$ is faithful (Exercise 5.6(c)), and $Z_i = \{g : \rho_i(g) \text{ scalar}\}$ is a normal subgroup (the preimage under $\rho_i$ of the scalars, which form a normal — indeed central — subgroup of the image) containing $g_C \neq e$ : $Z_i = G$ . Then $\rho_i(G)$ is abelian and faithful, making $G$ abelian: contradiction. No simple nonabelian group has a conjugacy class of prime-power size $> 1$ .

12. Let $G$ be simple of order $p^aq^b$ . If $b = 0$ ( $G$ a $p$ -group): $Z(G) \neq \{e\}$ (Theorem 1.12) is normal, so $Z(G) = G$ : abelian. Otherwise take $Q$ a Sylow $q$ -subgroup and $z \in Z(Q) \setminus\{e\}$ (Theorem 1.12 again). Then $Q \subseteq Z_G(z)$ , so the class of $z$ has size $[G : Z_G(z)]$ dividing $[G : Q] = p^a$ : a power of $p$ . If the size is $1$ , $z \in Z(G)$ : the center is a nontrivial normal subgroup, so $Z(G) = G$ , abelian. If the size is $p^k > 1$ : question 11 forbids it for simple nonabelian $G$ . Either way a simple group of order $p^aq^b$ is abelian ( $\cong \Z/p\Z$ ).

13. Induction on $\abs G$ ( $\abs G = 1$ : solvable). If $G$ is simple, question 12 makes it abelian, hence solvable. Otherwise pick $N$ a nontrivial proper normal subgroup: $\abs N$ and $\abs{G/N}$ are again of the form $p^{a'}q^{b'}$ and smaller, so $N$ and $G/N$ are solvable by induction, and $G$ is solvable (Proposition 1.29). — With three primes, the key step fails: the index of a Sylow subgroup is no longer a prime power, so the class of a central element of a Sylow need not have prime-power size. And some failure is inevitable: $A_5$ , of order $2^2\cdot3\cdot5$ , is simple and not solvable.

14. The classes and sizes are Exercise 1.11(a). The $S_5$ -class of $c$ has size $24$ ; if it stayed one $A_5$ -class, the orbit–stabilizer count would give $\abs{Z_{A_5}(c)} = 60/24$ , not an integer — concretely, $Z_{S_5}(c) = \langle c\rangle$ has order $5$ and sits inside $A_5$ , so the $A_5$ -class of $c$ has size $60/5 = 12$ : the $S_5$ -class splits in two ( $c$ and $c^2$ are conjugate in $S_5$ by an odd permutation only).

15. One linear character: a degree- $1$ representation factors through $G/D(G)$ , and $D(A_5) = A_5$ ( $A_5$ is simple nonabelian, and $D(A_5)$ is normal, nontrivial — $A_5$ is not abelian). So $n_1 = 1$ and $n_2^2 + n_3^2 + n_4^2 + n_5^2 = 59$ with each $n_i \geq 2$ . Squares available: $4, 9, 16, 25, 36, 49$ . A sum of four of them equal to $59$ : the largest must be $25$ ( $36 + 4 + 4 + 9 = 53 < 59$ fails to adjust: $36 + 16 + 4 + 4 = 60$ , $36 + 9 + 9 + 4 = 58$ , $36 + 16 + 9 + 4 = 65$ — no combination with $36$ or $49$ works), and $59 - 25 = 34 = 16 + 9 + 9$ (the only way: $16 + 16 + 4 = 36$ , $25 + 9 + 4 = 38$ , $25 + 4 + 4 = 33$ ): degrees $1, 3, 3, 4, 5$ .

16. $\langle\pi, \mathbf 1\rangle = \frac1{60}(5 + 15\cdot1 + 20\cdot2 + 0 + 0) = 1$ (one orbit — Burnside’s count), and $\langle\pi, \pi\rangle = \frac1{60}(25 + 15 + 80 + 0 + 0) = 2$ : $\pi$ contains the trivial character once, and its other constituent is a single irreducible. Hence $\chi_4 = \pi - \mathbf 1$ is irreducible, of degree $4$ , with values $4, 0, 1, -1, -1$ .

17. On pairs, an element fixes $\{i,j\}$ iff it fixes or swaps $i, j$ : the counts are $10$ ( $e$ ), $2$ ( $t = (1\,2)(3\,4)$ fixes $\{1,2\}, \{3,4\}$ ), $1$ ( $(1\,2\,3)$ fixes $\{4,5\}$ ), $0, 0$ . Then $\langle\pi_{10}, \pi_{10}\rangle = \frac1{60}(100 + 15\cdot4 + 20\cdot1) = 3$ : three irreducible constituents, each once. And $\langle\pi_{10}, \mathbf 1\rangle = \frac1{60}(10 + 30 + 20) = 1$ , $\langle\pi_{10}, \chi_4\rangle = \frac1{60}(40 + 0 + 20\cdot1\cdot1 + 0 + 0) = 1$ : so $\pi_{10} = \mathbf 1 + \chi_4 + \chi$ with $\chi$ irreducible of degree $10 - 1 - 4 = 5$ and values $\chi_5 = \pi_{10} - \mathbf 1 - \chi_4 = (5, 1, -1, 0, 0)$ .

18. Write $a, a'$ for the values of $\chi_2, \chi_3$ at $t$ , and $b, b'$ at $s = (1\,2\,3)$ ; all four are real ( $t$ and $s$ are conjugate to their inverses). Column $t$ against column $e$ : $1 + 3a + 3a' + 4\cdot0 + 5\cdot1 = 0$ , so $a + a' = -2$ ; column $t$ with itself: $1 + a^2 + a'^2 + 0 + 1 = \frac{60}{15} = 4$ , so $a^2 + a'^2 = 2$ ; hence $(a + a')^2 = 4 = 2 + 2aa'$ gives $aa' = 1$ and $a = a' = -1$ . Column $s$ against $e$ : $1 + 3(b + b') + 4 - 5 = 0$ gives $b + b' = 0$ ; column $s$ with itself: $1 + b^2 + b'^2 + 1 + 1 = \frac{60} {20} = 3$ gives $b = b' = 0$ .

19. Column $c$ against column $e$ : $1 + 3(x + y) + 4(-1) + 5\cdot0 = 0$ , so $x + y = 1$ . Column $c$ against column $c^2$ (distinct classes, orthogonal): $1 + xy + yx + 1 + 0 = 0$ , so $xy = -1$ . Thus $x, y$ solve $T^2 - T - 1 = 0$ : $\{x, y\} = \{\varphi, \bar\varphi\}$ with $\varphi = \frac{1 + \sqrt5}2$ . Consistency: $x^2 + y^2 = (x+y)^2 - 2xy = 3 = \frac{60}{12} - 2$ — matching the self-column identity $1 + x^2 + y^2 + 1 + 0 = 5$ . The table:

	$e$	$15\,t$	$20\,s$	$12\,c$	$12\,c^2$
$\chi_1$	$1$	$1$	$1$	$1$	$1$
$\chi_2$	$3$	$-1$	$0$	$\varphi$	$\bar\varphi$
$\chi_3$	$3$	$-1$	$0$	$\bar\varphi$	$\varphi$
$\chi_4$	$4$	$0$	$1$	$-1$	$-1$
$\chi_5$	$5$	$1$	$-1$	$0$	$0$

20. $\norm{\chi_2}^2 = \frac1{60}\bigl(9 + 15\cdot1 + 0 + 12\varphi^2 + 12\bar\varphi^2\bigr) = \frac{9 + 15 + 36}{60} = 1$ , using $\varphi^2 + \bar\varphi^2 = (\varphi + \bar\varphi)^2 - 2\varphi\bar\varphi = 1 + 2 = 3$ . Similarly $\langle\chi_2, \chi_3\rangle = \frac1{60}(9 + 15 + 0 + 12(2\varphi\bar\varphi)) = \frac{9 + 15 - 24}{60} = 0$ . Frobenius divisibility: $1, 3, 3, 4, 5$ all divide $60$ . The irrational values $\varphi, \bar\varphi$ are the visible difference with $S_5$ : in $S_5$ every element is conjugate to all generators of its cyclic group with the same cycle type — in particular $c \sim c^2$ — forcing rational (indeed integer) character values; in $A_5$ the splitting of the five-cycles opens the door to $\Q(\sqrt5)$ .

21. A normal subgroup $N$ is a union of classes, contains $e$ , and $\abs N \mid 60$ . The candidate sums: $1{+}15 = 16$ , $1{+}20 = 21$ , $1{+}12 = 13$ , $1{+}24 = 25$ , $1{+}15{+}20 = 36$ , $1{+}15{+}12 = 28$ , $1{+}15{+}24 = 40$ , $1{+}20{+}12 = 33$ , $1{+}20{+}24 = 45$ , $1{+}12{+}12 = 25$ , $1{+}15{+}20{+}12 = 48$ , $1{+}15{+}20{+}24 = 60 - 12 = \dots$ listing all proper sub-sums containing $1$ : none of $13, 16, 21, 25, 28, 33, 36, 40, 45, 48, 13{+}\dots$ divides $60$ except $1$ itself: $N = \{e\}$ or $A_5$ . Simplicity, read off five numbers. And question 11’s criterion is visible too: the class sizes $15 = 3\cdot5$ , $20 = 4\cdot5$ , $12 = 4\cdot3$ are all composite of two primes — no prime-power class, exactly as Burnside’s criterion demands of a simple group.

22. A rotation of $\R^3$ by angle $\theta$ has eigenvalues $1, \eu^{\iu\theta}, \eu^{-\iu\theta}$ : trace $1 + 2\cos\theta$ . For $\theta = \frac{2\pi}5$ : $2\cos\frac{2\pi}5 = \frac{\sqrt5 - 1}2$ , so the trace is $1 + \frac{\sqrt5-1}2 = \frac{1+\sqrt5}2 = \varphi$ . The nontrivial element $\tau$ of $\operatorname{Gal}(\Q(\sqrt5)/\Q)$ applied entrywise to a character table sends characters to characters (it commutes with the defining algebra: $\tau\circ\chi$ is the character of the representation obtained by transporting matrices through $\tau$ on the entries, or abstractly: orthogonality relations are $\Q$ -rational, so $\tau$ permutes their solutions); $\tau$ fixes $\chi_1, \chi_4, \chi_5$ (rational values) and must therefore exchange $\chi_2$ and $\chi_3$ : the second degree- $3$ character carries the conjugated values, no matrix computed. Geometrically, the two representations are the icosahedral action and its composite with an outer automorphism of $A_5$ (conjugation by a transposition), which swaps the two classes of five-cycles.

23. For $z \in Z(G)$ , $\rho(z)$ commutes with every $\rho(g)$ , so by Schur’s lemma $\rho(z) = \lambda\, \mathrm{id}$ ; since $z$ has finite order, $\lambda$ is a root of unity, and $\abs{\chi(z)} = \abs\lambda\,n = n$ . Then

\abs G = \abs G\,\langle\chi, \chi\rangle = \sum_{g \in G}\abs{\chi(g)}^2 \geq \sum_{z \in Z(G)}\abs{\chi(z)}^2 = \abs{Z(G)}\,n^2,

i.e. $n^2 \leq [G : Z(G)]$ . A nonabelian group of order $8$ ( $D_4$ or $Q_8$ ) has five conjugacy classes, so five irreducible degrees with $\sum n_i^2 = 8$ ; the only way to write $8$ as a sum of five squares $\geq 1$ is $1 + 1 + 1 + 1 + 4$ : degrees $1, 1, 1, 1, 2$ . Both groups have center of order $2$ , and the degree- $2$ character attains equality: $2^2 = 4 = [G : Z(G)]$ — the bound is sharp. For $A_5$ , $Z = \{e\}$ and the bound reads $n^2 \leq 60$ : satisfied by $1, 3, 3, 4, 5$ with room to spare ( $25 \leq 60$ ), as it must be since equality would force (by the same chain) $\chi$ to vanish off the center.

24. $\rho(g)^m = \mathrm{id}$ for $m$ the order of $g$ , so $\rho(g)$ is annihilated by $X^m - 1$ , split with simple roots over $\C$ : diagonalizable, with eigenvalues $\lambda_1, \dots, \lambda_n$ roots of unity, in an eigenbasis $(e_i)$ . The products $e_ie_j$ ( $i \leq j$ ) form an eigenbasis of $\operatorname{Sym}^2V$ with eigenvalues $\lambda_i\lambda_j$ , and $e_i \wedge e_j$ ( $i < j$ ) one of $\Lambda^2V$ ; since

\sum_{i<j}\lambda_i\lambda_j = \frac{(\sum_i\lambda_i)^2 - \sum_i\lambda_i^2}2 = \frac{\chi(g)^2 - \chi(g^2)}2,

and the symmetric sum adds $\sum_i\lambda_i^2$ instead of subtracting it, both formulas follow. For $\chi_2 = (3, -1, 0, \varphi, \bar\varphi)$ on the classes $(e, (2,2)\text{-}, 3\text{-cycles}, c, c^2)$ : squaring sends the double transpositions to $e$ , the three-cycles to three-cycles, the class of $c$ onto that of $c^2$ and conversely ( $c^{-1} \sim c$ in $A_5$ , so $c^4 \sim c$ ). Hence $\chi_2(g^2)$ reads $(3, 3, 0, \bar\varphi, \varphi)$ , and, using $\varphi^2 = \varphi + 1$ , $\bar\varphi = 1 - \varphi$ :

\Lambda^2\chi_2 = (3, -1, 0, \varphi, \bar\varphi) = \chi_2, \qquad \operatorname{Sym}^2\chi_2 = (6, 2, 0, 1, 1) .

Decomposing the latter, with the class sizes $1, 15, 20, 12, 12$ : $\langle\cdot, \mathbf 1\rangle = \frac1{60}(6 + 15\cdot2 + 0 + 12 + 12) = 1$ ; $\langle\cdot, \chi_5\rangle = \frac1{60}(30 + 30) = 1$ ; $\langle\cdot, \chi_4\rangle = \frac1{60}(24 - 12 - 12) = 0$ ; $\langle\cdot, \chi_2\rangle = \frac1{60}(18 - 30 + 12(\varphi + \bar\varphi)) = 0$ , and likewise for $\chi_3$ . So $\operatorname{Sym}^2\chi_2 = \mathbf 1 + \chi_5$ (dimensions $6 = 1 + 5$ ) and $\chi_2\otimes\chi_2 = \mathbf 1 + \chi_2 + \chi_5$ (dimensions $9 = 1 + 3 + 5$ ). Geometry: the equivariant isomorphism $\Lambda^2\R^3 \to \R^3$ , $u \wedge v \mapsto u \times v$ , is exactly $\Lambda^2\chi_2 = \chi_2$ for a rotation group; the summand $\mathbf 1$ of the symmetric square is the invariant quadratic form $x^2 + y^2 + z^2$ , and $\chi_5$ lives on the five-dimensional space of traceless symmetric tensors (harmonic quadratics).

25. Column of $c$ against column of $c^2$ :

1\cdot1 + \varphi\bar\varphi + \bar\varphi\varphi + (-1)(-1) + 0 = 1 - 1 - 1 + 1 + 0 = 0,

as orthogonality demands for distinct classes ( $\varphi \bar\varphi = -1$ ). Column of $c$ against itself: $1 + \varphi^2 + \bar\varphi^2 + 1 + 0 = 1 + 3 + 1 = 5 = 60/12 = \abs{Z_{A_5}(c)}$ . Regular character $\sum_in_i\chi_i$ on the four nonidentity columns:

1 - 3 - 3 + 0 + 5 = 0, \qquad 1 + 0 + 0 + 4 - 5 = 0,

on double transpositions and three-cycles, and on the class of $c$ (that of $c^2$ is its Galois conjugate):

1 + 3\varphi + 3\bar\varphi - 4 + 0 = 1 + 3 - 4 = 0,

using $\varphi + \bar\varphi = 1$ . The table passes every audit: it is the character table of $A_5$ .

5.1 Representations, Maschke, Schur

5.2 Characters and orthogonality

5.3 Character tables

5.4 Exercises

5.5 Problem: Burnside’s paqbp^aq^bpaqb theorem

5.5 Problem: Burnside’s $p^aq^b$ theorem