--- title: "Representations of Finite Groups" book: "University Mathematics — Year 3" subject: math language: en chapter: 5 exercises: 12 source: https://one-course.com/books/math/5/en/chapter/5-representations-of-finite-groups --- # Chapter 5 — Representations 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](#def-b3-representations-rep) theory*, is astonishingly effective for finite groups over $\C$: every [representation](#def-b3-representations-rep) splits into [irreducible](#def-b3-representations-rep) ones (Maschke), the [irreducible](#def-b3-representations-rep) ones are pinned down by their *[characters](#def-b3-representations-character)* (traces), and the [characters](#def-b3-representations-character) satisfy orthogonality relations that make computations mechanical. The chapter builds this calculus and its first masterpieces — [character tables](#def-b3-representations-table) 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](#def-b3-representations-rep)* 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](#def-b3-representations-rep)*, of degree $n - 1$. **Theorem 5.3 (Maschke).** Every invariant subspace $W$ of a [representation](#def-b3-representations-rep) $(V, \rho)$ admits an invariant complement. Consequently every [representation](#def-b3-representations-rep) is a direct sum of [irreducible](#def-b3-representations-rep) 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](#def-b3-representations-rep). ∎ **Theorem 5.4 (Schur’s lemma).** Let $f \colon V \to W$ be a morphism of *[irreducible](#def-b3-representations-rep)* [representations](#def-b3-representations-rep). 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](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#def-b3-galois-closure)); $f - \lambda\,\mathrm{id}$ 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](https://one-course.com/books/math/5/en/chapter/1-group-theory#ex-b3-groups-actions), 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](#def-b3-representations-character) 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](#def-b3-representations-rep). 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](#def-b3-representations-rep)*. Then $$ \langle\chi_\sigma, \chi_\rho\rangle = \begin{cases} 1 & \text{if } \rho \cong \sigma,\\ 0 & \text{otherwise:} \end{cases} $$ the [irreducible](#def-b3-representations-rep) [characters](#def-b3-representations-character) 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](#def-b3-representations-rep) ($V_i \not\cong V_j$). Then $m_i = \langle \chi_{V_i}, \chi_V\rangle$: the multiplicities — hence the [representation](#def-b3-representations-rep) up to isomorphism — are determined by the [character](#def-b3-representations-character). Moreover $\langle \chi_V, \chi_V\rangle = \sum_i m_i^2$; in particular $V$ is [irreducible](#def-b3-representations-rep) iff $\langle\chi_V, \chi_V\rangle = 1$. **Proof.** $\chi_V = \sum m_i\chi_{V_i}$ ([Proposition 5.6](#prop-b3-representations-charbasics)); take inner products with each $\chi_{V_i}$ and use orthonormality. Two [representations](#def-b3-representations-rep) with equal [characters](#def-b3-representations-character) have equal multiplicities, hence are isomorphic. ∎ **Theorem 5.10 (The regular representation).** Let $\chi_1, \dots, \chi_r$ be the distinct [irreducible](#def-b3-representations-rep) [characters](#def-b3-representations-character), of degrees $n_i = \chi_i(e)$. The [regular representation](#ex-b3-representations-examples) 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](#def-b3-representations-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](#def-b3-representations-rep) [characters](#def-b3-representations-character) form an orthonormal *basis* of $\mathcal{CF}(G)$: the number of [irreducible representations](#def-b3-representations-rep) equals the number of [conjugacy classes](https://one-course.com/books/math/5/en/chapter/1-group-theory#ex-b3-groups-actions) 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](#def-b3-representations-rep) $(V,\rho)$, 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](#def-b3-representations-character)). If $\rho$ is [irreducible](#def-b3-representations-rep) 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](#def-b3-representations-rep), hence (direct sums) on *every* [representation](#def-b3-representations-rep) — 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](https://one-course.com/books/math/5/en/chapter/1-group-theory#ex-b3-groups-actions). ∎ **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](#thm-b3-representations-orthogonality) 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](#def-b3-representations-rep) have degree $1$; the number of degree-$1$ [characters](#def-b3-representations-character) of any $G$ is $[G : D(G)]$ (they are the [characters](#def-b3-representations-character) of the abelianization). (b) If $N \trianglelefteq G$, the [irreducible representations](#def-b3-representations-rep) of $G/N$ lift (compose with $G \to G/N$) to exactly those [irreducible representations](#def-b3-representations-rep) 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](#ex-b3-representations-examples) 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](#def-b3-representations-rep) are morphisms $G \to \C^\times$ with abelian target: they factor through $G^{\mathrm{ab}} = G/D(G)$ ([Exercise 1.9](https://one-course.com/books/math/5/en/chapter/1-group-theory#exo-b3-groups-9)), and distinct [characters](#def-b3-representations-character) 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](#def-b3-representations-rep) trivial on $N$ factors through the quotient ([Theorem 1.3](https://one-course.com/books/math/5/en/chapter/1-group-theory#thm-b3-groups-firstiso)). ∎ ## 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](#def-b3-representations-rep) [characters](#def-b3-representations-character), columns by [conjugacy classes](https://one-course.com/books/math/5/en/chapter/1-group-theory#ex-b3-groups-actions) (with their sizes displayed). Rows are orthonormal for the weighted product, columns orthogonal ([Corollary 5.12](#cor-b3-representations-column)): the table is severely overdetermined, which is what makes it computable. **Example 5.15 (The table of S3S_3S3​).** Classes: $e$ (size 1), transpositions (3), $3$-cycles (2); so $r = 3$ [irreducibles](#def-b3-representations-rep), 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](#def-b3-representations-rep). **Example 5.16 (The table of S4S_4S4​).** Classes: $e$ [1], transpositions [6], double transpositions [3], $3$-cycles [8], $4$-cycles [6]: five [irreducibles](#def-b3-representations-rep), $\sum n_i^2 = 24$ 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](#prop-b3-representations-onedim)(b), $V$ the Klein group); degree $3$: the standard [representation](#def-b3-representations-rep) 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](#exo-b3-representations-3)’s warmup. **Method 5.17.** To build a [character table](#def-b3-representations-table): (1) list [conjugacy classes](https://one-course.com/books/math/5/en/chapter/1-group-theory#ex-b3-groups-actions) and sizes; (2) count degree-$1$ [characters](#def-b3-representations-character) 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](#def-b3-representations-rep): lift from quotients, subtract $\mathbf 1$ from permutation [characters](#def-b3-representations-character) (check $\langle\chi,\chi\rangle = 1$), multiply known [characters](#def-b3-representations-character) 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.](https://one-course.com/images/onecourse/chapters/math-5/b3-representations/fig-7b5373eb2027.svg) *The [regular representation](#ex-b3-representations-examples) 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](#def-b3-representations-rep) [characters](#def-b3-representations-character) 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](#def-b3-representations-table) of $\Z/4\Z$. (b) Verify both orthogonality relations on it — and recognize the matrix: where has this book seen it before? **Solution of Exercise 5.1.** (a) $\Z/n\Z$ is abelian: all [irreducibles](#def-b3-representations-rep) have degree $1$ ([Proposition 5.13](#prop-b3-representations-onedim)), 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](#def-b3-representations-character) $\chi_k(\bar m) = \eu^{2\iu\pi km/n}$. 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](#thm-b3-representations-orthogonality) generalizes the inversion formula of the DFT. **Exercise 5.2 ★.** Let $G$ act on a finite set $X$ and $\chi$ be the [character](#def-b3-representations-character) of the permutation [representation](#def-b3-representations-rep) $\C^X$. (a) Show $\chi(g) = \abs{\operatorname{Fix}_X(g)}$ and $\langle \mathbf 1, \chi\rangle = \#\{\text{orbits}\}$ — Burnside’s counting lemma ([Exercise 1.5](https://one-course.com/books/math/5/en/chapter/1-group-theory#exo-b3-groups-5)) is a [character](#def-b3-representations-character) computation. (b) Suppose the action is transitive, so $\chi = \mathbf 1 + \psi$. Show that $\psi$ is [irreducible](#def-b3-representations-rep) iff the action is *$2$-transitive* (transitive on ordered pairs of distinct points). *(Compute $\langle\chi,\chi\rangle$ as the number of [orbits](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-action) on $X \times X$.)* (c) Conclude that the standard [representation](#def-b3-representations-rep) of $S_n$ ($n \geq 2$) is [irreducible](#def-b3-representations-rep). **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](https://one-course.com/books/math/5/en/chapter/1-group-theory#exo-b3-groups-5)) — equivalently, this computes the multiplicity of the trivial [representation](#def-b3-representations-rep), whose isotypic space is the space of $G$-invariant vectors, of dimension the number of [orbits](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-action) (one indicator per [orbit](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-action)). (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](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-action); there is exactly one other [orbit](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-action) iff $G$ is transitive on distinct pairs: $\langle\psi,\psi\rangle = 1$ iff $2$-transitive ([Corollary 5.9](#cor-b3-representations-multiplicity)). (c) $S_n$ is $2$-transitive on $\{1,\dots,n\}$ (send any distinct pair anywhere): $\psi = \chi_{\mathrm{std}}$ is [irreducible](#def-b3-representations-rep). **Exercise 5.3 ★.** Rebuild the table of $S_3$ from scratch following [Method 5.17](#met-b3-representations-table), then verify two row and two column orthogonality relations in the $S_4$ table of [Example 5.16](#ex-b3-representations-s4). Decompose the permutation [character](#def-b3-representations-character) of $S_4$ acting on $\{1,2,3,4\}$ and the [character](#def-b3-representations-character) $\chi_{\mathrm{std}}^2$ (pointwise square) into [irreducibles](#def-b3-representations-rep). **Solution of Exercise 5.3.** $S_3$: three classes, $\sum n_i^2 = 6 = 1 + 1 + 4$; the two degree-$1$ [characters](#def-b3-representations-character) 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](#ex-b3-representations-s3). $S_4$ 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](#def-b3-representations-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](#def-b3-representations-table) of $D_4$ and of $Q_8$. Conclude that two non-isomorphic groups can have identical [character tables](#def-b3-representations-table) — what group-theoretic data does the table nevertheless capture in this pair (orders of [centers](https://one-course.com/books/math/5/en/chapter/1-group-theory#ex-b3-groups-actions), abelianizations, number of involutions)? Which of these does it *fail* to capture? **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](https://one-course.com/books/math/5/en/chapter/1-group-theory#pb-b3-groups-1)). The table *does* capture: $\abs G$, class sizes, the [center](https://one-course.com/books/math/5/en/chapter/1-group-theory#ex-b3-groups-actions) ($\{g : \abs{\chi_i(g)} = n_i\ \forall i\}$: order $2$ in both), the abelianization, the whole lattice of [normal subgroups](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-normal) (kernels and intersections, [Exercise 5.6](#exo-b3-representations-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](#def-b3-representations-table). **Exercise 5.5 ★★.** [Character table](#def-b3-representations-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](https://one-course.com/books/math/5/en/chapter/1-group-theory#ex-b3-groups-actions) in $S_4$ and $A_4$, as in [Exercise 1.11](https://one-course.com/books/math/5/en/chapter/1-group-theory#exo-b3-groups-11)). (b) Find the three degree-$1$ [characters](#def-b3-representations-character) (via $A_4/V \cong \Z/3\Z$) and the degree-$3$ [character](#def-b3-representations-character) (restrict $\chi_{\mathrm{std}}$ from $S_4$), and assemble the table. (c) Read off the [normal subgroups](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-normal) of $A_4$ from the table (kernels $\{g : \chi(g) = \chi(e)\}$ and their intersections). **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](#def-b3-representations-character) ($\omega = \eu^{2\iu\pi/3}$; the $3$-cycle classes map to $\bar1, \bar2$); the restriction of $\chi_{\mathrm{std}}$ stays [irreducible](#def-b3-representations-rep) ($\langle\chi,\chi\rangle = \frac1{12}(9 + 3 \cdot 1 + 0 + 0) = 1$): $$ \begin{array}{c|cccc} A_4 & e\,[1] & (1\,2)(3\,4)\,[3] & (1\,2\,3)\,[4] & (1\,3\,2)\,[4]\\ \hline \mathbf 1 & 1 & 1 & 1 & 1\\ \chi_\omega & 1 & 1 & \omega & \omega^2\\ \chi_{\bar\omega} & 1 & 1 & \omega^2 & \omega\\ \chi_3 & 3 & -1 & 0 & 0 \end{array} $$ (c) Kernels: $\ker\chi_\omega = \ker\chi_{\bar\omega} = V$; $\ker\chi_3 = \{e\}$ (no other entry has modulus $3$). The [normal subgroups](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-normal) are the intersections of kernels ([Exercise 5.6](#exo-b3-representations-6)): $\{e\}$, $V$, $A_4$ — in particular $A_4$ has no [normal subgroup](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-normal) 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](#def-b3-representations-rep) ([Proposition 5.6](#prop-b3-representations-charbasics), equality case). (b) Show that every [normal subgroup](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-normal) of $G$ is an intersection of kernels of [irreducible](#def-b3-representations-rep) [characters](#def-b3-representations-character). *(Represent $G/N$ faithfully: its [regular representation](#ex-b3-representations-examples).)* (c) Deduce: $G$ is simple iff $\ker\chi_i = \{e\}$ for every nontrivial [irreducible](#def-b3-representations-rep) $\chi_i$ — simplicity is readable from the [character table](#def-b3-representations-table). **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](#prop-b3-representations-charbasics)) with $n\lambda = n$: $\rho(g) = \mathrm{id}$. The converse is clear. (b) Let $N \trianglelefteq G$. The [regular representation](#ex-b3-representations-examples) of $G/N$ is faithful; decompose it into [irreducibles](#def-b3-representations-rep) of $G/N$ and lift them to $G$ ([Proposition 5.13](#prop-b3-representations-onedim)(b)): [irreducible](#def-b3-representations-rep) [characters](#def-b3-representations-character) $\chi_{i_1}, \dots$ 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](#def-b3-representations-rep) $\chi$, $\ker\chi \trianglelefteq G$ is not $G$ (an [irreducible representation](#def-b3-representations-rep) trivial on all of $G$ is the trivial [character](#def-b3-representations-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](#def-b3-representations-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](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-normal) 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](#def-b3-representations-rep) 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](#def-b3-representations-character) of degree $p$. *(Use $[G : D(G)]$ and $\sum n_i^2 = \abs G$; here $D(G) = Z(G)$ has order $p$.)* **Solution of Exercise 5.7.** Order $8$ nonabelian: the number of degree-$1$ [characters](#def-b3-representations-character) 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](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-derived) of $p$-groups ([Example 1.30](https://one-course.com/books/math/5/en/chapter/1-group-theory#ex-b3-groups-solvableexamples)). So the pattern is $(1,1,1,1,2)$. Faithfulness of $\chi_5$: the four degree-$1$ [characters](#def-b3-representations-character) 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](#ex-b3-representations-examples) 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](#def-b3-representations-character) 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](#pb-b3-representations-1), question 8) hence $n_i \in \{p\}$ ($n_i = p^2$ would exceed: $p^4 > p^3 - p^2$): exactly $p - 1$ [characters](#def-b3-representations-character) of degree $p$. **Exercise 5.8 ★★★.** For finite groups $G, H$: show that the [class functions](#def-b3-representations-character) $\chi(g)\psi(h)$ on $G \times H$, for $\chi, \psi$ [irreducible](#def-b3-representations-rep) [characters](#def-b3-representations-character) of $G, H$, are exactly the [irreducible](#def-b3-representations-rep) [characters](#def-b3-representations-character) of $G \times H$. *(Orthonormality is a direct computation; completeness by counting classes.)* Deduce the [character table](#def-b3-representations-table) of $\Z/2\Z \times \Z/2\Z$ and re-derive [Proposition 5.13](#prop-b3-representations-onedim)(a) for finite abelian groups via the structure theorem. **Solution of Exercise 5.8.** Define, for [representations](#def-b3-representations-rep) $\rho, \sigma$ of $G, H$ in $V, W$, the [representation](#def-b3-representations-rep) $\rho \boxtimes \sigma$ 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](#def-b3-representations-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](#def-b3-representations-rep) ([Corollary 5.9](#cor-b3-representations-multiplicity)). These are $r_Gr_H$ distinct [irreducible](#def-b3-representations-rep) [characters](#def-b3-representations-character); 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](#thm-b3-representations-numberirr)). For $(\Z/2\Z)^2$: the four sign [characters](#def-b3-representations-character) $(\pm1)\otimes(\pm1)$ — the table of [Exercise 5.4](#exo-b3-representations-4)’s top-left block. A finite abelian group is a product of cyclic groups ([Corollary 3.13](https://one-course.com/books/math/5/en/chapter/3-modules-over-a-principal-ideal-domain#cor-b3-modules-abelian)); its [irreducible](#def-b3-representations-rep) [characters](#def-b3-representations-character) are products of the cyclic ones ([Exercise 5.1](#exo-b3-representations-1)): all of degree $1$. **Exercise 5.9 ★★★.** The [character table](#def-b3-representations-table) of $A_5$ (classes of sizes $1, 15, 20, 12, 12$ from [Exercise 1.11](https://one-course.com/books/math/5/en/chapter/1-group-theory#exo-b3-groups-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](#exo-b3-representations-6)(c) with simplicity, no other $n_i = 1$)*. (b) Construct the degree-$4$ [character](#def-b3-representations-character) (permutation action on $5$ points) and the degree-$5$ [character](#def-b3-representations-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](#exo-b3-representations-6)(c)). **Solution of Exercise 5.9.** (a) $D(A_5) = A_5$ (simple nonabelian), so the only degree-$1$ [character](#def-b3-representations-character) is $\mathbf 1$ ([Proposition 5.13](#prop-b3-representations-onedim)). 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](#def-b3-representations-rep). Action on the six Sylow $5$-subgroups: an involution fixes exactly $2$ (the [normalizers](https://one-course.com/books/math/5/en/chapter/1-group-theory#ex-b3-groups-actions) 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](#def-b3-representations-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](#def-b3-representations-rep). Two rows $(3, a, b, c, d)$, $(3, a', b', c', d')$ remain. Column norms ([Corollary 5.12](#cor-b3-representations-column)): 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](#exo-b3-representations-6)(c), $A_5$ is simple. **Exercise 5.10 ★★.** Let $\rho$ be an [irreducible representation](#def-b3-representations-rep) 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](#def-b3-representations-character)*), and deduce $\abs{\chi(z)} = n$ for central $z$. Application: if $G$ has a faithful [irreducible representation](#def-b3-representations-rep), then $Z(G)$ is cyclic. **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](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#thm-b3-galois-cyclic)). **Exercise 5.11 ★★.** (Isotypic projections) Let $(V, \rho)$ be a [representation](#def-b3-representations-rep) of $G$ and $\chi_i$ an [irreducible](#def-b3-representations-rep) [character](#def-b3-representations-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](#def-b3-representations-rep) subrepresentation $W \subseteq V$ of [character](#def-b3-representations-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](#def-b3-representations-rep) subrepresentations of [character](#def-b3-representations-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](#def-b3-representations-rep). (c) For the [regular representation](#ex-b3-representations-examples) of $S_3$ and the sign [character](#def-b3-representations-character) $\varepsilon$, write out $p_\varepsilon$ explicitly as an element of the group algebra and check $p_\varepsilon^2 = p_\varepsilon$ by hand. **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](#def-b3-representations-character)): $p_i$ commutes with the action. On an [irreducible](#def-b3-representations-rep) $W$ of [character](#def-b3-representations-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](#def-b3-representations-rep) (Maschke): $p_i$ acts as identity on the summands of [character](#def-b3-representations-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](#def-b3-representations-rep) 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](#ex-b3-representations-examples) is the line spanned by $\sum_g\varepsilon(g)e_g$: the sign [representation](#def-b3-representations-rep) appears with multiplicity $1$, as the general theory demands.) **Exercise 5.12 ★★.** (Reading a table) The [character table](#def-b3-representations-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](#def-b3-representations-character) (one trivial; the other takes value $-1$ exactly on the classes of sizes $6$ and $6$), then the degree-$2$ [character](#def-b3-representations-character) via column orthogonality with the identity column, then the two degree-$3$ [characters](#def-b3-representations-character) 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](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-normal) via [Exercise 5.6](#exo-b3-representations-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](#def-b3-representations-character) factor through the quotient?)*. **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](#def-b3-representations-character) is $\chi_2 = \varepsilon$ with values $1, -1, 1, -1, 1$. For the degree-$2$ [character](#def-b3-representations-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](#def-b3-representations-character) times a linear one is again [irreducible](#def-b3-representations-rep) — 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](#exo-b3-representations-6)(b) these are *all* the [normal subgroups](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-normal) of $S_4$. (c) The [characters](#def-b3-representations-character) 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](#def-b3-representations-rep) 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](#def-b3-representations-character)), $G/V_4 \cong S_3$: 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](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-derived) — a statement about abstract groups whose only known proofs for half a century went through [character](#def-b3-representations-character) theory. This problem builds the proof in full, assembling [Chapter 1](https://one-course.com/books/math/5/en/chapter/1-group-theory#ch-b3-groups) ([solvability](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-derived)), [Chapter 3](https://one-course.com/books/math/5/en/chapter/3-modules-over-a-principal-ideal-domain#ch-b3-modules) ([finitely generated](https://one-course.com/books/math/5/en/chapter/3-modules-over-a-principal-ideal-domain#def-b3-modules-free) $\Z$-modules) and this chapter. Throughout, $\chi_1, \dots, \chi_r$ are the [irreducible](#def-b3-representations-rep) [characters](#def-b3-representations-character) of $G$, $n_i = \chi_i(e)$. **Part I — [Algebraic](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#def-b3-galois-algebraic) integers.** An *[algebraic](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#def-b3-galois-algebraic) integer* is a root of a *monic* polynomial of $\Z[X]$. 1. Show that $\alpha$ is an [algebraic](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#def-b3-galois-algebraic) integer iff the ring $\Z[\alpha]$ is a [finitely generated](https://one-course.com/books/math/5/en/chapter/3-modules-over-a-principal-ideal-domain#def-b3-modules-free) $\Z$ -module. 2. Deduce that the [algebraic](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#def-b3-galois-algebraic) integers form a subring of $\C$ . *(If $\Z[\alpha], \Z[\beta]$ are [finitely generated](https://one-course.com/books/math/5/en/chapter/3-modules-over-a-principal-ideal-domain#def-b3-modules-free), so is $\Z[\alpha, \beta]$, and submodules of [finitely generated](https://one-course.com/books/math/5/en/chapter/3-modules-over-a-principal-ideal-domain#def-b3-modules-free) $\Z$-modules are [finitely generated](https://one-course.com/books/math/5/en/chapter/3-modules-over-a-principal-ideal-domain#def-b3-modules-free), by [Theorem 3.5](https://one-course.com/books/math/5/en/chapter/3-modules-over-a-principal-ideal-domain#thm-b3-modules-submodule) and a presentation argument — or directly: a submodule of $\Z^n$ is free of rank $\leq n$.)* 3. Show that a *rational* [algebraic](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#def-b3-galois-algebraic) integer is an integer. *(Rational root theorem.)* 4. Show that every [character](#def-b3-representations-character) value $\chi(g)$ is an [algebraic](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#def-b3-galois-algebraic) integer. **Part II — The class-sum relations.** Fix an [irreducible](#def-b3-representations-rep) $(V, \rho)$ of degree $n$ and [character](#def-b3-representations-character) $\chi$. For a [conjugacy class](https://one-course.com/books/math/5/en/chapter/1-group-theory#ex-b3-groups-actions) $C$, let $S_C = \sum_{g \in C}\rho(g) \in \mathcal L(V)$. 5. Show that $S_C$ is equivariant, hence $S_C = \omega_C\,\mathrm{id}$ with $$\omega_C = \frac{\abs C\,\chi(g_C)}{n} \qquad (g_C \in C \text{ any representative}).$$ 6. 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 $\omega_C\,\omega_{C'} = \sum_{C''} a_{CC'C''}\,\omega_{C''}$ . 7. Conclude that each $\omega_C$ is an [algebraic](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#def-b3-galois-algebraic) integer. *(The $\Z$-module generated by $1$ and the $\omega_C$ is a [finitely generated](https://one-course.com/books/math/5/en/chapter/3-modules-over-a-principal-ideal-domain#def-b3-modules-free) 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.)* 8. Deduce *[Frobenius](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#thm-b3-galois-finitefields) divisibility* : $n_i$ divides $\abs G$ for every [irreducible](#def-b3-representations-rep) degree. *(Compute $\frac{\abs G}{n} = \frac{\abs G}{n} \langle\chi,\chi\rangle = \sum_C \omega_C\, \overline{\chi(g_C)}$: an [algebraic](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#def-b3-galois-algebraic) integer that is rational.)* **Part III — Burnside’s simplicity criterion.** 9. Let $\chi$ be [irreducible](#def-b3-representations-rep) 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](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#def-b3-galois-algebraic) integer. 10. Suppose moreover $0 < \abs{\chi(g_C)} < n$ . Show this is impossible: the [algebraic](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#def-b3-galois-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](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#def-b3-galois-algebraic) integer with $0 < \abs N < 1$ — justify each assertion, quoting [Theorem 4.23](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#thm-b3-galois-cyclotomicirred) for the [Galois group](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#def-b3-galois-galois) and the fact that $\sigma$ permutes roots of unity. Conclude: either $\chi(g_C) = 0$ or $\rho(g_C)$ is scalar ( [Proposition 5.6](#prop-b3-representations-charbasics) ). 11. (Burnside’s criterion) Let $C \neq \{e\}$ be a [conjugacy class](https://one-course.com/books/math/5/en/chapter/1-group-theory#ex-b3-groups-actions) 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](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#def-b3-galois-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](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-normal); use faithfulness from [Exercise 5.6](#exo-b3-representations-6))* . Conclude: *no simple nonabelian group has a [conjugacy class](https://one-course.com/books/math/5/en/chapter/1-group-theory#ex-b3-groups-actions) of prime power size $> 1$.* **Part IV — The theorem.** 12. 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](https://one-course.com/books/math/5/en/chapter/1-group-theory#ex-b3-groups-actions) of a Sylow $q$ -subgroup ( [Theorem 1.12](https://one-course.com/books/math/5/en/chapter/1-group-theory#thm-b3-groups-pfixed) ) and consider the size of its [conjugacy class](https://one-course.com/books/math/5/en/chapter/1-group-theory#ex-b3-groups-actions) $[G : Z_G(z)]$ , a power of $p$ (why?); apply question 11. 13. Conclude by induction on $\abs G$ : *every group of order $p^aq^b$ is [solvable](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-derived)* (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](#def-b3-representations-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](https://one-course.com/books/math/5/en/chapter/1-group-theory#exo-b3-groups-11). 14. Recall from [Exercise 1.11](https://one-course.com/books/math/5/en/chapter/1-group-theory#exo-b3-groups-11) the five [conjugacy classes](https://one-course.com/books/math/5/en/chapter/1-group-theory#ex-b3-groups-actions) 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. 15. Show that the [irreducible](#def-b3-representations-rep) 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](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#prop-b3-galois-perfect) ( $D(A_5) = A_5$ ), so the trivial [character](#def-b3-representations-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)* . 16. Let $\pi$ be the permutation [character](#def-b3-representations-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](#def-b3-representations-rep) of degree $4$ , with values $4, 0, 1, -1, -1$ . 17. 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](#def-b3-representations-rep) $\chi_5$ of degree $5$ with values $5, 1, -1, 0, 0$ . 18. The two remaining [irreducibles](#def-b3-representations-rep) $\chi_2, \chi_3$ 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. 19. 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](#def-b3-representations-table) of $A_5$. 20. 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](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#thm-b3-galois-finitefields) divisibility (question 8) for all five degrees. Where in the table do you *see* a difference with $S_5$ , all of whose [character](#def-b3-representations-character) values are rational integers? 21. Deduce from the table alone that $A_5$ is simple: a [normal subgroup](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-normal) is a union of [conjugacy classes](https://one-course.com/books/math/5/en/chapter/1-group-theory#ex-b3-groups-actions) 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 $> 1$ . 22. (Icosahedral coda) $A_5$ is the rotation group of the icosahedron, and the degree- $3$ [representations](#def-b3-representations-rep) 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](#def-b3-representations-character) must carry the conjugate value: the [Galois group](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#def-b3-galois-galois) of $\Q(\sqrt5)/\Q$ acts on the whole [character table](#def-b3-representations-table) (entrywise), permuting the [irreducible](#def-b3-representations-rep) [characters](#def-b3-representations-character) . **Part VI — Complements: a central bound and the tensor square.** 23. (Sharper than a divisibility) Let $\chi$ be [irreducible](#def-b3-representations-rep) 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](#def-b3-representations-rep) degrees $1, 1, 1, 1, 2$ *(five [conjugacy classes](https://one-course.com/books/math/5/en/chapter/1-group-theory#ex-b3-groups-actions); write $8$ as a sum of five squares)* and attain equality $4 = [G : Z(G)]$; check the bound on $A_5$, whose [center](https://one-course.com/books/math/5/en/chapter/1-group-theory#ex-b3-groups-actions) is trivial. 24. (Tensor square of $\chi_2$) For $g$ of finite order, $\rho(g)$ is diagonalizable with root-of-unity eigenvalues; deduce the [character](#def-b3-representations-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 $\R^3$. 25. (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](#def-b3-representations-character) $\sum_i n_i\chi_i(g) = 0$ on each of the four nonidentity columns of the table. **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](https://one-course.com/books/math/5/en/chapter/3-modules-over-a-principal-ideal-domain#def-b3-modules-free) $\Z$-module $\Z[\alpha, \beta]$, hence [finitely generated](https://one-course.com/books/math/5/en/chapter/3-modules-over-a-principal-ideal-domain#def-b3-modules-free) ($\Z[\alpha,\beta]$, 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](https://one-course.com/books/math/5/en/chapter/3-modules-over-a-principal-ideal-domain#thm-b3-modules-submodule), and its image generates). By question 1, $\alpha + \beta$ and $\alpha\beta$ are [algebraic](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#def-b3-galois-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](#prop-b3-representations-charbasics)), each an [algebraic](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#def-b3-galois-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](#def-b3-representations-character) 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](https://one-course.com/books/math/5/en/chapter/3-modules-over-a-principal-ideal-domain#def-b3-modules-free), and $\omega_C M \subseteq M$ for each $C$. In particular $\Z[\omega_C] \subseteq M$ is [finitely generated](https://one-course.com/books/math/5/en/chapter/3-modules-over-a-principal-ideal-domain#def-b3-modules-free) (submodule, as in question 2), and question 1 makes $\omega_C$ an [algebraic](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#def-b3-galois-algebraic) integer. **8.** For an [irreducible](#def-b3-representations-rep) $\chi$ 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](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#def-b3-galois-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](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#def-b3-galois-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](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#thm-b3-galois-cyclotomicirred)), 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](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#def-b3-galois-algebraic) integer (it has the same [minimal polynomial](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#def-b3-galois-algebraic) as $\alpha$). The product $P = \prod_\sigma \sigma(\alpha)$ is fixed by the whole [Galois group](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#def-b3-galois-galois), hence rational ([Theorem 4.21](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#thm-b3-galois-fundamental)(1)), and it is an [algebraic](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#def-b3-galois-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](#prop-b3-representations-charbasics)). **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](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#def-b3-galois-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](#exo-b3-representations-6)(c)), and $Z_i = \{g : \rho_i(g) \text{ scalar}\}$ is a [normal subgroup](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-normal) (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](https://one-course.com/books/math/5/en/chapter/1-group-theory#ex-b3-groups-actions) 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](https://one-course.com/books/math/5/en/chapter/1-group-theory#thm-b3-groups-pfixed)) is normal, so $Z(G) = G$: abelian. Otherwise take $Q$ a Sylow $q$-subgroup and $z \in Z(Q) \setminus\{e\}$ ([Theorem 1.12](https://one-course.com/books/math/5/en/chapter/1-group-theory#thm-b3-groups-pfixed) 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](https://one-course.com/books/math/5/en/chapter/1-group-theory#ex-b3-groups-actions) is a nontrivial [normal subgroup](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-normal), 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](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-simple) of order $p^aq^b$ is abelian ($\cong \Z/p\Z$). **13.** Induction on $\abs G$ ($\abs G = 1$: [solvable](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-derived)). If $G$ is simple, question 12 makes it abelian, hence [solvable](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-derived). Otherwise pick $N$ a nontrivial proper [normal subgroup](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-normal): $\abs N$ and $\abs{G/N}$ are again of the form $p^{a'}q^{b'}$ and smaller, so $N$ and $G/N$ are [solvable](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-derived) by induction, and $G$ is [solvable](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-derived) ([Proposition 1.29](https://one-course.com/books/math/5/en/chapter/1-group-theory#prop-b3-groups-derived)). — With three primes, the key step fails: the index of a [Sylow subgroup](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-sylow) 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](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-derived). **14.** The classes and sizes are [Exercise 1.11](https://one-course.com/books/math/5/en/chapter/1-group-theory#exo-b3-groups-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](#def-b3-representations-character): a degree-$1$ [representation](#def-b3-representations-rep) 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](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-action) — Burnside’s count), and $\langle\pi, \pi\rangle = \frac1{60}(25 + 15 + 80 + 0 + 0) = 2$: $\pi$ contains the trivial [character](#def-b3-representations-character) once, and its other constituent is a single [irreducible](#def-b3-representations-rep). Hence $\chi_4 = \pi - \mathbf 1$ is [irreducible](#def-b3-representations-rep), 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](#def-b3-representations-rep) 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](#def-b3-representations-rep) 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](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#thm-b3-galois-finitefields) 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](#def-b3-representations-character) values; in $A_5$ the splitting of the five-cycles opens the door to $\Q(\sqrt5)$. **21.** A [normal subgroup](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-normal) $N$ 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](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-simple). **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](#def-b3-representations-table) sends [characters](#def-b3-representations-character) to [characters](#def-b3-representations-character) (it commutes with the defining algebra: $\tau\circ\chi$ is the [character](#def-b3-representations-character) of the [representation](#def-b3-representations-rep) 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](#def-b3-representations-character) carries the conjugated values, no matrix computed. Geometrically, the two [representations](#def-b3-representations-rep) 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](https://one-course.com/books/math/5/en/chapter/1-group-theory#ex-b3-groups-actions), so five [irreducible](#def-b3-representations-rep) 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](https://one-course.com/books/math/5/en/chapter/1-group-theory#ex-b3-groups-actions) of order $2$, and the degree-$2$ [character](#def-b3-representations-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](https://one-course.com/books/math/5/en/chapter/1-group-theory#ex-b3-groups-actions). **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