We introduce the notion of character tables—which serve both to structure information and to facilitate calculations—through an extended example.
Example 8.1 (Character table of \(\mathcal{A}_{4}\))
Recall the action of \(\mathcal{A}_{4}\) on the tetrahedron, which induces an action on \(\mathbb{C}^3\) (Example 3.50): \[ \begin{aligned} (1\ 2\ 3) & \mapsto A \coloneqq \begin{pmatrix} 0&0&1\\1&0&0\\0&1&0 \end{pmatrix}, \text{ and} \\ (1\ 2)(3\ 4) & \mapsto B \coloneqq \begin{pmatrix} -1&0&0\\0&-1&0\\0&0&1 \end{pmatrix} \end{aligned} \] We are interested in the character of this representation. Rather than compute matrix representatives for all \(12\) elements of \(\mathcal{A}_{4}\), we will use the fact that characters are class functions to simplify our analysis.
First, we need to determine the conjugacy classes of \(\mathcal{A}_{4}\). These are inherited from \(\mathcal{S}_{4}\), namely the set of \(3\)-cycles and \((2,2)\)-cycles (recall Example 3.20), where the former splits into two conjugacy classes represented by \((1\ 2\ 3)\) and \((1\ 3\ 2)\)—this happens because we have fewer permutations to conjugate by!
To see this more carefully, one can use Theorem 3.8: \[ \begin{aligned} 24 & = |\mathcal{S}_{4}| = |\operatorname{C}_{\mathcal{S}_{4}}({(1\ 2\ 3)})|\ \overbrace{|\operatorname{cl}_{\mathcal{S}_{4}}({(1\ 2\ 3)})|}^{\text{8 $3$-cycles}} = 3 \cdot 8 \\ & = |\operatorname{C}_{\mathcal{S}_{4}}({(1\ 2)(3\ 4)})|\ \underbrace{|\operatorname{cl}_{\mathcal{S}_{4}}({(1\ 2)(3\ 4)})|}_{\text{3 $(2,2)$-cycles}} = 8 \cdot 3 \end{aligned} \] Note that we have \[ \begin{aligned} \operatorname{C}_{\mathcal{S}_{4}}({(1\ 2\ 3)}) & = \langle (1\ 2\ 3) \rangle \subseteq \mathcal{A}_{4} \\ \operatorname{C}_{\mathcal{S}_{4}}({(1\ 2)(3\ 4)}) & = \langle (1\ 2)(3\ 4), (1\ 3)(2\ 4), (1\ 2) \rangle \not \subseteq \mathcal{A}_{4}. \end{aligned} \] This means that \[ \begin{aligned} \operatorname{C}_{\mathcal{A}_{4}}({(1\ 2\ 3)}) & = \langle (1\ 2\ 3) \rangle = \operatorname{C}_{\mathcal{S}_{4}}({(1\ 2\ 3)}), \text{ but} \\ \operatorname{C}_{\mathcal{A}_{4}}({(1\ 2)(3\ 4)}) & = \langle (1\ 2)(3\ 4), (1\ 3)(2\ 4) \rangle \subset \operatorname{C}_{\mathcal{S}_{4}}({(1\ 2)(3\ 4)}). \end{aligned} \] We thus conclude: \[ \begin{aligned} |\operatorname{cl}_{\mathcal{A}_{4}}({(1\ 2\ 3)})| & = [\mathcal{A}_{4}:\langle (1\ 2\ 3) \rangle] = \tfrac{12}{3} = 4, \text{ and} \\ |\operatorname{cl}_{\mathcal{A}_{4}}({(1\ 2)(3\ 4)})| & = [\mathcal{A}_{4}:\langle (1\ 2)(3\ 4), (1\ 3)(2\ 4) \rangle] = \tfrac{12}{4} = 3, \end{aligned} \] i.e., the first conjugacy class has split in two. To find a representative for the other \(\mathcal{A}_{4}\) conjugacy class, i.e., the set \(\operatorname{cl}_{\mathcal{S}_{4}}({(1\ 2\ 3)}) \setminus \operatorname{cl}_{\mathcal{A}_{4}}({(1\ 2\ 3)})\), we can simply conjugate \((1\ 2\ 3)\) by an odd permutation. We choose \((2\ 3)\) for simplicity, so that \((1\ 3\ 2) = (1\ 2\ 3)^2\) is our representative, and note that \(A^2 = \begin{psmallmatrix}0&1&0\\0&0&1\\1&0&0\end{psmallmatrix}\).
We can arrange the representatives of our conjugacy classes, the value that the representation takes on them, and the resulting trace \(\chi\) as follows: \[ \begin{array}{r|rrrr} \text{class size} & 1 & 4 & 4 & 3 \\ \text{representative} & \mathbb{I}& (1\ 2\ 3) & (1\ 3\ 2) & (1\ 2)(3\ 4) \\ \hline \text{matrix} & \mathbb{I}& A & A^2 & B \\ \chi & 3 & 0 & 0 & -1. \end{array} \] Note that the matrices in the table above are not equal for every value in the conjugacy class, but the traces are.
To check that this representation is irreducible, we compute (by Remark 7.1): \[ \langle \chi, \chi \rangle = \frac{1}{12} \left(1 \cdot 3^2 + 4 \cdot 0^2 + 4 \cdot 0^2 + 3 \cdot (-1)^2 \right) = 1. \]
At this point in the course, we have also computed the \(1\)-dimensional representations of \(\mathcal{A}_{4}\) as lifts from its abelianization (see Exercise C.1 and Exercise G.2): indeed, these are all the distinct irreducible representations of \(\mathcal{A}_{4}\), since we can only have as many irreducibles as there are conjugacy classes. We can list all characters in a single table:
\[ \begin{array}{r|rrrr} \text{size} & 1 & 4 & 4 & 3 \\ \text{class} & \mathbb{I}& (1\ 2\ 3) & (1\ 3\ 2) & (1\ 2)(3\ 4) \\ \hline \text{trivial} & 1 & 1 & 1 & 1 \\ U_{1,3} & 1 & \zeta & \zeta^2 & 1 \\ U_{2,3} & 1 & \zeta^2 & \zeta & 1 \\ \text{tetrahedral} & 3 & 0 & 0 & -1, \end{array} \] where \(\zeta\) is a non-trivial third root of unity.
The previous analysis leads us to a means of bookkeeping irreducible representations of a finite group via its character table. This is a diagram, understood as a sort of decorated matrix with conjugacy classes listed across its top (usually depicted via a representative \(g\) and often augmented above with the size of the class if \(G\) is non-abelian) and the distinct irreducible representations \(G\) listed down the left. The contents of the table are the values each character \(\chi\) takes on the conjugacy classes.
If \(g_1, \dots, g_r \in G\) represent the conjugacy classes of \(G\) and \(V_1, \dots, V_d\) stand for the isomorphism classes of irreducible representations, then a character table for \(G\) might take the form: \[ \begin{array}{r|rrr} \text{size} & |\operatorname{cl}_{}({g_1})| & \cdots & |\operatorname{cl}_{}({g_r})| \\ \text{class} & g_1 & \cdots & g_r \\ \hline V_1 & \chi_{1}(g_1) & \cdots & \chi_{1}(g_r) \\ \vdots & \vdots & \ddots & \vdots \\ V_d & \chi_{d}(g_1) & \cdots & \chi_{d}(g_r) \\ \end{array} \] We will soon see that \(r = d\) holds in general—this is Theorem 8.1.
The character orthonormality as in Equation 7.3 is called the row orthogonality of the character table. We emphasize that the rows themselves, understood as vectors, are not usually orthogonal using the standard Hermitian dot product. In Example 8.1, for example, we see that the dot product of the first and last rows is \[ \begin{pmatrix}1\\1\\1\\1\end{pmatrix}^* \begin{pmatrix}3\\0\\0\\-1\end{pmatrix} = 2 \not = 0 \] To correctly interpret row orthogonality, we must use the inner product of Equation 7.4. We can write this as a weighted dot product (in the style of Theorem 2.12) using a diagonal matrix: \[ \begin{pmatrix}\chi_i(g_1)\\\vdots\\\chi_i(g_r)\end{pmatrix}^* \begin{pmatrix} \frac{|\operatorname{cl}_{}({g_1})|}{|G|} & \cdots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \cdots & \frac{|\operatorname{cl}_{}({g_r})|}{|G|} \end{pmatrix} \begin{pmatrix}\chi_i(g_1)\\\vdots\\\chi_i(g_r) \end{pmatrix} = \left\{ \begin{array}{rr} 1 & \text{if } i = j \\ 0 & \text{if } i \not = j \\ \end{array} \right. \] If we write \(M\) is the character table regarded as a matrix, then row orthogonality can be written as \[ \overline{M} \operatorname{Diag}(\tfrac{|\operatorname{cl}_{}({g_1})|}{|G|}, \dots, \tfrac{|\operatorname{cl}_{}({g_r})|}{|G|}) M^T = \mathbb{I} \tag{8.1}\]
Example 8.2 (Character table of \(\mathcal{S}_{3}\))
We have identified three distinct irreducible representations of the symmetric group \(\mathcal{S}_{3}\): the trivial representation \(U\), the alternating representation \(U'\), and the standard representation \(V\). Note, for example, the orthogonality relation \[ \langle \chi_{U'}, \chi_V \rangle = \tfrac{1}{6}(1 \cdot 1 \cdot 2 + 3 \cdot (-1) \cdot 0 + 2 \cdot 1 \cdot (-1)) = 0. \] Since \(\mathcal{S}_{3}\) has three conjugacy classes, these must be all its irreducible representations.
\[ \begin{array}{r|rrr} \text{size} & 1 & 3 & 2 \\ \text{class} & \mathbb{I}& (1\ 2) & (1\ 2\ 3) \\ \hline U & 1 & 1 & 1 \\ U' & 1 & -1 & 1 \\ V & 2 & 0 & -1, \end{array} \]
Example 8.3 (Character table of \(\mathbb{Z}/{n}\mathbb{Z}\))
The representations of Example 4.2 give all \(n\) irreducibles of \(\mathbb{Z}/{n}\mathbb{Z} \cong \mathcal{C}_{n}\) that we are guaranteed by Corollary 7.7.
\[ \begin{array}{r|rrrr} \text{element} & 0 & 1 & \cdots & n-1 \\ \hline U_{0,n} & 1 & 1 & \cdots & 1 \\ U_{1,n} & 1 & e^{\frac{2\pi i}{n}} & \cdots & e^{\frac{2\pi i}{n}(n-1)} \\ \vdots & \cdots & \cdots & \ddots & \vdots \\ U_{n-1,n} & 1 & e^{\frac{2\pi i}{n}(n-1)} & \cdots & {\frac{2\pi i}{n} (n-1)^2}, \end{array} \]
We also mention a result to be proved in the homework, and formally state the implicit principle mentioned in Example 8.1:
Proposition 8.1 Let \(G\) and \(H\) be finite groups with irreducible representations \(V_1,\dots,V_r\) and \(W_1,\dots,W_s\), respectively. Then the irreducible representations of \(G \times H\) are exactly those of the form \[ \{ V_i \otimes W_j : 1 \leq i \leq r \text{ and } 1 \leq j \leq s \}. \]
Proposition 8.2 Let \(G\) be a finite group. The \(1\)-dimensional irreducible representations of \(G\) all lift from irreducible representations of \({G}^{\mathrm{Ab}}\); in particular, \(G\) has exactly \(|{G}^{\mathrm{Ab}}|\) many distinct \(1\)-dimensional representations.
The latter follows immediately from Exercise C.1 and Corollary 7.7. We mention two more examples:
Example 8.4 (Character table of \(\mathcal{S}_{4}\))
First we use Proposition 8.2 to determine the \(1\)-dimensional representations, using that \[ {\mathcal{S}_{4}}^{\mathrm{Ab}} = \mathcal{S}_{4} / \mathcal{A}_{4} \cong \mathcal{C}_{2}. \] The cyclic group of order \(2\) has two irreducible representations, \(U_{0,1}\) and \(U_{1,1}\), which lift to the trivial \(U\) and alternating \(U'\) representations, respectively. Next, we consider the standard representation \(V\). Recalling that the permutation representation \(\mathcal{S}_{4} ⟳\mathbb{C}^4\) decomposes as \(U \oplus V\), we have \(\chi_V = \chi_{\mathbb{C}^4} - 1\); thanks to Exercise A.6, we know that \[ \chi_{\mathbb{C}^4}(\sigma) = \left|\operatorname{Fix}(\sigma)\right|. \] Since the conjugacy classes of \(\mathcal{S}_{4}\) are given by cycle types (Example 3.20), we have the partially completed character table: \[ \begin{array}{r|rrrrr} \text{size} & 1 & 6 & 8 & 6 & 3 \\ \text{class} & \mathbb{I}& (1\ 2) & (1\ 2\ 3) & (1\ 2\ 3\ 4) & (1\ 2)(3\ 4) \\ \hline U & 1 & 1 & 1 & 1 & 1 \\ U' & 1 & -1 & 1 & -1 & 1 \\ V & 3 & 1 & 0 & -1 & -1 \\ \end{array} \] Where we confirm that \(V\) is irreducible using \[ \langle \chi_V, \chi_V \rangle = \frac{1}{24} \left( 1 \cdot 3^2 + 6 \cdot 1^2 + 8 \cdot 0^2 + 6 \cdot (-1)^2 + 3 \cdot (-1)^2 \right) = 1. \] Now we appeal to a general trick: that the tensor product of a \(1\)-dimensional representation with an irreducible representation must also be irreducible (see Exercise F.5). In particular, setting \(V' = V \otimes U'\) yields a new representation whose character is given by \(\sigma \mapsto \chi_V(\sigma) \operatorname{sgn}(\sigma)\). Since \[ |\mathcal{S}_{4}| - (1^2 + 1^2 + 3^2 + 3^2) = 4, \] and we have exhausted all \(1\)-dimensional representations, we know that a \(2\)-dimensional representation \(W\) remains to be identified. We can compute its character without knowing anything about what \(W\) actually looks like. Indeed, we know that \(\chi_W\) is orthogonal to the other four irreducible characters, i.e., \[ \operatorname{Span}\{ \chi_W \} = \operatorname{Span}\{ \chi_U, \chi_{U'}, \chi_V, \chi_{V'} \}^\perp, \] under the inner product Equation 7.4, since \[ \dim \mathscr{C}({\mathcal{S}_{4}}) = 5 = \# \text{ number of conjugacy classes of } \mathcal{S}_{4}. \] We can compute this orthogonal complement using Equation 8.1: writing \(M\) for the partially completed character table, we want to find a vector orthogonal to its rows as weighted by the conjugacy classes.
partial_table = Matrix([
[1, 1, 1, 1, 1],
[1, -1, 1, -1, 1],
[3, 1, 0, -1, -1],
[3, -1, 0, 1, -1],
])
weights = diagonal_matrix([1, 6, 8, 6, 3]) / 24
(partial_table.conjugate() * weights).right_kernel().basis()[(1, 0, -1/2, 0, 1)]This result is well-defined up to scaling, so we multiply by \(2\) to complete the character table.
\[ \begin{array}{r|rrrrr} \text{size} & 1 & 6 & 8 & 6 & 3 \\ \text{class} & \mathbb{I}& (1\ 2) & (1\ 2\ 3) & (1\ 2\ 3\ 4) & (1\ 2)(3\ 4) \\ \hline U & 1 & 1 & 1 & 1 & 1 \\ U' & 1 & -1 & 1 & -1 & 1 \\ V & 3 & 1 & 0 & -1 & -1 \\ V' & 3 & -1 & 0 & 1 & -1 \\ W & 2 & 0 & -1 & 0 & 2 \\ \end{array} \]
In general, whenever \(H \leq G\), every \(G\)-representation \(\rho: G \to \operatorname{GL}(V)\) is also an \(H\) representation via \[ H \hookrightarrow G \stackrel{\rho}{\to} \operatorname{GL}(V), \] where we often abuse notation and simply write \(\rho\) for the restriction \(\rho|_H\) (cf. Exercise C.3). Distinct representations of \(G\) may restrict to the same representation of \(H\); further, irreducible representations of \(G\) need not remain irreducible after restriction to \(H\).
Example 8.5 Both irreducibles \(V\) and \(V'\) of \(\mathcal{S}_{4}\) restrict to the tetrahedral representation of \(\mathcal{A}_{4}\). Moreover, the restricted representation \(W\) is no longer irreducible! We compute, as functions on \(\mathcal{A}_{4}\): \[ \begin{aligned} \langle \chi_{\text{trivial}}, \chi_W|_{\mathcal{A}_{4}} \rangle & = \frac{1}{12} \left( 1 \cdot 1 \cdot 2 + 4 \cdot 1 \cdot (-1) + 4 \cdot 1 \cdot (-1) + 3 \cdot 1 \cdot (2) \right) = 0, \\ \langle \chi_{U_{1,3}}, \chi_W|_{\mathcal{A}_{4}} \rangle & = \frac{1}{12} \left( 1 \cdot 1 \cdot 2 + 4 \cdot \zeta^2 \cdot (-1) + 4 \cdot \zeta \cdot (-1) + 3 \cdot 1 \cdot (2) \right) = 1, \\ \langle \chi_{U_{2,3}}, \chi_W|_{\mathcal{A}_{4}} \rangle & = \frac{1}{12} \left( 1 \cdot 1 \cdot 2 + 4 \cdot \zeta \cdot (-1) + 4 \cdot \zeta^2 \cdot (-1) + 3 \cdot 1 \cdot (2) \right) = 1, \end{aligned} \] where we recall that \(\overline{\zeta} = \zeta^2\) and \(\zeta + \zeta^2 = -1\). Hence \(W = U_{1,3} \oplus U_{2,3}\) as \(\mathcal{A}_{4}\)-representations.
Next we complete our analysis of irreducible characters of a finite group \(G\), and give a more general projection formula for sending a represention \(V\) onto its isotypic factors. The main idea in both instances will emphasize our previous theme of “averaging” over \(G\). To construct this theory, we first flesh out more of the details regarding complex-valued functions on \(G\).
Note that \(\dim \mathscr{L}^{2}({G}) = |G|\) and, more specifically, there is an isomorphism \[ \Phi: \mathscr{L}^{2}({G}) \to \mathbb{C}[G] \quad \text{ via } \quad \Phi(f) = \sum_{h \in G} f(h) e_h \] with inverse \(\Psi: \mathbb{C}[G] \to \mathscr{L}^{2}({G})\) defined by \[ \Psi\left(\sum_{h \in G} \alpha_h e_h\right)(g) = \alpha_g. \] This correspondence carries the subspace of class functions \(\mathscr{C}({G})\) to the center of the group algebra, as to be verified in Homework. Accordingly, both of these subspaces have dimension equal to the number of conjugacy classes of \(G\). Note that the above is even an isomorphism of \(G\)-representations if we define an action on \(f \in \mathscr{L}^{2}({G})\) by \[ (g \cdot f)(h) \coloneqq f(g^{-1} h). \]
Lemma 8.1 Let \(f \in \mathscr{L}^{2}({G})\). Given a representation \(\rho: G \to \operatorname{GL}(V)\), define an operator on \(V\) via \[ \hat{f}(\rho) \coloneqq \sum_{h \in G} f(h) \rho(h): V \to V. \] Then \(\hat{f}(\rho)\) is an intertwiner for all representations \(V\) if and only if \(f \in \mathscr{C}({G})\).
The reverse implication we leave for Exercise G.5. For the forward direction, we introduce a common theme for the proofs ahead: to see that a property holds over all representations of \(G\), check that it holds on the regular representation \(\mathbb{C}[G]\). Since this representation contains every irreducible, if the operations of the proof respect decompositions into subrepresentations, then we can conclude the result in a general context.
Proof (Proof of forward direction). First consider \(\rho: G \to \operatorname{GL}(\mathbb{C}[G])\) the regular representation. If we suppose that \(\hat{f}(\rho)\) is an intertwiner, then we compute \(\hat{f}(\rho)(e_e)\) in two different ways: \[ \begin{split} \hat{f}(\rho)(g^{-1} \cdot e_g) & = \sum_{h \in G} f(h) h \cdot e_e = \sum_{h \in G} f(h) e_h \\ & \stackrel{\text{req}}{=} g^{-1} \cdot \hat{f}(\rho)(e_g) = \sum_{h \in G} \underbrace{f(h) e_{g^{-1}hg} = \sum_{h \in G}}_{\text{re-index } h} f(ghg^{-1}) e_{h}. \end{split} \] By equating basis vectors, we see that \(f(h) = f(ghg^{-1})\) holds, so \(f \in \mathscr{C}({G})\).
Theorem 8.1 (Fundamental theorem of character theory) The irreducible characters of \(G\) are an orthonormal basis for \(\mathscr{C}({G})\).
Proof. We will prove that any class function orthogonal to the irreducible characters of \(G\) is identically zero. To that end, take \(f \in \mathscr{C}({G})\), enumerate the irreducible representations of \(G\) as \(V_1, \dots, V_r\), and suppose that \(\langle \chi_{V_i}, f \rangle = 0\) holds for all \(i\); by linearity, this means \(\langle \chi_V, f \rangle = 0\) for all representations \(\rho: G \to \operatorname{GL}(V)\). We claim this also forces \(\hat{f}(\rho) = 0\).
First, consider when \(V\) is irreducible. Schur’s Lemma (Lemma 5.1) tells us that \(\hat{f}(\rho) = \lambda \mathbb{I}\) for some \(\lambda \in \mathbb{C}\). Taking the trace gives: \[ \lambda = \frac{1}{\dim V} \operatorname{Tr}( \hat{f}(\rho) ) = \frac{1}{\dim V} \sum_{h \in G} f(h) \chi_{V}(h) = \frac{|G|}{\dim V} \langle \chi_{V^*}, f \rangle = 0, \] where we use the fact that \(\overline{\chi_V} = \chi_{V^*}\), together with the fact that \(\chi_V\) is irreducible if and only if \(\chi_{V^*}\) is irreducible.
Now we claim that \(\hat{f}(\rho) = 0\) holds for any representation \(V\). If we decompose into irreducibles, \[ V = V_1^{\oplus a_1} \oplus \cdots V_r^{\oplus a_r}, \] then expanding with respect to this decomposition (writing \(\rho_i: G \to \operatorname{GL}(V_i)\) for the \(i\)th irreducible) gives \[ \hat{f}(\rho) = \sum_{i=1}^r \hat{f}(\rho_i)^{\oplus a_i}, \] and from the previous step every \(\hat{f}(\rho_i) = 0\), so \(\hat{f}(\rho) = 0\).
Continuing our previous theme, consider \(\rho: G \to \operatorname{GL}(\mathbb{C}[G])\) the regular representation. As elements of \(\operatorname{End}(\mathbb{C}[G])\), the linear operators \(\rho(g)\) are linearly independent (Exercise C.2) and so \(\hat{f}(\rho) = 0\) means that \(f\) is identically zero.
Corollary 8.1 There are as many distinct irreducible representations of \(G\) as there are conjugacy classes of \(G\).
Corollary 8.2 \(G\) is abelian if and only if its irreducible representations are all \(1\)-dimensional.
The orthogonality of the rows of the character table is equivalent to an orthogonality relationship for the columns.
Theorem 8.2 Let \(G\) be a finite group. Then \[ \sum_{\chi} \overline{\chi(g)} \chi(h) = \left\{ \begin{array}{ll} |C_G(g)| & \text{if $g$ and $h$ are conjugate} \\ 0 & \text{otherwise,} \end{array} \right. \] where the sum ranges over all irreducible representations of \(G\).
Proof. Consider the matrix version of (Equation 8.1). If we take the transpose, we see that \(M\) (the character table regarded as a matrix) is invertible with \[ M^{-1} = \operatorname{Diag}(\tfrac{|\operatorname{cl}_{}({g_1})|}{|G|}, \tfrac{|\operatorname{cl}_{}({g_r})|}{|G|}) M^*. \tag{8.2}\] Therefore we can compute \[ M^* M = \operatorname{Diag}(\tfrac{|G|}{|\operatorname{cl}_{}({g_1})|}, \tfrac{|G|}{|\operatorname{cl}_{}({g_r})|}) \] which is the desired column orthogonality in matrix form.
Generalizing Theorem 7.2, these relations are especially useful to manually complete partial character tables.
Example 8.6 (Character table of \(\mathcal{Q}_{8}\))
The quaternion group \(\mathcal{Q}_{8}\) has the commutator subgroup \(\langle -1 \rangle\); its abelianization is isomorphic to \(\mathcal{V}\). Therefore \(\mathcal{Q}_{8}\) has four \(1\)-dimensional representations: the trivial representation, and one with each subgroup \(\langle i \rangle\), \(\langle j \rangle\), and \(\langle k \rangle\) as its kernel. Moreover, because \[ 8 - (1^2+1^2+1^2+1^2) = 4, \] the only remaining irreducible representation is of degree \(2\). We have: \[ \begin{array}{r|rrrrr} \text{size} & 1 & 1 & 2 & 2 & 2 \\ \text{class} & 1 & -1 & i & j & k \\ \hline \text{trivial} & 1 & 1 & 1 & 1 & 1 \\ \text{$\langle i \rangle$ kernel} & 1 & 1 & 1 & -1 & -1 \\ \text{$\langle j \rangle$ kernel} & 1 & 1 & -1 & 1 & -1 \\ \text{$\langle k \rangle$ kernel} & 1 & 1 & -1 & -1 & 1 \\ \text{mystery} & 2 & \alpha & \beta & \gamma & \delta, \end{array} \] From column orthogonality, comparing the first with each column that follows, we have: \[ \begin{aligned} 1 \cdot 1 + 1 \cdot 1 + 1 \cdot 1 + 1 \cdot 1 + 2\alpha & = 0 \\ 1 \cdot 1 + 1 \cdot 1 + 1 \cdot (-1) + 1 \cdot (-1) + 2\beta & = 0 \\ 1 \cdot 1 + 1 \cdot (-1) + 1 \cdot 1 + 1 \cdot (-1) + 2\gamma & = 0 \\ 1 \cdot 1 + 1 \cdot (-1) + 1 \cdot (-1) + 1 \cdot 1 + 2\delta & = 0, \end{aligned} \] i.e., \(\alpha = 2\) and \(\beta = \gamma = \delta = 0\). Note that this mystery representation can be realized, for example, by the homomorphism \(\mathcal{Q}_{8}\to \operatorname{GL}_2(\mathbb{C})\) given by \[ i \mapsto \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix} \quad \text{ and } \quad j \mapsto \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}. \]
\[ \begin{array}{r|ccccc} \text{size} & 1 & 1 & 2 & 2 & 2 \\ \text{class} & 1 & -1 & i & j & k \\ \hline \text{trivial} & 1 & 1 & 1 & 1 & 1 \\ \text{$i$-kernel} & 1 & 1 & 1 & -1 & -1 \\ \text{$j$-kernel} & 1 & 1 & -1 & 1 & -1 \\ \text{$k$-kernel} & 1 & 1 & -1 & -1 & 1 \\ \text{2‑dimensional} & 2 & -2 & 0 & 0 & 0, \end{array} \]
Example 8.7 (Character table of \(\operatorname{Aff}_{1}({\mathbb{F}_5})\))
For any field \(\mathbb{F}\), consider the affine group \[ \operatorname{Aff}_{n}({\mathbb{F}}) \coloneqq \left\{ \begin{pmatrix} A & x \\ 0 & 1 \end{pmatrix} : A \in \operatorname{GL}_n(\mathbb{F}), x \in \mathbb{F}^n \right\} \leq \operatorname{GL}_{n+1}(\mathbb{F}). \] This group is so named because it encodes all invertible affine transformations of \(n\)-space: we think of an element \(M = \begin{psmallmatrix} A & x \\ 0 & 1 \end{psmallmatrix}\) as sending \(y \in \mathbb{F}^n\) to \(My \coloneqq Ay + x \in \mathbb{F}^n\). One can check that matrix multiplication corresponds to the composition of affine transformations.
The group \(\operatorname{Aff}_{1}({\mathbb{F}_5})\) has twenty elements, consisting of matrices \(\begin{psmallmatrix} \alpha & \beta \\ 0 & 1 \end{psmallmatrix}\) with \(\alpha, \beta \in \mathbb{F}_5 \coloneqq \mathbb{Z}/5\mathbb{Z}\) and \(\alpha \not = 0\). The determinant homomorphism is \[ \begin{aligned} \operatorname{Aff}_{1}({\mathbb{F}_5}) & \to \mathbb{F}_5^\times \cong \mathcal{C}_{4} \\ \begin{pmatrix} \alpha & \beta \\ 0 & 1 \end{pmatrix} & \mapsto \alpha, \end{aligned} \] This map has has kernel \[ K \coloneqq \left\{ \begin{pmatrix} 1 & \beta \\ 0 & 1 \end{pmatrix} : \beta \in \mathbb{F}_5 \right\} \unlhd\operatorname{Aff}_{1}({\mathbb{F}_5}). \] In fact this map \(\operatorname{Aff}_{1}({\mathbb{F}_5})/K \cong \mathcal{C}_{4}\) is the abelianization, allowing us to deduce four rows of the character table of \(\operatorname{Aff}_{1}({\mathbb{F}_5})\) once we know its conjugacy classes. We have \[ \begin{pmatrix} \gamma & \delta \\ 0 & 1 \end{pmatrix} \begin{pmatrix} \alpha & \beta \\ 0 & 1 \end{pmatrix} \begin{pmatrix} \gamma & \delta \\ 0 & 1 \end{pmatrix}^{-1} = \begin{pmatrix} \alpha & \beta \gamma - \alpha \delta + \beta \\ 0 & 1 \end{pmatrix}, \] from which we can deduce conjugacy classes representatives. In particular, note that conjugate elements must have the same determinant! With some work, one can show the classes are: \[ \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 3 & 0 \\ 0 & 1 \end{pmatrix}, \text{ and } \begin{pmatrix} 4 & 0 \\ 0 & 1 \end{pmatrix} \] of cardinality \(1\), \(4\), \(5\), \(5\), and \(5\), respectively. This is great news—we only have one more irreducible, call it \(V\), to identify! From column orthogonality:
\[ \begin{array}{r|rrrrr} \text{size} & 1 & 4 & 5 & 5 & 5 \\ \text{class} & \begin{psmallmatrix} 1 & 0 \\ 0 & 1 \end{psmallmatrix} & \begin{psmallmatrix} 1 & 1 \\ 0 & 1 \end{psmallmatrix} & \begin{psmallmatrix} 2 & 0 \\ 0 & 1 \end{psmallmatrix} & \begin{psmallmatrix} 3 & 0 \\ 0 & 1 \end{psmallmatrix} & \begin{psmallmatrix} 4 & 0 \\ 0 & 1 \end{psmallmatrix} \\ \hline \text{trivial} & 1 & 1 & 1 & 1 & 1 \\ U_{1,4} & 1 & 1 & i & -1 & -i \\ U_{2,4} & 1 & 1 & -1 & 1 & -1 \\ U_{3,4} & 1 & 1 & -i & -1 & i \\ V & 4 & -1 & 0 & 0 & 0, \end{array} \]
We conclude by remarking that, while the character table is a useful invariant of finite groups, it does not determine a group up to isomorphism. For example, the groups \(\mathcal{Q}_{8}\) and \(\mathcal{D}_{8}\) have the same table, but are not isomorphic (\(\mathcal{Q}_{8}\) has 6 elements of order 4 while \(\mathcal{D}_{8}\) has 2):
Example 8.8 (Character table of \(\mathcal{D}_{8}\))
We use \({\mathcal{D}_{8}}^{\mathrm{Ab}} = \mathcal{D}_{8}/\langle r^2 \rangle \cong \mathcal{V}\) together with column orthogonality.
\[ \begin{array}{r|rrrrr} \text{size} & 1 & 1 & 2 & 2 & 2 \\ \text{class} & e & r^2 & r & s & sr \\ \hline \text{trivial} & 1 & 1 & 1 & 1 & 1 \\ \text{$\langle r \rangle$-kernel} & 1 & 1 & 1 & -1 & -1 \\ \text{$\langle s, r^2 \rangle$-kernel} & 1 & 1 & -1 & 1 & -1 \\ \text{$\langle sr, r^2 \rangle$-kernel} & 1 & 1 & -1 & -1 & 1 \\ \text{standard} & 2 & -2 & 0 & 0 & 0, \end{array} \]
Nonetheless, the character table is an essential tool in the classification of finite groups. We will soon see how we can use the tools of character theory to construct tables, in addition to a number of results allowing us to deduce properties of \(G\) from its representation theory.
To summarize, the fundamental theorem together with the orthogonality relations allow us to decompose representations as linear combinations of irreducibles. Indeed, these relations provide the clues necessary to complete a character table with explicitly writing down representations!