Remark: The text says to use Frobenius reciprocity, but it seems obvious that it means we should use the Frobenius formula from section 5.9. One might wonder which of the three \(\mathbb{Z}_2\)-subgroups of \(S_3\) to use in (1), and similarly for (2) and (3). However, if we look at the Frobenius character formula, we see that replacing \(H\) with a conjugate subgroup doesn't change the result; in (1), the three subgroups of \(G\) isomorphic to \(\mathbb{Z}_2\) are all conjugate to each other, so we must get the same decomposition regardless of which one we choose for the computation. A similar statement holds for (2) and (3)
Problem 5.11.1 The group \(A_5\) has the property that any two isomorphic subgroups are conjugate, so in each of (a) to (e), all possible induced representations from a subgroup of the given isomorphism class are isomorphic. In all of the below, let \(G = A_5\). We assume the ground field is \(\mathbb{C}\).
We'll set \(H\) to be the subgroup generated by \((1\ 2)(3\ 4)\). \(H\) has two irreps, which were previously denoted \(\mathbb{C}_+\) and \(\mathbb{C}_-\). We will also use that notation here. The character of either of the two possible induced representations vanishes at \(g \in G\) whenever \(g\) isn't conjugate to any elements of \(H\), namely, when \(g\) is a 3-cycle or a 5-cycle. By Remark 5.8.3, both induced representations have dimension 30, so \(\chi(1) = 30\) in both cases. The double transpositions of \(A_5\) form a single conjugacy class, so let \(g = (1\ 2)(3\ 4)\); computing \(\chi(g)\) in \(\Ind^G_H \mathbb{C}_+\) and \(\Ind^G_H \mathbb{C}_-\) will then complete the calculations of their characters.
In \(\Ind^G_H \mathbb{C}_+\), \(\chi(g)\) is \(1/|H|\) times the order of the centralizer of \(g\), since \(g\) is not conjugate to the identity. By the orbit-stabilizer theorem, the order of the centralizer is the order of \(A_5\) divided by the size of its conjugacy class, or \(60 \div 15\), where we have used the previously stated fact that all double transpositions are conjugate to each other. Thus, \(\chi(g) = 2\) in \(\Ind^G_H \mathbb{C}_+\). In \(\Ind^G_H \mathbb{C}_-\), we must then have \(\chi(g) = -2\) since \(\chi((1\ 2)(3\ 4)) = -1\) in \(\mathbb{C}_-\). We therefore have the following characters:
\(A_5\) 1 (123) (12)(34) (12345) (13245) # 1 20 15 12 12 \(\Ind^G_H \mathbb{C}_{+}\) 30 0 2 0 0 \(\Ind^G_H \mathbb{C}_{-}\) 30 0 −2 0 0 Using orthogonality of characters, we find \begin{align*} \Ind^G_H \mathbb{C}_+ &\cong \mathbb{C} \oplus \mathbb{C}^3_+ \oplus \mathbb{C}^3_- \oplus 2\mathbb{C}^4 \oplus 3\mathbb{C}^5 \\ \Ind^G_H \mathbb{C}_- &\cong 2\mathbb{C}^3_+ \oplus 2\mathbb{C}^3_- \oplus 2\mathbb{C}^4 \oplus 2\mathbb{C}^5 \end{align*} where we have labelled the irreps of \(A_5\) as they were labelled in section 4.8.
We'll set \(H\) to be the subgroup generated by \(g = (1\ 2\ 3)\). \(H\) has three irreps; denote these by \(\mathbb{C}\), \(\mathbb{C}_\omega\), and \(\mathbb{C}_{\omega^2}\), where \(g\) respectively acts as 1, \(\omega\), and \(\omega^2\), with \(\omega = \exp(2\pi i/3)\). By Remark 5.8.3, the three induced representations will each have dimension 20, and therefore \(\chi(1) = 20\). The characters of the double transpositions and 5-cycles will vanish in all three induced representations since such elements of \(A_5\) are never conjugate to \(g\), a 3-cycle. As all 3-cycles in \(A_5\) are conjugate to each other, we only need to compute \(\chi(g)\) in each of the three induced representations in order to fill out the character table.
In \(\Ind^G_H \mathbb{C}\), \(\chi(g)\) will be \(1/|H|\) times the number of elements of \(G\) that either fix \(g\) under conjugation or take \(g\) to \(g^2\). By the orbit-stabilizer theorem, since the conjugacy class of \(g\) contains 20 elements, there are 3 elements that fix \(g\) (namely, the 3 elements of \(H\) itself) and 3 elements that take \(g\) to \(g^2\), namely \((1\ 2)(4\ 5), (2\ 3)(4\ 5), (3\ 1)(4\ 5)\). Thus, \(\chi(g) = 2\). In \(\Ind^G_H \mathbb{C}_{\omega}\), the sum in the Frobenius formula of section 5.9 includes \(\omega\) twice and \(\omega^2\) twice, so \(\chi(g) = (1/3)(3\omega + 3\omega^2) = -1\). In \(\Ind^G_H \mathbb{C}_{\omega^2}\), we get result, \(\chi(g) = (1/3)(3\omega^2 + 3\omega) = -1\), the same result. We now write out the characters of the induced representations:
\(A_5\) 1 (123) (12)(34) (12345) (13245) # 1 20 15 12 12 \(\Ind^G_H \mathbb{C}\) 20 2 0 0 0 \(\Ind^G_H \mathbb{C}_{\omega}\) 20 −1 0 0 0 \(\Ind^G_H \mathbb{C}_{\omega^2}\) 20 −1 0 0 0 from which we obtain the decompositions \begin{align*} \Ind^G_H \mathbb{C} &\cong \mathbb{C} \oplus \mathbb{C}^3_+ \oplus \mathbb{C}^3_- \oplus 2\mathbb{C}^4 \oplus \mathbb{C}^5 \\ \Ind^G_H \mathbb{C}_{\omega} \cong \Ind^G_H \mathbb{C}_{\omega^2} &\cong \mathbb{C}^3_+ \oplus \mathbb{C}^3_- \oplus \mathbb{C}^4 \oplus 2\mathbb{C}^5 \end{align*}
Set \(H = \langle g \rangle\) with \(g = (1\ 2\ 3\ 4\ 5)\). Write \(\zeta = \exp(2\pi i/5)\). There are 5 irreps of \(H\), which we'll denote by \(\mathbb{C}\) and \(\mathbb{C}_{\zeta^i}\) for \(i = 1, 2, 3, 4\), using similar notation to that of part (b). The dimension of each induced representation is 12 and all induced characters vanish for double transpositions and 3-cycles. The 5-cycles in \(A_5\) split into two conjugacy classes, one containing \(g\) and \(g^4\), the other containing \(g^2\) and \(g^3\). Thus, we will need to compute \(\chi(g)\) and \(\chi(g^2)\) separately for each of the induced representations.
Start with \(\Ind^G_H \mathbb{C}\). The conjugacy class of \(g\) has size 12, so there are 5 elements of \(A_5\) that commute with \(g\), and another 5 that take \(g\) to \(g^4\) under conjugation. Thus \(\chi(g) = (1/5)(5 + 5) = 2\). Likewise, there are 5 elements that commute with \(g^2\) and another 5 that take \(g^2\) to \(g^3\), so \(\chi(g^2) = 2\) as well. In \(\Ind^G_H \mathbb{C}_{\zeta}\) and \(\Ind^G_H \mathbb{C}_{\zeta^4}\), we obtain \(\chi(g) = (1/5)(5\zeta + 5\zeta^4) = 2\cos(2\pi/5) = \frac{\sqrt{5}-1}{2}\), and \(\chi(g^2) = (1/5)(5\zeta^2 + 5\zeta^3) = 2\cos(4\pi/5) = \frac{-\sqrt{5}-1}{2}\). In \(\Ind^G_H \mathbb{C}_{\zeta^2}\) and \(\Ind^G_H \mathbb{C}_{\zeta^3}\), we obtain \(\chi(g) = (1/5)(5\zeta^2 + 5(\zeta^2)^4) = \frac{-\sqrt{5}-1}{2}\) and \(\chi(g^2) = (1/5)(5(\zeta^2)^2 + 5(\zeta^2)^3) = \frac{\sqrt{5}-1}{2}\). In tabular format:
\(A_5\) 1 (123) (12)(34) (12345) (13245) # 1 20 15 12 12 \(\Ind^G_H \mathbb{C}\) 12 0 0 2 2 \(\Ind^G_H \mathbb{C}_{\zeta^{\pm 1}}\) 12 0 0 \(\frac{\sqrt{5}-1}{2}\) \(\frac{-\sqrt{5}-1}{2}\) \(\Ind^G_H \mathbb{C}_{\zeta^{\pm 2}}\) 12 0 0 \(\frac{-\sqrt{5}-1}{2}\) \(\frac{\sqrt{5}-1}{2}\) yielding the decompositions \begin{align*} \Ind^G_H \mathbb{C} &\cong \mathbb{C} \oplus \mathbb{C}^3_+ \oplus \mathbb{C}^3_- \oplus \mathbb{C}^5 \\ \Ind^G_H \mathbb{C}_{\zeta^{\pm 1}} &\cong \mathbb{C}^3_+ \oplus \mathbb{C}^4 \oplus \mathbb{C}^5 \\ \Ind^G_H \mathbb{C}_{\zeta^{\pm 2}} &\cong \mathbb{C}^3_- \oplus \mathbb{C}^4 \oplus \mathbb{C}^5 \end{align*}
Let \(H\) be the \(A_4\)-subgroup acting on 1, 2, 3, 4 (i.e., all elements of \(H\) fix 5). There are 4 irreps of \(A_4\), which were assigned the names \(\mathbb{C}\), \(\mathbb{C}_{\epsilon}\), \(\mathbb{C}_{\epsilon^2}\), and \(\mathbb{C}^3\) in section 4.8 and whose characters were given. In each of the representations \(\mathbb{C}_{\epsilon}\) and \(\mathbb{C}_{\epsilon^2}\), the elements of the normal \(V_4\)-subgroup of \(A_4\) act trivially, and some other element \(g = (1\ 2\ 3)\) is then chosen to act as the scalar \(\epsilon\) or \(\epsilon^2\), respectively (here, \(\epsilon = \exp(2\pi i/3)\). In the 3-dimensional representation, the double transpositions have trace −1 while the 3-cycles have trace 0. This gives us all the information we need to compute the induced characters, which have dimension \(|A_5|/|A_4| = 5\) when induced from the 1-dimensional representations of \(H\), and dimension 15 when induced from the 3-dimensional representation.
As always, the character of \(\Ind^G_H \mathbb{C}\) will be the easiest to compute. It vanishes at all the 5-cycles, since no 5-cycle can be conjugate to any element of \(H\). Taking \(g = (1\ 2\ 3)\) as above, there are 3 elements of \(A_5\) that commute with \(g\) and 3 elements mapping it to each of its 7 conjugates in \(A_4\) (namely, the other 3-cycles in \(A_4\)), so the Frobenius formula of section 5.9 gives \(\chi(g) = (1/12)(8 \times 3) = 2\). (The remaining 12 conjugates lie outside \(H\).) Set \(g' = (1\ 2)(3\ 4)\); there are 4 elements of \(A_5\) that commute with \(g'\), and 4 taking it to each of its two conjugates in \(A_4\), so \(\chi(g') = (1/12)(4 + 4 + 4) = 1\).
To compute \(\Ind^G_H \mathbb{C}_\epsilon\), we use the fact that \(g\) belongs to a conjugacy class of \(H\) that contains four 3-cycles, with the other four 3-cycles belonging to another conjugacy class; but all eight of these 3-cycles are conjugate to each other in \(A_5\). The sum in the Frobenius formula of section 5.9 therefore includes 12 terms with value \(\epsilon\) (since there are 4 elements of \(H\) with trace \(\epsilon\) and 3 elements of \(G\) taking \(g\) to each of these under conjugation) and another 12 with value \(\epsilon^2\), giving \(\chi(g) = (1/12)(12\epsilon + 12\epsilon^2) = -1\). Since all double transpositions have trace 1 in \(\mathbb{C}_\epsilon\), it follows that \(\chi(g') = 1\) here as in the previous paragraph. A similar calculation, only exchanging \(\epsilon\) with \(\epsilon^2\), shows that \(\Ind^G_H \mathbb{C}_{\epsilon^2}\) is isomorphic to \(\Ind^G_H \mathbb{C}_\epsilon\).
Finally, in \(\Ind^G_H \mathbb{C}^3\), all terms in \(\chi(g)\) vanish since the 3-cycles are trace-free in \(\mathbb{C}^3\), and the 12 terms in \(\chi(g')\) each have value \(-1\), so \(\chi(g') = -1\). In tabular form:
\(A_5\) 1 (123) (12)(34) (12345) (13245) # 1 20 15 12 12 \(\Ind^G_H \mathbb{C}\) 5 2 1 0 0 \(\Ind^G_H \mathbb{C}_{\epsilon^{\pm 1}}\) 5 −1 1 0 0 \(\Ind^G_H \mathbb{C}^3\) 15 0 −1 0 0 yielding the decompositions \begin{align*} \Ind^G_H \mathbb{C} &\cong \mathbb{C} \oplus \mathbb{C}^4 \\ \Ind^G_H \mathbb{C}_{\epsilon^{\pm 1}} &\cong \mathbb{C}^5 \\ \Ind^G_H \mathbb{C}^3 &\cong \mathbb{C}^3_+ \oplus \mathbb{C}^3_- \oplus \mathbb{C}^4 \oplus \mathbb{C}^5 \end{align*}
The Klein four-group can be realized as a subgroup of \(G = A_5\) as \(H = \{1, x, y, z\}\) where \(x = (1\ 2)(3\ 4)\), \(y = (1\ 3)(2\ 4)\), and \(z = (1\ 4)(2\ 3)\). By Theorem 5.6.1, all irreps of \(H\) are tensor products of \(\mathbb{C}_+\) and \(\mathbb{C}_-\); so there are 4 such irreps, each one-dimensional, and we denote them by \(\mathbb{C}\) (the trivial representation), the non-trivial \(\mathbb{C}_x\) where \(x\) and the identity act as the scalar \(+1\) while \(y, z\) act as \(-1\), and the similarly defined \(\mathbb{C}_y\) and \(\mathbb{C}_z\). The four induced representations of \(G\) all have dimension 15 and the induced characters must be nonzero only at the identity and at the double transpositions. In each induced representation, \(\chi(x) = \chi(y) = \chi(z)\) since they are all conjugate to each other.
The element \(x\) belongs to a conjugacy class of \(A_5\) of size 15, so its centralizer has order 4. There are 4 elements of \(A_5\) that, under the action of conjugation, take \(x\) to each of its conjugates \(y, z\) in \(H\). Therefore, in \(\Ind^G_H \mathbb{C}\), \(\chi(x) = (1/4)(4 + 4 + 4) = 3\).
When computing \(\chi(x)\) in \(\Ind^G_H \mathbb{C}_x\), we keep in mind that each of the 4 elements that commute with \(x\) contributes \(+1\) to the sum, while each of the 4 elements that take \(x\) to each of \(y\) and \(z\) contributes \(-1\). Therefore, \(\chi(x) = (1/4)(4 - 4 - 4) = -1\). A similar calculation shows that \(\chi(x) = -1\) in the representations induced from \(\mathbb{C}_y\) and \(\mathbb{C}_z\). Thus, we have the following characters:
\(A_5\) 1 (123) (12)(34) (12345) (13245) # 1 20 15 12 12 \(\Ind^G_H \mathbb{C}\) 15 0 3 0 0 \(\Ind^G_H \mathbb{C}_{xyz}\) 15 0 −1 0 0 yielding the decompositions \begin{align*} \Ind^G_H \mathbb{C} &\cong \mathbb{C} \oplus \mathbb{C}^4 \oplus 2\mathbb{C}^5 \\ \Ind^G_H \mathbb{C}_{xyz} &\cong \mathbb{C}^3_+ \oplus \mathbb{C}^3_- \oplus \mathbb{C}^4 \oplus \mathbb{C}^5 \end{align*}