Brian Bi
$\DeclareMathOperator{\Ind}{Ind} \DeclareMathOperator{\Res}{Res}$

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 $$\mathbb{C}$$.

1. 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)
#120151212
$$\Ind^G_H \mathbb{C}_{+}$$3002 00
$$\Ind^G_H \mathbb{C}_{-}$$300−2 00

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.

2. 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)
#120151212
$$\Ind^G_H \mathbb{C}$$2020 00
$$\Ind^G_H \mathbb{C}_{\omega}$$20−1 000
$$\Ind^G_H \mathbb{C}_{\omega^2}$$20−1 000

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*}

3. 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)
#120151212
$$\Ind^G_H \mathbb{C}$$1200 22
$$\Ind^G_H \mathbb{C}_{\zeta^{\pm 1}}$$120 0$$\frac{\sqrt{5}-1}{2}$$ $$\frac{-\sqrt{5}-1}{2}$$
$$\Ind^G_H \mathbb{C}_{\zeta^{\pm 2}}$$120 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*}

4. 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)
#120151212
$$\Ind^G_H \mathbb{C}$$521 00
$$\Ind^G_H \mathbb{C}_{\epsilon^{\pm 1}}$$5 −1100
$$\Ind^G_H \mathbb{C}^3$$150−1 00

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*}

5. 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)
#120151212
$$\Ind^G_H \mathbb{C}$$1503 00
$$\Ind^G_H \mathbb{C}_{xyz}$$150 −100

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*}