MATH 126 SOLUTION SET 3

NILS R. BARTH

1. χ(g) = tr(ρ(g)) = ∑ _{i} λ_{i}, where λ_{i} are the characteristic values of ρ(g). Now |∑ _{i} λ_{i}| ≤ ∑ _{i} |λ_{i}|, with equality iff all the λ are equal. Since all the λ_{i} are roots of unity^{1}, |λ_{i}| = 1, so |∑ _{i} λ_{i}| ≤ ∑ 1 = d, with equality iff λ_{i} = λ, which is the case iff ρ(g) = λI.

2. You should not use contradiction here; it is straight-forward. ρ(G) = ρ(G/H) (because π is surjective), and since there are no subspaces invariant under the latter, there are none under the former.

3. Associative: The tensor product is associative; you can take this as known, but should mention it^{2}. Also, for characters the tensor product corresponds to the standard product on C, which is associative.

Identity: The trivial character is the identity:

(1 ⊗ χ)(g) = 1χ(g) = χ(g) (3.1)

Composition: The tensor product of two representations is a representation, and if the two repre-

sentations are one-dimensional, their product will be too, since 1 · 1 = 1. Inverse: Since GL(C) = C^{∗}, χ^{−1}(g) = χ(g)^{−1} is well-defined^{3}, and

χ^{−1}(gh) = χ(gh)^{−1} = (χ(g)χ(h))^{−1} = χ(h)^{−1}χ(g)^{−1} = χ(g)^{−1}χ(h)^{−1} (3.2)

since GL(C) is commutative, so this defines a one-dimensional character such that χ^{−1}(g)⊗χ(g) = 1, so it is the inverse of χ. Note that since C is commutative, this group is in fact commutative.

4. This solution was given (in different forms) by several students; I have rewritten it a bit.

C[G] can be view as a C[G] module in more than one way: if we define g(h) = gh, we get the regular representation. However, if we define g(h) = ghg^{−1} we get another representation^{4}, which we shall call the conjugation representation. This is a kind of permutation representation, so its character is the number of fixed points, so π(g) = \frac{|G|}{|C_{g} |} (by orbit-stabilizer), where C_{g} denotes the conjugacy class of g and π is the character of our representation. Now given an irreducible character (indeed, any character) χ, ⟨π, χ⟩ is a non-negative integer. Also,

⟨π, χ⟩ = \frac{1}{|G|} ∑_{g∈G} \frac{|G|}{|C_{g}|}χ(g) = ∑_{g∈G} \frac{1}{|C_{g}|}χ(g) (3.3)

which is just the sum of χ’s row in the character table. Thus, this sum is a non-negative integer, as desired.

5. Again, culled from students.

(A) Every index two subgroup is normal^{5}, thus we can quotient out by H and get G/H = C_{2} = {1, −1}. By question 4, we can sum across conjugacy classes and get a non-negative integer. However,

χ([g]) = / \begin{matrix}1 & [g] ⊂ H \\ −1 & [g] ⊄ H \end{matrix} (3.4)

Received by the editors October 26, 1998.

1Since ρ(g) has finite order, since G is finite, remember?

2This follows immediately from category theory; both (A⊗B)⊗C and A⊗(B⊗C) are trilinear, and since the tensor product is the solution to a universal mapping property, it is unique up to a unique isomorphism. Thus, (A ⊗ B) ⊗ C ≅ A ⊗ (B ⊗ C) by a unique isomorphism.

3Since one dimensional characters are never zero.

4This is easily verified to be a representation.

5Because G = H ∐ sH = H ∐ Hs, so sH = Hs, so H is normal.

so this sum is the number of conjugacy classes in H minus the number of conjugacy classes not in H. Thus, there are at least as many of the former as the latter. In particular, [S_{n} : A_{n} ] = 2, so this holds for A_{n} (even permutations) having more classes than S_{n} \ A_{n} (odd permutations).

(B) Again, quotient out by H to get π: G ) G/H = C_{3} = {1, ζ_{3}, ζ_{3}^{2}}. So H = π^{−1}(1) is a union of conjugacy classes (the conjugacy classes which get mapped to 1), and similarly for sH = π^{−1}(ζ_{3}), tH = π^{−1}(ζ_{3}^{2}). Summing across conjugacy classes as in question 4, we get a non-negative integer. In particular, this is real, so sH, tH have the same number of conjugacy classes (since we need the ζ, ζ^{2} to cancel). Also, it is non-negative, and since ζ_{3} + ζ_{3}^{2} = −1, we see that H has at least as many conjugacy classes as sH, tH.

6. I did parts A, B, D, E, F, H and students did C, G (though I polished the solutions).

1 C ι Δ ∇ H H′
|C_{g}| 1 6 7 7 7 7 7
χ_{1} 1 1 1 1 1 1 1
χ_{2} 1 1 −1 ω^{4} ω^{2} ω^{1} ω^{5}
χ_{3} 1 1 1 ω^{2} ω^{4} ω^{2} ω^{4}
χ_{4} 1 1 −1 1 1 ω^{3} ω^{3}
χ_{5} 1 1 1 ω^{4} ω^{2} ω^{4} ω^{2}
χ_{6} 1 1 −1 ω^{2} ω^{4} ω^{5} ω^{1}
χ_{7} 6 −1 0 0 0 0 0

Character Table 1: G

(A) This character table is obtained as follows:

• χ_{2}(G) forms a group which contains ω, so it must contain ω^{5}. Thus, χ_{2}(H) = ω^{5}

• Tensoring χ_{2} with itself yields more representations, which are χ_{3}, …, χ_{6}.^{6}

• χ_{7} must have zeros except for 1, C, since otherwise tensoring with χ_{2} would yield another simple character. It remains to determine χ_{7}(1), χ_{7}(C). Because χ_{2} gives a quotient G ) C_{6}, we know that 6| |G|. Also, χ_{7}(1)^{2} + 6 = |G|, so 6|χ_{7}(1). However, we know that the dimension of a representation divides the order of the group, so χ_{7}(1)|χ_{7}(1)^{2} +6, so χ_{7}(1)|6. Since χ_{7}(1) is a positive integer, χ_{7}(1) = 6. Lastly, by column orthogonality of the first and second columns, 6 + χ_{7}(1)χ_{7}(C) = 0, so χ_{7}(1)χ_{7}(C) = −6, so χ_{7}(C) = −1.

(B) The number of elements in each conjugacy class can be determined by multiplying the character table by its conjugate transpose^{7} and looking at the diagonal entries, which give us the order of the stabilizers of elements in each class. These are included in the character table above.

(C) Clearly, 1 has order 1. From part D we see that the elements of C have order 7. Elements of H, H′ must have order divided by 6, so order 6, 42. The latter is impossible, as G isn’t cyclic, so they have order 6. The orders of the rest follow from the determination of Sylow subgroups in part G. To summarize:

Conjugacy Class 1 C ι Δ ∇ H H′ ; Order of Elements 1 7 2 3 3 6 6

Table 1. Order of Elements

(D) The commutator subgroup is the kernel of the one-dimensional characters, so it’s 1 ∐ C = C_{7},

since this is the only group of order 7. (E) χ_{3} gives us a map G ↠ C_{3}, so the kernel of this map has order 42/3 = 14. We’ll denote this group

by ℌ to avoid confusion with the conjugacy class H.

6We use ω^{3} instead of −1 to make the group structure visible.

7Also a good way to check your work.

(F) ℌ is a non-commutative (since we haven’t quotiented out by the commutator) group of order

14 = 2 · 7, so ℌ = D_{7}. (G) By the Sylow theorems, we know that we have subgroups of order 2, 3, 7. Let s_{p} be the number of

Sylow subgroups of order p. By the Third Sylow Theorem: s_{p}| |G|/p = 42/p, since |G| = 2 · 3 · 7; and s_{p} ≡ 1 mod p. From these we immediately obtain s_{3}, s_{7}. s_{7} = 1 so there is 1 7-Sylow group (namely 1 ∐ C). s_{3} = 7 (these correspond to elements in Δ and their square in ∇, since elements of C, ι, H, H′ cannot have the correct order). Lastly, we must have some 2-Sylow group, and the only elements with order 2 possible are the remaining ones, namely ι. These are all conjugate, so there are 7 2-Sylow groups. To summarize (and then a bit):

Subgroup  1 1 ∐ C {1, i} , i ∈ ι {1, δ, δ^{2}} {1, h, …, h^{5}} {1, h′, …, (h′)^{5}} ; Number of Conjugate 1  1  7  7  7  7

Table 2. Sylow and other cyclic subgroups, in conjugate groups

where δ ∈ Δ, δ^{2} ∈ ∇, h ∈ H, h′ ∈ H′ (H) The restrictions of one-dimensional representations for G are irreducible representations for H

(either χ′_{1} or χ′_{2}, depending on where they send ι), so it suffices to decompose χ_{7}. The character table for D_{7} is (from a previous homework):

1 {c^{1}, c^{6}} {c^{2}, c^{5}} {c^{3}, c^{4}} ι
χ′_{1}  1  1  1  1  −1
χ′_{2}  1  1  1  1  1
χ′_{3}  2 ζ^{1} + ζ^{6} ζ^{2} + ζ^{5} ζ^{3} + ζ^{4}  0
χ′_{4}  2 ζ^{2} + ζ^{5} ζ^{3} + ζ^{4} ζ^{1} + ζ^{6}  0
χ′_{5}  2 ζ^{3} + ζ^{4} ζ^{1} + ζ^{6} ζ^{2} + ζ^{5}  0
Res_{H}^{G}(χ_{7}) 6  −1  −1  −1  0

Character Table 2: D_{7} (with an extra character)

where ζ is a primitive seventh root of unity, and we are considering D_{7} ◁ G, D_{7} = 1 ∐ C ∐ ι. You could compute inner products of these to conclude that Res_{H}^{G}(χ_{7}) = χ′_{3} ⊕ χ′_{4} ⊕ χ′_{5}, or you could observe^{8} that this works, and since simple characters form a basis for class functions, this is the unique solution.

8You should notice this, since the n-th roots of unity sum to zero. Plus, you should conjecture that it would be symmetric.