1

(a) Compute the character table of the Heisenberg group H(F₃)

by which we see that

represent the conjugacy classes of H(F₃). Thus there are 11 irreducible characters. We will denote the conjugacy classes by C₀, C₁, C₂, C₀₁, C₀₂, C₁₀, C₁₁, C₁₂, C₂₀, C₂₁, and C₂₂ where

We will denote the eleven characters by χ₁ through χ₁₁, with χ₁ being the trivial character. With this notation, we have the following steps of reasoning about the values of the character table.

1. The trivial character takes all elements to the identity. So we have:

2. Since the degrees of the each character must divide the order of the group (which is 27) and the sum of the squares of the degrees must equal the order of the group, then the degrees of the irreducible characters have to be 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 3 as 9(1²) + 2(3²) = 27. This gives us the entire first column of the table.
3. Define the maps φ : H(F₃) → F₃ and ϕ : H(F₃) → F₃ by

   Therefore kerφ = ∪ and kerϕ = ∪ , which implies H(F₃)∕kerφF₃ and H(F₃)∕kerϕF₃. This, in turn, demands that the non-trivial degree-1 characters can be determined from the non-trivial degree-1 characters on F₃. Define χ₂ to be the character corresponding to φ and the character on F₃ mapping iζⁱ where ζ = e is the third root of unity. Similarly let χ₃ be the character corresponding to ϕ and the character on F₃ mapping iζⁱ. We thus have:

   

4. Given that χ₂ and χ₃ are non-trivial characters of dimension one, their tensor yields another non-trivial character of dimension one. We can then again tensor the resulting one dimensional character with other one-dimensional characters to get more one dimensional characters, and then repeat. In this vein we define:
   
   and therefore we are only left with finding the values of χ₁₀ and χ₁₁. This gives us:

5. For each conjugacy class which is not one of C₀, C₁, or C₂, the current values of χ₁ through χ₉, when paired with themselves and multiplied by the size of the conjugacy class (all have size 3), gives the size of the group. Therefore the values of χ₁₀ and χ₁₁ must all be zero on these conjugacy classes. We are now left with finding the values of χ₁₀ and χ₁₁ on C₁ and C₂.

   Now because

   

   and χ(x) = χ(x^-1) for any x ∈ G, then both χ₁₀(x₁) = χ₁₀(x₂) and χ₁₁(x₁) = χ₁₁(x₂) for all x₁ ∈ C₁, x₂ ∈ C₂. Furthermore, the orthogonality relations tell us that for x₁ ∈ C₁ and x₂ ∈ C₂ (denoting χ₁₀(x₁), χ₁₁(x₂) by α, β respectively)

   1. Pairing the first and second column yields α + β = -3
   2. Pairing the first and third column yields α + β = -3
   3. Pairing the second to last row with itself yields αα = 9, i.e. |α| = 3
   4. Pairing the last row with itself yields ββ = 9, i.e. |β| = 3

   By the above constraints, we can see that χ₁₀(x₁) = χ₁₁(x₂) = 3ζ is a viable solution. This yields

   

Bringing together all character values from above, we obtain the entire character table, seen in Table 1.1.

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Table 1.1: Character table of H(F₃)

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(b) Extra Credit: Compute the character table of H(F_p) for larger primes

Therefore there are 29 irreducible characters. The degrees of each of these characters must divide |G| = 125, i.e. be 1, 5, 10, or 25, and the sum of their squares must be 29. The only set of values that satisfy such contraints are 25 1’s and 4 5’s. This gives us the first column, and of course we have all 1’s in the first row for the trivial representation. Now if we take φ and ϕ to be as they are defined in the previous subsection, the quotients of G by their respective kernels each result in F₅. Thus each 1-degree character is made up of fifth roots of unity, which we will denote by ζ. Hence we can set χ₂ to the character arising from the character on F₅ which takes iζⁱ, and similarly with χ′ and ϕ. This is similar to our method above. We can then set the rest of the degree-1 characters to be, i.e. alterating “tensoring” by χ₂ and χ₃ so that

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Table 1.2: Character table of H(F₅)

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2

(a) Find a non-commutative group G ⊂ S₇ with 21 elements.

Since Aut(C₃) is isomorphic to C₂, the identity map is the only homomophism in Aut(C₃)^C₇. Hence any C₃ ⋊ C₇ would just be isomorphic to the direct product, thereby making G commutative give that the components C₃ and C₇ are commutative. Therefore we must find a non-trivial homomorphism φ : C₃ → Aut(C₇) which will then make C₇ ⋊ _φC₃ into a semi-direct product. The automorphisms on C₇ are defined by their mapping of 1, so define f_n ∈ Aut(C₇) by the map mnm for m ∈ C₇. Because φ needs to be a homomorphism, we must have

(2.1)

For our group we will select n = 2

(b) Determine the conjugacy classes of G

from which we obtain the conjugacy classes:

(c) Compute the character table for G

1. χ₁ is the trivial representation so

2. Given the hint in the problem statement, there are three characters of degree 1. Since the degree of the characters need to divde the order of the group and the squares of their sums need to be 21 (in this case), the first column of the character table is 1,1,1,3,3
3. Define the homomorphism ϕ : G → C₃ by ϕ(m,n) = m. Then kerϕ = {(m,0)|n ∈ C₇} implying that we can obtain degree-1 characters from the degree-1 characters of C₃. We will assign these to χ₂ and χ₃ yielding:

   where ζ is the third root of unity, e.

4. Since 7(1 + ζζ + ζ²ζ²) = 7(3) = 21, then on C₁ and C₂, χ₃ and χ₄ both have a value of zero.
5. Since (2,0)^-1 = (φ(0)(5),0) = (5,0) and because (5,0) ∈ C₀₁ while (2,0) ∈ C₀₂, then the value of χ₄ on C₀₁ and C₀₂ must be conjugates of each other. By the same reasoning, the same is true for χ₅. If we let α be the value of χ₄ on C₀₁, pairing the fourth row with itself yields αα = 2 and similarly if we let β be the value of χ₅ on C₀₂. Therefore the magnitude of α and β must be . We quickly realize that α = ζ and β = ζ² is a viable solution, yielding:

Therefore the whole character table for our G can be seen in table 2.3

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Table 2.3: Character table of G = C₇ ⋊ _φC₃

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(d) Extra Credit

3 Characters of symmetric and alternating products

(3.2)

and

(3.3)

(a) Show that the character of (S²(V ),S²ρ) at x ∈ G is

then by equations 3.2 and 3.3

(b) Show that the character of (Λ²(V ), Λ²ρ) at x ∈ G is

then by equations 3.2 and 3.3

(c) Extra Credit

4

(a)

(b)

(c)

(d)

5 Extra Credit