February 27, 2014

http://coursework.tylerlogic.com/courses/upenn/math503/homework05
Let a,b,c,x,y,z ∈ F

by which we see that

represent the conjugacy classes of H(F_{3}). Thus there are 11 irreducible characters. We will denote the conjugacy classes by
C_{0}, C_{1}, C_{2}, C_{01}, C_{02}, C_{10}, C_{11}, C_{12}, C_{20}, C_{21}, and C_{22} where

We will denote the eleven characters by χ_{1} through χ_{11}, with χ_{1} being the trivial character. With this notation, we have
the following steps of reasoning about the values of the character table.

- The trivial character takes all elements to the identity. So we have:
- 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(1
^{2}) + 2(3^{2}) = 27. This gives us the entire first column of the table. - Define the maps φ : H(F
_{3}) → F_{3}and ϕ : H(F_{3}) → F_{3}byTherefore kerφ = ∪ and kerϕ = ∪ , which implies H(F

_{3})∕kerφF_{3}and H(F_{3})∕kerϕF_{3}. This, in turn, demands that the non-trivial degree-1 characters can be determined from the non-trivial degree-1 characters on F_{3}. Define χ_{2}to be the character corresponding to φ and the character on F_{3}mapping iζ^{i}where ζ = e^{}is the third root of unity. Similarly let χ_{3}be the character corresponding to ϕ and the character on F_{3}mapping iζ^{i}. We thus have: - Given that χ
_{2}and χ_{3}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:_{10}and χ_{11}. This gives us: - For each conjugacy class which is not one of C
_{0}, C_{1}, or C_{2}, the current values of χ_{1}through χ_{9}, 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 χ_{10}and χ_{11}must all be zero on these conjugacy classes. We are now left with finding the values of χ_{10}and χ_{11}on C_{1}and C_{2}.Now because

and χ(x) = χ(x

^{-1}) for any x ∈ G, then both χ_{10}(x_{1}) = χ_{10}(x_{2}) and χ_{11}(x_{1}) = χ_{11}(x_{2}) for all x_{1}∈ C_{1}, x_{2}∈ C_{2}. Furthermore, the orthogonality relations tell us that for x_{1}∈ C_{1}and x_{2}∈ C_{2}(denoting χ_{10}(x_{1}), χ_{11}(x_{2}) by α, β respectively)- Pairing the first and second column yields α + β = -3
- Pairing the first and third column yields α + β = -3
- Pairing the second to last row with itself yields αα = 9, i.e. |α| = 3
- Pairing the last row with itself yields ββ = 9, i.e. |β| = 3

By the above constraints, we can see that χ

_{10}(x_{1}) = χ_{11}(x_{2}) = 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.

We will compute the character table for H(F

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 F_{5}. Thus
each 1-degree character is made up of fifth roots of unity, which we will denote by ζ. Hence we can set χ_{2} to the character
arising from the character on F_{5} which takes iζ^{i}, 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 χ_{2} and χ_{3} so that

Since S

Since Aut(C_{3}) is isomorphic to C_{2}, the identity map is the only homomophism in Aut(C_{3})^{C7}. Hence any C_{3} ⋊ C_{7} would
just be isomorphic to the direct product, thereby making G commutative give that the components C_{3} and C_{7} are
commutative. Therefore we must find a non-trivial homomorphism φ : C_{3} → Aut(C_{7}) which will then make
C_{7} ⋊ _{φ}C_{3} into a semi-direct product. The automorphisms on C_{7} are defined by their mapping of 1, so define
f_{n} ∈ Aut(C_{7}) by the map mnm for m ∈ C_{7}. Because φ needs to be a homomorphism, we must have

| (2.1) |

For our group we will select n = 2

With the fomula from equation 2.1 in hand, the conjugation of (r,y) ∈ C

from which we obtain the conjugacy classes:

We will denote the characters by χ

- χ
_{1}is the trivial representation so - 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
- Define the homomorphism ϕ : G → C
_{3}by ϕ(m,n) = m. Then kerϕ = {(m,0)|n ∈ C_{7}} implying that we can obtain degree-1 characters from the degree-1 characters of C_{3}. We will assign these to χ_{2}and χ_{3}yielding:where ζ is the third root of unity, e

^{}. - Since 7(1 + ζζ + ζ
^{2}ζ^{2}) = 7(3) = 21, then on C_{1}and C_{2}, χ_{3}and χ_{4}both have a value of zero. - Since (2,0)
^{-1}= (φ(0)(5),0) = (5,0) and because (5,0) ∈ C_{01}while (2,0) ∈ C_{02}, then the value of χ_{4}on C_{01}and C_{02}must be conjugates of each other. By the same reasoning, the same is true for χ_{5}. If we let α be the value of χ_{4}on C_{01}, pairing the fourth row with itself yields αα = 2 and similarly if we let β be the value of χ_{5}on C_{02}. Therefore the magnitude of α and β must be . We quickly realize that α = ζ and β = ζ^{2}is a viable solution, yielding:

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

Let (V,ρ) be a finite dimensional representation of a finite group G. Let χ

| (3.2) |

and

| (3.3) |

The elements {e

The elements {e