## Math 503: Abstract Algebra

Homework 6 <me@tylerlogic.com>
March 19, 2014
http://coursework.tylerlogic.com/courses/upenn/math503/homework06

### 1

Since the conjugacy classes of S4 are
which we will represent by the elements (1), (12), (123), (12)(34), and (1234), respectively. Thus there are five irreducible characters of S4. Since the degrees of the irreducible characters must be factors of |S4| = 24, and their squares must sum to 24, then the only possible degrees are 1,1,2,3,3. Letting χ1 be the trivial character and χ2 be the character resulting from the sign homomorphism, we immediately have

Note that becuase each conjugacy class, C, has the property that x C implies x-1 C, then all values of the irreducible characters must be real, since χ(σ-1) = χ(σ) for any σ S4 and character χ. Furthermore, since each character value must be the sum of roots of unity, then each current unknown in the table above must be an integer. In light of this and the fact that

for any σ S4 and associated conjugacy class Cσ we get the following four equations

which yields the possible solution sets
Equation 1.2, solutions sets 1.6, and by pairing columns (123) and (1) reveals b3 = -1 and b4 = b5 = 0. Then by pairing (1) with (12), (1) with (1234), and (12) with (1234) we get
which, when considering equations 1.1, 1.4, 1.5, and 1.8, imply that either a3 = d3 = 0, a4 = d5 = 1, and a5 = d4 = -1 is the solution, or a3 = d3 = 0, a4 = d5 = -1, and a5 = d4 = 1 is the solution. We will just choose the former, as chosing one arbitrarily just dictates to which irreducible characters χ4 and χ5 will correspond. Finally, pairing columns (123) and (12)(34) demands that c3 = 2 (remember b3 = -1 and b4 = b5 = 0), and pairing columns (1) and (12)(34) therefore demands that c4 = c5 = -1. Hence the full character table of S4 is revealed in Table 1.1.

Table 1.1: Character table of S4

### 2

Let p be a prime, and H(Fp) be the Heisenberg group with p3 elememts.

Lemma 2.1. The center of H(Fp) is

and is, furthermore, the same as the commutator subgroup, (H(Fp)).

Proof. For any a,b,c,x,y,z Fp

and therefore for any to be in the center of H(Fp), cx - az must be zero, which, since x,z can be anything in Fp, means a = c = 0. Hence the center of H(Fp) is

Furthermore, because

then the elements of the center coincide with the elements of the commutator subgroup. __