## Math 503: Abstract Algebra

Homework 6
Lawrence Tyler Rush

<me@tylerlogic.com>

March 19, 2014

http://coursework.tylerlogic.com/courses/upenn/math503/homework06
### 1

Since the conjugacy classes of S_{4} are which we will represent by the elements (1), (12), (123), (12)(34), and (1234), respectively. Thus there are five irreducible
characters of S_{4}. Since the degrees of the irreducible characters must be factors of |S_{4}| = 24, and their squares must sum to
24, then the only possible degrees are 1,1,2,3,3. Letting χ_{1} be the trivial character and χ_{2} 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 ∈ S_{4} 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 ∈ S_{4} and associated conjugacy class C_{σ} 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 b_{3} = -1 and b_{4} = b_{5} = 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 a_{3} = d_{3} = 0, a_{4} = d_{5} = 1, and a_{5} = d_{4} = -1 is
the solution, or a_{3} = d_{3} = 0, a_{4} = d_{5} = -1, and a_{5} = d_{4} = 1 is the solution. We will just choose the former, as chosing one
arbitrarily just dictates to which irreducible characters χ_{4} and χ_{5} will correspond. Finally, pairing columns
(123) and (12)(34) demands that c_{3} = 2 (remember b_{3} = -1 and b_{4} = b_{5} = 0), and pairing columns (1) and
(12)(34) therefore demands that c_{4} = c_{5} = -1. Hence the full character table of S_{4} is revealed in Table 1.1.

Table 1.1: Character table of S_{4}

### 2

Let p be a prime, and H(F_{p}) be the Heisenberg group with p^{3} elememts.
Lemma 2.1. The center of H(F_{p}) is

and is, furthermore, the same as the commutator subgroup, (H(F_{p}))′.

Proof. For any a,b,c,x,y,z ∈ F_{p}

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

Furthermore, because

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