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 S₄ are

{(1)} {(12),(13),(14),(23),(24),(34)} {(123),(132),(124),(142),(134),(143),(234),(243)} {(12)(34),(13)(24),(14)(23)} {(1234),(1243),(1324),(1342),(1423),(1432)}

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

| size: |1 6 8 3 6 -class:-|(1)--(12)-(123)--(12)(34)-(1234)- χ1 |1 1 1 1 1 χ2 |1 - 1 1 1 - 1 χ3 |2 a3 b3 c3 d3 χ4 |3 a4 b4 c4 d4 χ5 3 a5 b5 c5 d5

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

∑ χ2i(σ) = |S4|∕|Cσ| i

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

a2+ a2 + a2 = 2 (1.1) 32 42 52 b3 + b4 + b5 = 1 (1.2) c23 + c24 + c25 = 6 (1.3) d23 +d24 + d25 = 2 (1.4)

which yields the possible solution sets

a3,a4,a5 → 0,±1, ±1 (1.5) b3,b4,b5 → 0,0,±1 (1.6) c3,c4,c5 → ±1, ±1,±2 (1.7) d3,d4,d5 → 0,±1, ±1 (1.8)

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

2a3 + 3a4 + 3a5 = 0 2d3 + 3d4 + 3d5 = 0 a3d3 + a4d4 + a5d5 = - 2

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.

size: |1 6 8 3 6 class: |(1) (12) (123) (12)(34) (1234) ---χ1--|1----1-----1------1-------1---- χ2 |1 - 1 1 1 - 1 χ3 |2 0 - 1 2 0 χ4 |3 1 0 - 1 - 1 χ5 |3 - 1 0 - 1 1

Table 1.1: Character table of S₄

2

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

Lemma 2.1. The center of H(F_p) is

(( ) ) { 1 0 b } Z(H (Fp)) = ( 1 0) ( 1 )b∈F p

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

Proof. For any a,b,c,x,y,z ∈ F_p

( 1 x y ) ( 1 a b) ( 1 x y )- 1 ( 1 a cx - az + b ) ( 1 z ) ( 1 c) ( 1 z ) = ( 1 c ) 1 1 1 1

and therefore for any ( )
1 a b
( 1 c )
1 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

( ( ) ) { 1 0 b } Z(H (Fp)) = ( ( 1 0 ) ) 1 b∈Fp

Furthermore, because

( ) ( ) ( ) -1( ) -1 ( ) 1 x y 1 a b 1 x y 1 a b 1 0 cx- az ( 1 z) ( 1 c ) ( 1 z ) ( 1 c) = ( 0 1 0 ) 1 1 1 1 0 0 1

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