Math 502: Abstract Algebra

Homework 4
Lawrence Tyler Rush
<me@tylerlogic.com>
January 5, 2014
http://coursework.tylerlogic.com/courses/upenn/math502/homework04

1


(a) Prove that there exists a unique group homomorphism sgn : Sn μ2


Homomorphic Let σ,τ each be elements of Sn. Then we have
sgn(στ)f (x ,...,x ) =   f (x     ,...,x     )
        n  1     n       n  (στ)(1)     (στ)(1)
                    =   fn(xσ(τ(1)),...,xσ(τ(1)))
                    =   sgn(σ)fn(xτ(1),...,xτ(1))
                    =   sgn(σ)fn(xτ(1),...,xτ(1))
                    =   sgn(σ) sgn(τ)f (x,...,x )
                                    n  1     n
and because is a field, we can cancel fn(x1,,xn) on both sides leaving us with sgn(στ) = sgn(σ)sgn(τ).

Uniqueness For later contradiction, let g : Sn μ2 be another group homomorphism not equal to sgn that satisfies the same properties of the sgn function. Then for σ Sn, we have

fn(xσ(1),...,xσ(n)) = g(σ)fn(x1,...,xn)

but there must exist at least one τ where sgn(τ)g(τ). So without loss of generality, let sgn(τ) = 1 and g(τ) = -1. However, then we would have

fn(xσ(1),...,x σ(n)) = sgn(σ )fn(x1,...,xn) = fn(x1,...,xn)

and

fn(xσ(1),...,xσ(n)) = g(σ)fn(x1,...,xn) = - fn(x1,...,xn )

which is not possible with our definition of fn. Hence we’ve reached a contradiction, and thus g and sgn must be one in the same.

(b)


(c)


The following proof is derived from [DF04, pg. 109]. It’s just so slick.

We will first prove that the transposition (1,2) has negative sign and then move on to the general case. According to the definition of the sgn function we have that for σ = (1,2),

                               fn(x σ(1),xσ(2),...,xσ(n))  =  sgn(σ )fn(x1,x2,...,xn)
                                Π1≤i<j≤n(xσ(i) - xσ(j)) =  sgn(σ )Π1 ≤i<j≤n(xi - xj)
Π2≤j≤n(xσ(1) - xσ(j))Π3≤j≤n(xσ(2) - x σ(j))Π3≤i<j≤n(xi - xj) = sgn(σ )Π2 ≤j≤n (x1 - xj)Π3≤j≤n (x2 - xj
                                                         Π       (x - x )
                                                          3≤i<j≤n  i   j
               Π2≤j≤n (xσ(1) - xσ(j))Π3 ≤j≤n(x σ(2) - xσ(j)) = sgn(σ )Π2 ≤j≤n (x1 - xj)Π3≤j≤n (x2 - xj)
   (xσ(1) - x σ(2))Π3≤j≤n (xσ(1) - xσ(j))Π3 ≤j≤n(x σ(2) - xσ(j)) = sgn(σ )(x1 - x2)Π3≤j≤n(x1 - xj)Π3Ȧ
                (x2 - x1)Π3≤j≤n(x2 - xj)Π3≤j≤n(x1 - xj) = sgn(σ )(x1 - x2)Π3≤j≤n(x1 - xj)Π3≤j≤n(x2 - xj)
                                            (x - x )  =  sgn(σ )(x - x )
                                              2   1             1   2
which indicates that sgn((1 2)) = -1.

Now let (i,j) be an abitrary transposition in Sn. Then we have that (i,j) = (1,i)(2,j)(1,2)(1,i)(2,j). This then yields the following

sgn((i,j)) =   sgn((1,i)(2,j)(1,2)(1,i)(2,j))

          =   sgn((1,i))sgn((2,j))sgn((1,2))sgn((1,i))sgn((2,j))
          =   sgn((1,i))2sgn((2,j))2 sgn((1,2))
          =   - 1
where we make use of the fact that sgn is a homomorphism and the commutativity of μ2.

(d)


We can see that a cycle of length m has that (1,2,,m) = (1,2)(2,3)⋅⋅⋅(m- 1,m). Therefore, using the fact that the sgn function is a homomorphism and that transpositions have sign of -1, we obtain
sgn((1,2,...,m )) = sgn ((1,2)(2,3)⋅⋅⋅(m - 1,m )) = sgn((1,2))sgn((2,3))⋅⋅⋅sgn((m - 1,m )) = (- 1)m-1

(e)


(f) Extra Credit


(g) Extra Credit


2


For a matrix A GL3(), left LA : 3 3 denote the linear transformation of left multiplication by A.

(a)


First, we see that any vecter v in C that is on the “surface” of C, i.e. any vecter with at least one coordinate of 1, must have that for A G, LA(v) must also be on the surface of C. The reason being because if it weren’t, then there would be some scalar α > 1 with LA(αv) C, but this can’t happen as it contradicts A(C) = C.

Not the least important of the vectors on the surface of C are e1,e2,e3. Since e1,e2,e3 each have a length of which is the shortest possible length of a vector on the surface of C, a length of one, then their images under LA must have length of at least one since they must be on the surface. On the other hand, because e1 + e2 + e3 is also on the surface of C then an increase in any one of the sizes of e1, e2 or e3 under LA will result in LA(e1 + e2 + e3) being outside of C. Hence LA will map each of e1, e2 and e3 to one of e1,±e2,±e3}. However, because A GL3(), then LA is an isomorphism and will map e1, e2 and e3 to distinct elements and furthermore to a basis. Therefore LA preserves the size and angles between e1, e2 and e3, which since those vectors are a basis for 3 makes A orthogonal, and thus G O3.

Now I G since it preserves C. If A,B G, then A(C) = C and B(C) = C, implying that (BA)(C) = B(A(C)) = B(C) = C and therefore AB G. And finally, since A(C) = C then C = A-1(C), and thus A-1 G. These three things, combined with the fact G O3 proven above makes G a subgroup of O3.

(b)


Let A G. Since LA maps {e1,e2,e3} to a basis of elements of e1,±e2,±e3} as per above, then LA maps e1 to one of the 6 elements of e1,±e2,±e3}, maps e2 to one of the four elements of e1,±e2,±e3}-{±LA(e1)} and e3 to two of elements of e1,±e2,±e3}-{±LA(e1),±LA(e2)}. Hence G has 6(4)(2) = 48 elements and consists of
(            |             )      (             |            )
{(  a      ) ||             }      { (  a      ) ||            }
 (     b   ) ||a,b,c ∈ {1,- 1} ⋃     (       b ) ||a,b,c ∈ {1,- 1}
(         c  |             )      (      c      |            )
                                  ( (         ) ||            )
                              ⋃   { (    a    ) ||            }
                                  (    b        ||a,b,c ∈ {1,- 1})
                                  ( (       c ) |            )
                              ⋃   {      a      ||            }
                                  ( (       b ) ||a,b,c ∈ {1,- 1})
                                  (    c        ||            )
                                  { (       a ) ||            }
                              ⋃     (  b      ) ||a,b,c ∈ {1,- 1}
                                  (      c      |            )
                                  ( (         ) ||            )
                              ⋃   { (       a ) ||            }
                                  (      b      ||a,b,c ∈ {1,- 1})
                                       c

(c)


We will abuse notation here and consider S3 and Perm({L1,L2,L3}) to be one in the same. They are, after all, isomorphic.

π is surjective Let σ S3 and define A to be the matrix with ith column eσ(i). Therefore according to the previous part of this problem, A G. Thus Aei = eσ(i) which implies that A(Li) = Lσ(i) and therefore π(A) = σ.

Ker(π) The kernel of the π will be the set of all matrices A such that A(Li) = Li. Therefore A will need to be such that Aei = ±ei. Hence

        ( (         ) |            )
        {    a        ||            }
Ker(π) =  (    b    ) ||a,b,c ∈ {- 1,1}
        (         c   |            )

(d)


Denote the set of diagonals of the problem’s statement by D = {d1,d2,d3,d4}. We will again abuse notation here and consider S4 and Perm({d1,d2,d3,d4}) to be one in the same.

We will use the following for vectors to help in our discussion here.

 v1 = e1 + e2 + e3
v2 = - e1 + e2 + e3
 v3 = e1 - e2 + e3
v = - e - e + e
 4    1   2    3
These vectors are the vectors contained in the diagonals of D where e3 is always positive. With these vectors, we have a basis B = {v1,v3,v4} and can write the standard basis E = {e1,e2,e3} as
       1
e1  =   2(v3 - v4)                                  (2.1)
       1
e2  =   2(v1 - v3)                                  (2.2)
       1
e3  =   2(v1 + v4)                                  (2.3)
which gives rise to
            (       1  1 )
[id]BE   =  1 (   1  - 1   )                               (2.4)
          2    - 1     1
          (            )
          (  1   1  - 1)
[id]EB   =     1  - 1 - 1                                  (2.5)
             1   1    1
where [id]BE and [id]E B are, respectively, the change of basis matrix from E to B and from B to E (yes it looks backwards).

ϕ is surjective Let σ S4. We then aim to find a matrix A G such that Avi = vσ(i). So let’s define a linear transformation T : 3 3 by its transformation of the basis B: T(vi) = vσ(i). By equations 2.1 through 2.3 we have that

          1
T(e1) =   2(vσ(3) - vσ(4))
          1
T(e2) =   2(vσ(1) - vσ(3))
          1
T(e3) =   2(vσ(1) + vσ(4))
leaving us with the following matrix representation of T in the standard basis
    1(                                    )
A = 2  vσ(3) - vσ(4) vσ(1) - vσ(3) vσ(1) +vσ(4)

The definition of T informs us that A G and that A(di) = dσ(i), thus implying that ϕ(A) = σ.

Ker(ϕ) We know that every element of the kernel will have that di = di, which implies that e1 + e2 + e3, being contained in a diagonal, will need to be mapped to itself or -e1 - e2 - e3. Hence

        ( (         )  (              ))
        {   1             - 1          }
Ker(ϕ) = ( (   1    ) ,(      - 1     ))
                  1               - 1

(e)


Is det |G = sgn π This is not true. According to part (c) of this problem, A = (           )
  - 1
(      1    )
          1 is in the kernel of π, meaning that sgnπ(A) = 1. However, A has a determinant of -1.

Is sgn π = sgn ϕ

Is det |G = sgn ϕ This is not true. According to part (d) of this problem, A = (                )
   - - 1
(        - 1     )
              - 1 is contained in the kernel of ϕ, meaning that sgn(ϕ(A)) = 1. However, A has a determinant of -1.

(f) Extra Credit


3


(a)


The dihedral group D8 is indeed isomorphic to a subgroup of G. Since G is the group of symmtries of a cube, then all elements which, say, keep e1 fixed will be symmetries of the unit square contained in the plane containing e2 and e3. Call this square S. We can use this in our definition of an injective group homomorphism ϕ : D8 G.

We can think of ϕ as the map that takes s (where D8 = s,r) to the symmetry that reflects S across the plane containing e1 and e3; i.e. take s to the matrix representation of the linear transformation defined by e1↦→e1, e2↦→- e2, and e3↦→e3. Similarly ϕ will take r to the matrix representation of rotation about e3 by π∕2.

This high-level definition of ϕ can be made concrete by the formula

        (  1        ) ℓ( 1         )i
ϕ(sℓri) = (    - 1   )  (       - 1 )
                  1         1

for all sri D8. By noticing that

(           ) (           )   (          ) (           )
   1            1               1             1
(       - 1 ) (    - 1    ) = (    - 1   ) (         1 )
      1                 1               1        - 1

and

(          ) -1   (          )
( 1        )      ( 1        )
        - 1     =           1
     1                 - 1

we can see that the following implies that ϕ is a group homomorphism

                 (  1        )ℓ ( 1        ) i( 1         )k ( 1        ) j
ϕ(sℓri)ϕ(skrj)  =  (     - 1   )  (       - 1 ) (    - 1    )  (       - 1 )
                           1         1                  1         1
                 (           )ℓ+k (          ) i(           )j
                    1               1              1
              =  (     - 1   )    (        1 )  (       - 1 )
                           1           - 1           1
                 (  1        )ℓ+k ( 1        ) -i(  1        )j
              =  (     - 1   )    (       - 1 )  (       - 1 )
                           1           1               1
                 (           )ℓ+k (          ) j-i
                    1               1
              =  (     - 1   )    (       - 1 )
                           1           1
              =  ϕ(sℓ+krj-i)
              =  ϕ(sℓskr- irj)
                    ℓ ik j
              =  ϕ(s rs r )

Now assume that some element sri is mapped by ϕ to the identity of G. Then we would have that

                                 (  1      )
                      ϕ(sℓri)  =  (     1   )
                                          1
(          ) ℓ(           )i     (         )
  1              1                  1
(    - 1   )  (       - 1 )   =  (     1   )
         1         1                      1
              (  1        )i     (  1        ) ℓ
              (       - 1 )   =  (     - 1   )
                   1                       1
which implies that = i = 0 since
(  1        )2      ( 1          )
(       - 1 )   =   (    - 1     )
      1                      - 1
(           )3      (          )
   1                  1
(       - 1 )   =   (        1 )
      1                  - 1
(  1        )4      ( 1       )
(       - 1 )   =   (    1    )
      1                     1
shows that the order of ( 1        )
(       - 1 )
     1 is four. Hence the kernel is trivial, implying that ϕ is injective.

The fact that ϕ : D8 G is an injective group homomorphism implies that D8 is isomorphic to a subgroup of G, namely ϕ(D8).

(b) Extra Credit


4


(a) Show that dim(V ) = 2 and {1,j} is a -basis of V


Let a,b,c,d . Then since
1(a+ bi)+ j(c - di) = a + bi+ cj - dji = a+ bi+ cj + dk

we see that {1,j} spans V .

Not if we were to have 1(a + bi) + j(c + di) = 0, then a + bi + cj -dk = 0 which implies that a = b = c = d = 0. Hence 1 and j are linearly independent.

With these two results, {1,j} is a basis for V , and thus dim(V ) = 2.

(b) Show for x that “left multiplication by x” is an endomorphism on V .


By the following, it can be seen that “left multiplication by x” is an endomorphism, where v,w V The distributive law on is what gives us that α(x) is an endomorphism, i.e.
α (x)(v+ w ) = x(v+ w) = xv+ xw = α(x)v+ α (x)w

(c) Show that α : End(V ) is a ring homomorphism.


By the following two equations, we have that α is a ring homomorphism
α(x+ y)(v) = (x+ y)(v) = xv+ yv = α(x)(v) +α (y)(v) = (α(x)+ α(y))(v)

α (xy)(v) = (xy)(v) = x(y(v)) = α (x)(α(y)(v)) = (α(x)∘ α(y))(v)

(d) Prove that α is injective


Let x H be such that α(x) = idV . Then for all v V xv = α(x)(v) = idV (v) = v, but the only element of with this property is the multiplicative identity, so x must be the identity. Hence the kernel of α is trivial and thus α is injective.

(e) Write down the matrix representation of α(i), α(j), and α(k) with respect to the basis {1,j}


Set B to the basis {1,j}. We will use the notation []B to denote “representation in the basis B”. With that said
                                                    (      )
[α(i)]B  =   ([α(i)(1)]B   [α(i)(j)]B ) = ( [i]B [ij]B ) =   i  0
                                                      0  i           (        )
            (                    )   (       2   )   (            )    0  - 1
[α(j)]B  =    [α(j)(1)]B  [α (j)(j)]B   =   [j]B  [j]B   =  [j]B   [- 1]B   =   1    0
            (                     )  (            )  (            )   ( 0  - i )
[α(k)]B  =    [α(k)(1)]B  [α(k)(j)]B   =   [k]B  [kj]B   =   [k]B   [- i]B   =   i   0

(f)


Because is a -vector space of dimension four ({1,i,j,k} is an -basis) and because α is injective as per part (d) of this problem, then as a linear transformation, α has a trivial kernel. Therefore the rank+nullity theorem informs us that the image of α has dimension of four since dim() = 4. However, because dim(End(V )) = 4 as well, we must have α() = End(V ).

References


[DF04]   D.S. Dummit and R.M. Foote. Abstract Algebra. John Wiley & Sons Canada, Limited, 2004.