Math 502: Abstract Algebra

Homework 5
Lawrence Tyler Rush
<me@tylerlogic.com>
January 5, 2014
http://coursework.tylerlogic.com/courses/upenn/math502/homework05

1

Let V be a vector space over a field F and W1,W2 be F-vector subspaces of V .

(a) Show dimF V ( ∕Wi) < ∞ implies dimF V ( ∕W1∩ W2) < ∞

Lemma 1.1. Let V be a vector space over a field F with subspace W. Then the F-linear transformation T : V →V∕ W defined by v↦→[v] is a surjection with kernel W. Furthermore, dimV∕ W = dimV - dimW.

Proof. From a group theoretic point of view, (V,+) is an abelian group, and therefore we know T to be the surjective canonical map from the group V to the cosets of its normal subgroup W with kernel W. Thus the rank+nullity Theorem yields dimV = dimkerT + dimT(V ) = dimW + dimV∕ W, i.e. dimV∕ W = dimV - dimW. __

First note that if V is finite dimensional, then this problem is trivial. So assume that V is infinite-dimensional. Then Lemma 1.1 informs us that

dim V - dimWi = dim V∕Wi < ∞
(1.1)

for each Wi. Noting that

dim V - dimW2 < ∞ =⇒ dim(W1 + W2 )- dim W2 < ∞

since W2 ⊂ W1 + W2 ⊂ V , we then have that the right hand side of the last line of

dim V - dim (W1 ∩W2 ) = dim V - (dim W1 + dim W2 - dim(W1 + W2 )) = (dim V - dim W1)+ (dim(W1 + W2 )- dim W2 )
is the sum of two finite numbers. Hence dimV - dim(W1 ∩W2) is finite, but according to Lemma 1.1 this is the dimension of V∕ (W1 ∩ W2).

(b)

Analogous Statement: If G is a group with H1,H2 ≤ G such that the index of Hi in G is finite for each Hi, then the index of H1 ∩ H2 in G is also finite.

Let G be a group with subgroups of finite index H1 and H2. Then we know that the values of

|G| |G| ∕|H1| and ∕|H2|

are both finite. Since |H1 ∪H2|≥|Hi| from a set-theoretic perspective, then we also know |G| ∕ |H1 ∪ H2| = n for some finite n. This yields

---|G|--- = n |H1 ∪ H2| |G| |H1-|+-|H2-|- |H1-∩-H2| = n |H1-|+-|H2-|- |H1-∩-H2| = 1∕n |G| |H1-|+ |H2|- |H1-∩-H2| = 1∕n |G| |G | |G| |H1-| |H2| |H1-∩-H2|- |G| + |G | = 1∕n + |G |
Since the left hand side of the above equation is finite, then so must be the right. Hence |H1∩H2| |G| must be finite, and therefore H1 ∩ H2 has finite index in G.

(c) Extra Credit

(d) Extra Credit

2

Let R be a ring and G a group.

(a) Show that the group ring R[G] is a ring.

Has Zero Element The 0 here is that of the ring R is

( ) ( ) ∑ ∑ ∑ ∑ ax[x] + 0[x] = (ax + 0)[x] = ax[x] x∈G x∈G x∈G x∈G

(∑ ) (∑ ) ∑ ∑ 0[x] + ax[x] = (0+ ax)[x] = ax[x] x∈G x∈G x∈G x∈G

Addition is associative We get the following because of the associativity of addition on R.

( ∑ ∑ ) ∑ ∑ (∑ ) ax[x]+ bx[x] + cx[x] = (ax + bx)[x]+ cx[x] x∈G x∈G x∈G x∑∈G x∈G = ((ax + bx)+ cx)[x ] x∈G ∑ = (ax + (bx + cx))[x ] x∑∈G ∑ = ax[x]+ (bx + cx)[x] x∈G x(∈G ) ∑ ∑ ∑ = ax[x]+ bx[x]+ cx[x] x∈G x∈G x∈G

Addition is commutative We get the following by the commutative property of addition on R.

( ) ( ) ( ) ( ) ∑ ∑ ∑ ∑ ∑ ∑ ax[x ] + bx[x] = (ax + bx)[x] = (bx + ax)[x] = bx[x] + ax[x] x∈G x∈G x∈G x∈G x∈G x∈G

Additive elements have inverses The 0 here is that of the ring R is

(∑ ) ( ∑ ) ∑ ∑ ( ∑ ) ax[x] + - ax[x] = (ax - ax)[x] = (0)[x] = 0[x] x∈G x∈G x∈G x∈G x∈G

(∑ ) ( ∑ ) ∑ ∑ (∑ ) - ax[x] + ax[x ] = (- ax + ax)[x] = (0)[x ] = 0[x] x∈G x∈G x∈G x∈G x∈G

Multiplicative Identity The 1 here is that of the ring R is

( ) ( ) ∑ ∑ ∑ ∑ ∑ ax[x] (1[e]) = ( ax1) [z] = (az)[z] = ax[x] x∈G z∈G x,y∈G, xe=z z∈G x∈G

Multiplication is associative We get the following by the associative property of multiplication in R.

(( ) ( )) ( ) ( ( ) ) ( ) ∑ ax[x] ∑ bx[x] ∑ cx[x] = ( ∑ ( ∑ axby) [z]) ∑ cx[x] x∈G x∈G x∈G z∈G x,y∈G, xy=z x∈G ( ( ) ) ∑ ∑ ∑ = ( ( axby) ct) [w ] w∈G (z,t∈G, zt=w x,y∈G, xy=z ) ∑ ∑ ∑ = ( axbyct) [w ] w∈G z,t∈G, zt=wx,y∈G, xy=z ( ) = ∑ ( ∑ ∑ axbyct) [w ] w∈G x,z∈G, xz=w y,t∈G, yt=z ( ( ) ) ∑ ∑ ∑ = ( ax( byct) ) [w ] w∈G x,z∈G, x(z=w ( y,t∈G, yt=z ) ) ( ∑ ) ∑ ∑ = ax[x] ( ( byct) [z]) x∈G z∈G y,t∈G, yt=z ( ∑ ) (( ∑ ) ( ∑ )) = ax[x] bx[x] cx[x] x∈G x∈G x∈G

Multiplication distributes over addition

( ∑ ) ( ∑ ∑ ) ( ∑ ) ( ∑ ) ax[x] bx[x]+ cx[x] = ax[x] (bx + cx)[x] x∈G x∈G x∈G x∈G( x∈G ) ∑ ∑ = ( ax(by + cy)) [z] z∈G x,y∈G, xy=z ( ) = ∑ ( ∑ a b + a c) [z] z∈G x,y∈G, xy=z xy xy (( ) ( )) ∑ ∑ ∑ = (( axby) + ( axcy)) [z] z∈G ( x,y∈G, xy=z ) x,y∈(G, xy=z ) ∑ ∑ ∑ ∑ = ( axby) [z]+ ( axcy) [z] z∈G x,y∈G, xy=z z∈G x,y∈G, xy=z ( ∑ ) ( ∑ ) (∑ ) (∑ ) = ax[x] bx[x] + ax[x] cx[x] x∈G x∈G x∈G x∈G
( )( ) ( ) ( ) ∑ ∑ ∑ ∑ ∑ bx[x]+ cx[x] ax[x] = (bx + cx)[x] ax[x] x∈G x∈G x∈G x∈G( x∈G ) ∑ ∑ = ( (bx + cx)ay) [z] z∈G (x,y∈G, xy=z ) ∑ ∑ = ( bxay + cxay) [z] z∈G x,y∈G, xy=z (( ) ( )) = ∑ (( ∑ b a ) + ( ∑ c a )) [z] z∈G x,y∈G, xy=z x y x,y∈G, xy=z x y ( ) ( ) ∑ ∑ ∑ ∑ = ( bxay) [z]+ ( cxay) [z] z(∈G x,y∈G,) x(y=z ) (z∈G x,y∈G),( xy=z ) ∑ ∑ ∑ ∑ = bx[x ] ax[x] + cx[x] ax[x] x∈G x∈G x∈G x∈G

(b)

(c)

(d)

(e) Extra Credit

(f) Extra Credit

3

(a)

Let i : D3 → GLℝ(ℂ) be the inclusion map. Then define β : ℝ[D3] → Endℝ(ℂ) by
( ) ∑ ∑ xs[s] ↦→ z ↦→ xs(i(s)(z)) s∈D3 s∈D3

Outputs of β are indeed in End ℝ(ℂ) Let a,b ∈ ℝ and z,z′∈ ℂ

( ∑ ) ∑ β xs[s] (az + bz′) = xs(i(s)(az + bz′)) s∈D3 s∈D3 = ∑ xs(ai(s)(z) +bi(s)(z′)) s∈D3 ∑ ′ = (xsai(s)(z) +xsbi(s)(z)) s∑∈D3 ∑ = xsai(s)(z)+ xsbi(s)(z′) s∈D3 s∈D3 = a ∑ x i(s)(z)+ b ∑ x i(s)(z′) s∈D3 s s∈D3 s ( ) ( ) = aβ ∑ xs[s] (z) + bβ ∑ xs[s] (z′) s∈D3 s∈D3

β is a homomorphism over addition

( ) ( ) β ∑ xs[s]+ ∑ ys[s] (z) = β ∑ (xs + ys)[s] (z) s∈D3 s∈D3 s∈D3 ∑ = (xs + ys)(i(s)(z)) s∈∑D3 = (xs(i(s)(z))+ ys(i(s)(z))) s∈D3 = ∑ x(i(s)(z))+ ∑ y(i(s)(z)) s∈D s s∈D s (3 ) 3( ) = β ∑ x[s] (z)+ β ∑ y [s] (z) s∈D3 s s∈D3 s

β is a homomorphism over multiplication

(( ) ( )) ( ( ) ) ∑ ∑ ∑ ∑ β xs[s] ys[s] (z) = β( ( xsyt) [u]) (z) s∈D3 s∈D3 u∈(D3 s,t∈G, st=u ) ∑ ∑ = ( xsyt) i(u)(z) u∈D3 s,t∈G, st=u ( ) = ∑ ( ∑ xsyti(u )(z)) u∈D3 s,t∈G, st=u ( ) ∑ ∑ = ( xsyti(st)(z)) u∈D3( s,t∈G, st=u ) ∑ ∑ = ( xsyti(s)i(t)(z)) u∈D3 s,t∈G, st=u ( ) = ∑ ( ∑ xsi(s)(yti(t)(z))) u∈D3 s,t∈G, st=u ( ) ∑ ∑ = ( xsi(s)(yti(t)(z))) s∈D3 t,u∈G(, t-1u=s ) ∑ ∑ = xsi(s)( yti(t)(z)) s∈D3 t,u∈G, t-1u=s ( ∑ ) (∑ ) = β xs[s] ysi(s)(z) s∈D3 s∈G ( ∑ ) ( (∑ ) ) = β xs[s] β ys[s] (z) s∈D3 s∈G ( ( ∑ ) (∑ ) ) = β xs[s] β ys[s] (z) s∈D3 s∈G
This shows that β is a ring homomorphism.

Uniqiueness

(b)

(c) Extra Credit

4

Let ℍ be the ring of Hamiltonian quaternians and V = ℝi + ℝj + ℝk be the three-dimensional vector subspace spanned by the vectors i,j,k ∈ ℍ.

(a) Show uxu-1 ∈ V for all u ∈ ℍ× and x ∈ V

Let u ∈ ℍ× and x,y ∈ V . Since we have u0u-1 = 0 and u(xy)u-1 = ux(u-1u)yu-1 = (uxu-1)(uyu-1), then ℍ× acts on V by u ⋅ x = uxu-1. This induces a homomorphism φ : ℍ×→ Perm(V ) by φ(u) = (x↦→uxu-1). Hence, since the range of φ is Perm(V ), we have uxu-1 ∈ V for all u ∈ ℍ×.

(b) Show that α : ℍ×→ GLℝ(V ) defined by α(u)(v) = uvu-1 is a group homomorphism

The map α : ℍ×→ GLℝ(V ) defined by α(u)(v) = uvu-1 is simply the permutation representation of the action u ⋅ x = uxu-1, above. This is induced by the action, and is always a homomorphism.

(c) Determine the kernel of α

The kerα is the set {u ∈ ℍ×|α(u) = idV }; in other words it’s the set {u ∈ ℍ× | uvu-1 = v, ∀v ∈ V }. Let Z ⊂ ℍ× be the set of all invertible elements that commute with every element of ℍ. Then it’s clear that Z ⊂ kerα since an element u ∈ Z that commutes with every element of ℍ, will certainly have uvu-1 = v for all v ∈ V .

So now let u ∈ kerα. Then for any a+bi+cj +dk ∈ ℍ we have that u(a+bi+cj +dk)u-1 = uau-1+u(bi+cj +dk)u-1 = a+bi+cj +dk, since a ∈ ℝ commutes with all of ℍ. Thus u ∈ Z, implying that kerα ⊂ Z.

Putting together the results Z ⊂ kerα and kerα ⊂ Z from above, we get that kerα = Z, i.e kerα is the set of all units that commute with all elements of ℍ.

(d) Extra Credit

5

For the sake of clarity, we will use {0,1,2} as the set F3, with the necessary understanding that the elements aren’t really integers.

(a) Show that #GL2(F3) = 48

There are a total of 34 = 81 possible elements in M2(F3) and GL2(F3) will be the matrices with nonzero determinant, so we’ll subtract the number of matrices with determinant zero from 81. The matrices with zero determinant and independent of the non-zero entries are the zero matrix, the matrices with three zero entries
( α ) ( α ) ( ) ( ) α α

and the matrices with two non-diagonal zero entries

( ) ( ) ( ) ( ) α β α α β α β β

These account for 1, 4(2) = 8, and 4(22) = 16, respectively, of the matrices with zero determinant. Thus 25 of the matrices with zero determinant. The remaining such matrices are matrices with all non-zero entries where the product of the two diagonal entries are equal. In other words, a matrix

( ) a b c d
(5.2)

with a,b,c,d≠0 we’ll have zero determinant when ad = bc. Therefore the following “truth” table of possible values of ad-bc reveals that there are eight matrices in the form of 5.2 above that have zero determinant.

a d b c ad bc ad- bc -1--1--1--1--1---1----0---- 1 1 1 2 1 2 2 1 1 2 1 1 2 2 1 1 2 2 1 1 0 1 2 1 1 2 1 1 1 2 1 2 2 2 0 1 2 2 1 2 2 0 1 2 2 2 2 1 1 1 1 1 1 1 1 0 1 1 1 2 1 2 2 1 1 2 1 1 2 2 1 1 2 2 1 1 0 1 2 1 1 2 1 1 1 2 1 2 2 2 0 1 2 2 1 2 2 0 1 2 2 2 2 1 1

Thus adding 8 to the 25 we tallied initially, we have a total of 33, resulting in #GL2(F3) = 81 - 33 = 48.

(b) Find explicitly a 3-Sylow subgroup of GL2(F3)

Because of the following
( )2 ( ) ( )3 ( ) 1 1 = 1 2 and 1 1 = 1 1 1 1 1

then ( 1 1 ) 1 is an element of order 3. However, because #GL2(F3) = 48 = 3(24), then the subgroup generated by ( 1 1 ) 1 is a 3-Sylow subgroup. For the remaining parts of the problem, call it P3.

(c) Determine the normalizer of P3. How many 3-Sylow subgroups are there?

Let a,b,c,d ∈ F3 and set A = ( ) a b c d Then we have
( a b ) ( 1 1 )( a b ) ( ad- ac- bc a2 ) c d 1 c d = (ad - bc)-1 - c2 ac+ ad- bc

So in order for A to be in the normalizer of P3, -c2 must be zero, i.e. c must be zero. This yields

( ) ( ) - 1 ad - ac- bc a2 1 ad- 1 (ad- bc) - c2 ac+ ad - bc = 1

Finally we can see that for A to be in the normalizer of P3, ad-1 must be 1 or 2, since an element of the normalizer will permute the non-identity elements of P3. Furthermore, b is free to be anything in F3. This implies that

| | | | {( 1 b) || } ⋃ { ( 2 b )|| } ⋃ { ( 1 b ) || } ⋃ { ( 2 b )|| } NGL2 (F3)(P3) = 1 |b ∈ F3 2 |b ∈ F3 2 |b ∈ F3 1 |b ∈ F3

or for brevity’s sake:

| {( u b )|| ×} NGL2 (F3)(P3) = v |b ∈ F3, u,v ∈ F3

This means that the size of the normalizer of P3 is 2(2)(3) = 12. By Sylow’s Theorem’s, we know the number of 3-Sylow subgroups of GL2(F) to be

1.
congruent mod 3
2.
divide 16
3.
is equal the index of the normalizer of the 3-Sylow subgroup

These come by way of the fact that #GL2(F) = 48 and the size of a 3-Sylow subgroup in the case being 3. Hence the number of 3-Sylow subgroups is four as it is the only value satisfying the above three properties.

(d) Extra Credit: Find, explicitly, a 2-Sylow subgroup.

One such subgroup is the subgroup generated by
( 1 1 ) ( 1 ) 1 2 and 1

The entire subgroup is:

⟨( ) ( )⟩ { ( ) ( ) ( ) ( ) 1 1 , 1 = 1 , 1 , 1 1 , 2 , 1 2 1 ( 1 ) ( 1 ) ( 1 2 ) ( 2 ) 2 2 1 1 2 2 2 2 1 , 2 1 , 2 , 1 2 , ( 1 2 ) ( 2 1 ) ( 2 ) ( 1 ) 1 1 , 2 2 , 1 , 2 , ( ) ( ) ( ) ( ) } 1 2 , 2 , 2 1 , 1 2 2 1 1 1 2
This 2-Sylow subgroup was discovered via trial/error and intuition with the aide of the following sage [S+13] functions.
  IDENTITY = matrix.identity(GF(3), 2)  
 
  def translations(x, group):  
    """ Generates all possible translations of group by x """  
    unflattened_translations = []  
    for g in group:  
      xg = x*g  
      gx = g*x  
      if gx == xg:  
        unflattened_translations.append(gx)  
      else:  
        unflattened_translations.append((gx,xg))  
    return flatten(unflattened_translations)  
 
  def cyclic_group(x,y):  
    """  
    Generates a list of elements contained within the cyclic group generated by x  
    and y.  
    """  
    group = [IDENTITY]  
    stack = [x,y]  
    while len(stack) != 0:  
      m = stack.pop()  
      translations_by_m = translations(m, group)  
      for new_g in translations_by_m:  
        if new_g not in group:  
          group.append(new_g)  
          stack.append(new_g)  
    return group

(e) Extra Credit

(f)

References

[S+13]   W. A. Stein et al. Sage Mathematics Software (Version 5.11). The Sage Development Team, 2013. http://www.sagemath.org.