Diana Amador, Dolores Lopez, Sandra Alvarez, Dulce Valencia, and Cory Barr

Section 6  #46 and  55

 

 Section 6 #46 (Third Draft)

 

Let a and b elements of a group G. Show that if ab has finite order n, then ba also has order n.

 

Proof:

       Given a and b elements of G and ab has finite order n; (ab)^n = e

       Need to show n is the smallest positive integer

       such that (ba)^n = e.

 

      First show (ba)^n=e.

 

       Given (ab)^n = e; multiply b to the right and a to the left to both sides.

       implies b[(ab)^n]a = bea

       implies [(ba)^n+1] = ba

           ==> [(ba)^n](ba) = ba

           ==> [(ba)^n](ba)[(ba)^-1] = (ba)(ba)^-1

       implies [(ba)^n](e) = e

       implies [(ba)^n] = e

       Therefore (ba)^n = e

 

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Quimby writes: Good!

---

 

      Now show that there is no positive integer m

      such that m<n and (ba)^m = e.

 

      Suppose m is an element of the positive integers

      and m < n and (ba)^m = e.

      then (ba)^m = (ab)^m, since (ab)^n = (ba)^n

      implies  (ab)^m = e , since (ba)^m = e

      implies  (ab)^m = e = (ab)^n, since (ab)^n = e

      implies  (ab)^m = (ab)^n = e,

      implies  (ba)^m = (ba)^n = e, since (ab)^n = (ba)^n and (ab)^m = (ba)^m

      implies m = n ==>contradiction since we let m be lest that n

    

      Therefore n is the smallest interger

      Therefore ba has order n.///

 

 

Section 6 # 55 (THIRD DRAFT)

Show that Zp has no proper nontrivial subgroups if p is a prime number.

Proof:

        Suppose H is a subgroup of Z_p where p is prime and H is proper and nontrivial. By Thm. 6.6, H is cyclic because H is a subgroup of Z_p, a cyclic group. Then, H = <b> for some b in Z_p, by definiton of a cyclic subgroup. By Thm 6.14, H contains p/d elements where d is gcd(p,b).

Now, we know that d can only be p or 1, since p is prime.Then if d is p, p/p is 1. Hence, H contains one element which is the identity element and H would be the trivial subgroup of Zp by the definition of the trivial subgroup. But if d is 1, p/1 is p. In this case H has p elements, which would then be the improper subgroup of Z_p by the definition of the improper subgroup.But this contradicts our assumption of p being a prime number, then Z_p has no proper and nontrivial subgroup. Q.E.D.

 

(Quimby writes: Good!  The last sentence is still a run-on, but let's call this a final draft.)

 

Professor out of curiosity how should the last sentece look? I tried my best to fix it but I am a bad writer, Diana.

 

(Quimby writes: No worries---this class is almost as much about learning some tips for good writing as it is about learning mathematics.  It should be "This contradicts our assumption of p being a prime number.  Therefore Z_p has no proper and nontrivial subgroup."  Each of these is a complete sentence, with a subject, verb, and object.  That's why you can't smush them both into a single sentence.)

 

Section 8 #46, 47 and 52

 

#46

 

Show that Sn is a nonabelian group for n> or =3.

 

case 1: n=3

     Sn has 3! elements.

     let p1=( 1 2 3 ) and σ1=( 1 2 3 )

              ( 2 3 1 )           ( 1 3 2 )

 

     p1σ1=( 1 2 3 )( 1 2 3 ) = ( 1 2 3 )            σ1p1=( 1 2 3 )( 1 2 3 ) = ( 1 2 3 )       

             ( 2 3 1 )( 1 3 2 )    ( 2 1 3 )                   ( 1 3 2 )( 2 3 1 )    ( 3 2 1 )

 

      p1σ1 is not = to σ1p1

S3 is not cummalative. S3 is nonabelian.

 

case 2: n=4, 4>3

     Sn has 4! elements.

     let p1=( 1 2 3 4 ) and σ1=( 1 2 3 4 )

              ( 2 3 4 1 )           ( 1 2 4 3 )

 

     p1σ1=( 1 2 3 4 )( 1 2 3 4 ) = ( 1 2 3 4 )            σ1p1=( 1 2 3 4 )( 1 2 3 4 ) = ( 1 2 3 4 )       

             ( 2 3 4 1 )( 1 2 4 3 )    ( 2 3 1 4 )                   ( 1 2 4 3 )( 2 3 4 1 )    ( 2 4 3 1 )

 

      p1σ1 is not = to σ1p1

S4 is not cummalative. S4 is nonabelian.

Therefore, Sn is a nonabelian group for n> or =3.

 

 

Quimby writes: You have shown that S_n is nonabelian if n=3 or n=4.  What about n=5, 6, 7, 8, . . . ?  Here's a hint: Let f such that f(1)=2, f(2)=1, and f(x)=x if x>2.  (f switches 1 and 2.)  This f makes sense for any n greater than or equal to 3.  Now find g such that fg does not equal gf.

 

Click here for Section 8, number 46, draft 2.

 

 

 

#47

 

Show that if n > or = 3 then the only element of σ of Sn satisfying σγ = γσ for all γ is

 

σ = ι, the identity permutation.

 

 

 

Proof:

 

           Let  Ïƒ and γ  be elements of Sn

 

         and σ does not equal ι

 

         then since Sn is not abelian   σγ  does not equal γσ

 

 

 

Quimby writes: No!  The fact that S_n is not abelian means that there exist some σ and γ in S_n such that σγ does not equal γσ.  It does not mean that you can pick any old σ and γ at random and conclude that σγ does not equal γσ.  For example, if σ=γ, then this is false.  So, back to the drawing board on this one.  See below for a hint.

 

 

         however since it is a group there must exist an e*γ  = γ*e = γ  where e is an element of Sn

 

           therefore the only element that satisfies σγ = γσ for all γ is  Î¹ the identity permutation

 

Quimby writes: You have to show that if sg=gs for all g in G, then s=i.  Here's a hint:  Assume temporarily that s does not equal i.  Let's find g such that sg does not equal gs.  That will give a contradiction.  Since s does not equal i, there exists m such that s(m) does not equal m.  Divide into two cases.  Case 1: s(s(m)) does not equal m.  In this case, define g by g(m)=s(m), g(s(m))=m, and g(x)=x otherwise.  Then show that sg does not equal gs.  Case 2: m=s(s(m)).  I'll let you find g in this case.  You might work with the person doing #46 on this one.

 

 

 

 

 

#52  

Let G be a group.  Prove that the permutations r_a : G ® G, where r_a (x) = xa  for a ÃŽ G and x ÃŽ G, do form a group isomorphic to G.

 

 

Quimby writes: I'm confused.  What's "®"?  (I think it's "->".)  What's "ÃŽ"?  (I think it's "in.")  Don't just copy and paste---check to make sure it's coming out OK!

 

Proof:   We need to show that G is isomorphic to G.

 

 

 

Quimby writes: No!  Let S = {r_a | a in G}.  Find a group law on S that makes S into a group.  (Tricky point: ordinary function composition might not work.  I'm not sure, but you might have to define r_a * r_b as r_b composed with r_a for this to work.)  Then show that G is isomorphic to S.

 

            Let r_a : G ® G where r_a (x) = xa  for a,x ÃŽ G

            Now let f : G ® r_a  such that f(xy) = f(x) f(y) for all x, y ÃŽ G such that

 

 

Quimby writes: r_a is a function, not a set.  So you can't define a function from G to r_a.  Also, you can't define a function to have the property "f(xy) = f(x) f(y)."  You have to define the function f, then prove that "f(xy) = f(x) f(y)."  You might try the following.  Define f : G -> S by f(a)=r_a.  Then prove that "f(xy) = f(x) f(y)."

  f(a) = r_a  then  f(xy) = r_xy = r_xr_y = f(x) f(y) by Lemma 8.15

 

            So r_a (a^-1c) = a(a^-1c) = c for all c ÃŽ G and this implies r_a maps G onto G.

Now if we let r_a (x) = r_a (y), then ax = ay so x = y by cancellation.  So r_a is also a 1 – 1 and is a permutation of G.

 

Quimby writes:  Oops!  I think you're copying a little too closely from the proof of Cayley's Thm.  Remember, r_a (x)=xa, not ax.

 

 

Now we need to show that f is a 1 – 1.  So now lets assume that f(x) = f(y), then  r_x = r_y.  Then r_x(e) = r_y(e) so we have xe = ye Þ x = y. 

\ f is 1 – 1 .

 Now we need to show that f(xy) = f(x) f(y) where r_xy = r_xr_y.  Now we have r_xy(g) = (xy)g for any g ÃŽ G so, (r_xr_y)(g) = r_x(r_y(g)) = r_x(yg) = x(yg).  \  r_xy = r_xr_y .

 

Quimby writes: Try a second draft!  You've got the main idea here, that this is just like the proof of Cayley's Thm.  You just need to fix some of the details.

 

  -------------------------------------------------------------------------

 

Quimby writes: Steps 1 and 2 below look good!  You've still got some copy-and-paste issues.  If you're wrting in Word, you're welcome to just e-mail me the file.

 

 

H = { ι, (1,2), (3,4), (1,4,2,3), (1,3,2,4), (1,2)(3,4), (1,3)(2,4), (1,4)(2,3)}

Step 1 Cases:

Claim: H ≈ D₈

Step 2 Cases:

Proof:  Show H is isomorphic to D₈

Show that $_m : H ® D₈ such that f is a homomorphism and f is 1 – 1 and onto.

a = m( ι )

b = m(1,3,2,4)

c = m((1,2) (3,4))

d = m(1,4,2,3)

f = m(1,2)

g = m((1,3)(2,4))

h = m(3,4)

i = m(1,4)(2,3)

 

Quimby writes: Rewrite this as follows:

 

Define a function m : H -> D_8 by

 

m(i) = a

m( (1,3,2,4) ) = b

.

.

.

 

 

 

By inspection m is 1 – 1 and onto. 

Now show m is a homomorphism

Need to show m(xy) = m(x)m(y) "x, y ÃŽ H

Case a:

x = ι, y = ι

m( ι · ι ) = a

m( ι ) m( ι ) = a · a = a

  

x = ι, y = (1,3,2,4)

m(( ι ) · (1,3,2,4)) = b

m( ι ) m( 1,3,2,4 ) = a · b = b

 

 x = ι, y = ((1,2)(3,4))

m(( ι ) · (1,2)(3,4)) =  c

m( ι ) m((1,2)(3,4)) = a · c = c

  

x = ι, y = ((1,4,2,3))

m(( ι ) · (1,4,2,3)) = d

m( ι ) m(1,4,2,3) = a · d = d

 

x = ι, y = (1,2)

m(( ι ) · (1,2)) =  f

m( ι ) m((1,2)) = a · f = f

 

x = ι, y = (1,3)(2,4)

m(( ι ) · (1,3)(2,4)) =  g

m( ι ) m((1,3)(2,4)) = a · g = g

 

x = ι, y = (3,4)

m(( ι ) · ( 3,4)) =  h

m( ι ) m(3,4) = a · h = h

 

x = ι, y = (1,4)(2,3)

m(( ι ) · (1,4)(2,3)) =  i

m( ι ) m((1,4)(2,3)) = a · i = i

 

 

Quimby writes: Can you consolidate some of these cases?

 

 

Quimby writes: I seem to have inadvertently deleted some of your cases, but you can still see them by clicking on the "history" link below and going to a previous version of this page (The Feb. 9 version at 8:09 p.m.).  They looked good!  Your function works out.

 

Step 3

(Prove H is not isomorphic to any other group of 8 elements.)

 

Quimby writes: OK, as for the third part, how did you know that H was not isomorphic to C_8?  Perhaps you noticed that H is nonabelian, whereas C_8 is abelian.  In that case, prove the following lemma: If A is a nonabelian group and B is an abelian group, then A is not isomorphic to B.

 

Cory writes: As for part 3, I do not see the logical flaw or why the proof does not work.  As for H and C_8 specifically, if two groups are isomorphic, then the cardinality of the set of elements along the main diagonal must be equal.  (p->q)  As the proof shows, this is not the case for H and C_8.  (~q->~p).  Stated another way, if I were to code an algorithm to check isomorphism between two groups, as an initial step I could examine the multiplicities of the multiset of elements along the main diagonals and eliminate some non-isomorphic sets with 100% accuracy--and in just O(n) time.  For the groups in this problem, all are eliminated with no need to check to see if cyclic, abelian, etc.

 

Quimby writes: Ah!  I had not seen that "Step 3" was a link to a pdf.  I thought your team was stuck and needed a hint; hence my comment above.  The idea and intuition behind your proof is perfect.  What needs some touching up is the way you've written it.  Here are my comments on your Step 3:

 

- I'm a bit confused what you mean by "Let H_s be {s | there exists h in H and s=h^2 }".  There is no s in H.  Which element of H did you mean?  Actually, there is no s in C_8 either.  Did you mean the identity of each group?

 

- Once you clear that up, here's my main comment: Justify your claim that | H_s | = | (C_8)_s |.  At some point, you will need to temporarily assume that there exists an isomorphism f from H to C_8.  Then use the function f to justify this claim.

 

- Change "Let H_s be {s | there exists h in H and s=h^2 }" to "Let H_s be {s | there exists h in H such that s=h^2 }".

 

- Well, you learn something new every day!  I didn't know that a multiset was also called a bag.

 

- OK, let's see a 2nd draft of Step 3 . . .

 

OK... Click here for v 2.0