Quimby writes: What are the full names of your team members?

- Carlos Valadez

- Joaquin

- Martha

- Jomarie

- Wilbur

- Herbert Huang

Section 6 #46

Proof: Let ab have order n.  Then the cardinality of <ab> is n.  Let f: <ab> → f[<ab>] such that

f((ab)ⁱ) = (ab)ⁱa.  This function is onto because of the way that it has been defined.  It is also

one-to-one since (ab)^ja = (ab)^ka implies (ab)^j = (ab)^k by the cancellation property of groups.  Let

g: f[<ab>] → g[f[<ab>]] such that g((ab)^la) = a^-1(ab)^la.  This function onto and one-to-one

by analogous reasoning.  It is easy to verify that (ba)^p = a^-1(ab)^pa.  But this implies that

g[f [<ab>]] = <ba>.  Notice that g◦f is a bijection between <ab> and g[f [<ab>]] = <ba>.  We may

therefore conclude that <ab> and <ba> have the same cardinality//

Quimby writes: This is very clever and well-written, but still needs a little work.  I like the idea of defining a bijection between <ab> and <ba>.  I'm not sure, but I think that with a small change, it should work out.  The problem is that you cannot define a function by "Let f: <ab> → f[<ab>] such that f((ab)ⁱ) = (ab)ⁱa." since when you write "f[<ab>]" it implies that f has already been defined.  Instead, you might fix this in the following way.  Let S be the set of all xa in G such that x is in <ab>.  Define f from <ab> to S exactly as before.  Make similar changes for the function g.  Also, I know what you mean when you say "It is easy to verify that (ba)^p = a^-1(ab)^pa."  You should spell this out, though, so that I know that you know what you mean.

Section 6 #55

Proof: Suppose that p is prime and that Z modulo p has a proper nontrivial subgroup--call it H.  We

know by the definition of proper nontrivial subgroups that 1 < |H| < p.  Also, H must be cyclic since

Z modulo p is cyclic.  It follows from theorem 6.14 that there exists a positive integer d ≠ 1 such that

|H| = p/d, implying that p is not prime: in contradiction with our intial assumption.  It must therefore

be the case that Z modulo p has no proper nontrivial subgroups//

 Quimby writes: Good first draft---you got all the main ideas!  You can omit the "--call it."  Fix spelling of "initial."  Change the colon to a comma.  You next draft should be final.

Section 8 #46

Proof: Consider the permutations

   1  2  3  …  n-1   n

                                     σ =

                                           n  1  2  …  n-2  n-1

   1    2     3  …  n-1   n

     τ =

   n  n-1  n-2  …  2     1

Multiplying we obtain

   1      2      3  …  n-1   n

                                  στ =

  n-1  n-2  n-3  …    1    n

   1     2     3    …  n-1   n

                                   τσ =

   1     n    n-1  …   3     2

Notice that στ ≠ τσ.  This implies that S_n is nonabelian//

Quimby writes: Excellent!  Now fill in the details.  Justify your claim that στ ≠ τσ.  At some point, you will need to use the fact that n is at least 3.  Where?

Section 8 #47

??????

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.

Section 8 #52

Proof: The set {ρ_a | a ε G} is a group: associativity follows from the 

associativity of function composition, the identity is ρ_e, and the inverse

of ρ_a is ρ_a-1.  Let x, y ε G.  Let φ: G → {ρ_a | a ε G} such that φ(a) = ρ_a. 

Then φ(xy) = ρ_xy =  ρ_x ρ_y = φ(x) φ(y).  The groups {ρ_a | a ε G} and

G are therefore isomorphic//

Quimby writes: You claim that {ρ_a | a ε G} is a group.  Careful!  What is the group law?  Justify that it is closed under the group law.  Ordinary function composition might not work.  I'm not sure, but you might have to defineρ_a*ρ_b by ρ_b composed with ρ_a.

Assigment:

Let H be the subgroup of S_4 whose elements are {e, (1,2), (3,4), (1,2)(3,4), (1,3,2,4), (1,4,2,3), (1,4)(2,3), (1,3)(2,4)}. Then H is isomorphic to one of the five groups on the handout.

Step one: Determine which one H is isomorphic to.

Step two: Prove it.

Step three: Prove that H is not isomorphic to any of the others.

By constructing the table for "H," we see that H is isomorphic to either Q_8 or D_8.

--> We know this because Q_8 and D_8 are non-abelian. Checking the Cayley table for C_8, C_2 x C_4, and C_2xC_2xC_2, we see that these three are abelian.

(that's all I've got so far...)

Quimby writes: You do need to prove a lemma here.  Specifically, prove that if A is isomorphic to B and A is abelian, then B is abelian.