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

Jamen Chan

Marco Acuna

Martha Amaya

Marcia Japutra

Megan Wyrinski

 Cayley tables wiki problem

2.1.2007 wiki.doc

(Quimby writes: This looks really good!  Your function t (later named m) works out.  So you're done with steps 2 and 3.  (I'll give you a break on writing out the remaining cases in Step 2.  Here is your document with comments.)

note: we are confused about the third part.

(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.)

#46:   Let A & B in G. Let n be the finite order and be the smallest number in Z^+ such that (ab)^n = e

          Claim: (ba)^n=e

           (pf):  bababa.....ba= (bababa...ba)(e)            (by def of identity)

                                         = (bababa...ba)(bb^-1)    (by def of inverse)

                                         =  b(ababa...bab)b^-1      (by associativity)

                                         =  b(e)b^-1                       (by assumption)

                                         =  (be)b^-1                      (by associativity)

                                         =   bb^-1                          (by def of identity)

                                         =  e                                  (by def of inverse)

Quimby writes: Good first draft!  But you've still got some work to do.  What you've written does not yet show that ba has order n.  You must show that n is the smallest positive integer such that (ba)^n=e.  So far, you've shown only that n is some positive integer such that (ba)^n=e.

#55:  Let G=Zp, where p is prime. Let H be the subgroup of G.  (Quimby writes: Change "the" to "a".)

          Since G is generated by <1>, G is cyclic.

          By theorem 6.6, subgroup H is also cyclic.  (Quimby writes: don't need to write the word "subgroup.")

          Let n be the order of G, then n=/Zp/  (Quimby writes: why not just write "Let n=p"?)

          By Theorem 6.14, subgroup H contains n/d elements   (Quimby writes: What's d?  Introduce variables before using them.  Also, add a period.)

          Since n is a positive prime number and n/d in Z^+, d must be n or 1  (Add a period and the word "is.")

          When d=1, H=Zp => H is improper subgroup of G (by def of 5.5)  (Add the word "an."  Change "5.5" to the term whose def. you're using.)

          When d=n, H={0} =>H is trivial subgroup of G (by def of 5.5)  (Add the word "a."  Change "5.5" to the term whose def. you're using.)

          Therefore, Zp has no proper nontrivial subgroups.

Quimby writes: This is pretty good---you got all the main ideas.  Make the minor changes above for your next and probably final draft.

2nd draft:  section 6 #55.doc

(Quimby writes: I couldn't save this file.  Can you e-mail it to me?)

Section 8

#52

see attachment below

section 8 #52.doc

(Quimby writes: I couldn't save this file.  Can you e-mail it to me?)