The class performance on the final examination ranged from a low of 40 to a high of 146 on an examination with 150 maximum points (not including 15 bonus points available). The mean score of 96 (64%) was almost exactly the same as that for Exam 2, but seven students improved enough on the final to take advantage of the opportunity to improve midterm exam scores in the final grade calculation. For students whose percentage scores on Parts 1 and 2 of the final exceeded the mean of their two midterm examinations, the latter two grades were replaced by the average of the sum of the two midterm grades and the percent score on Parts 1 and 2 of the final exam.
Final course grades are available by clicking on the link Final Course Grades.
Late-breaking news about a study of the effects of cramming, as reported by the Hartford Courant Wednesday, May 15. Get a good night's sleep before the final exam!
Specially selected Review Problems for Final Exam preparation are now available.
Final Exam is scheduled for Thursday, May 17, from 8:00 to 10:00 AM in MSB 307.
Consultation hours prior to final exam, all in MSB 218*.
*In the event of a large turnout, the location will move to a classroom, and a note will be posted on the door of MSB 218 to specify which one.
The Exciting Summer Opportunities page has a new listing from People's Bank.
Two interactive Maple notebooks have been added to the course folder in MSB 203: Groups and Cosets and Normal Subgroups.
Note: The Course Information Sheet now shows the correct normal office hours.
Exercises 7 and 11(a), Section 8: the calculations in class for S3 enable you to show that S3 is not abelian and to determine its center. In doing the latter, it may be helpful to construct the Cayley multiplication table of S3.
Next, note that every element of S3, such as f = (1 2 3), is also in every Sn for n > 3: for those cases, such elements fix everything other than 1, 2 and 3. What do you need to do to show that Sn is not abelian?
Exercise 19, Section 5. Divide the problem into two pieces: (a) Show that xr and xs generate different subgroups unless r = s. How? Well, note that if they coincided, you could without loss take r and s to be positive, and you would have both xr in the subgroup generated by xs and xs in the subgroup generated by xr. Use that to show that r = s.
(b) Every subgroup H of an infinite cyclic group generated by x is generated by xr for some positive integer r. If H is the identity subgroup, then of course that's fine. Otherwise, H must have some proper powers of x in it. You can then show that it must have some POSITIVE powers of x in it. Consider the set of all such positive integer powers, and ask yourself what do we usually do next when a set of positive integers is not empty?
Exercise 21, Section 5. Let G be a cyclic group of order n. Find a condition on the integers r and s that is equivalent to the subgroup generated by xr being a subgroup of the subgroup generated by xs.
Comments. First, it is easy to dispose of the case r = 0. So suppose without losing any generality that 0 < r < n. The subgroup generated by xr will be contained in the subgroup generated by xs if and only if xr itself belongs to the subgroup generated by xs. That in turn is equivalent to xr = xls for some integer l chosen so that 0 < ls < n. Convert the last equation to a form for which Theorem 4.4(ii) is helpful.
Last updated by J. Hurley 5/16/2001.