Today was the day of the second algebra exam. I had thought that I was pretty well prepared for the exam going in, but I got in there and kind of panicked when the exam was pretty much nothing like I expected. I was pretty certain there would be something about quotient groups, but I was dead wrong there. Most of the test was pretty much just applications of Lagrange’s theorem and Lagrange’s obit-stabilizer theorem. All in all I think the test was a little bit easier than I had expected.
One of the problems I think I didn’t get full credit on defined G as a group of order 10, H a subgroup of order 5, and K a subgroup of order 2. The part of that I probably messed up was proving the intersection of H and K was the group containing just identity. He gave a hint to use Lagrange’s theorem which I promptly ignored. I just used the fact that the order of H was prime thus H is cyclic and I wrote out the elements of H and showed there was no element of order 2. Then I used the fact that the order of K was even to show that non-identity element of K had to be of order 2. Nothing really wrong with the argument I guess, but given the hint I think he wanted some sort of proof that a group of prime order was cyclic. At least I can’t see another way to apply Lagrange to this problem.
The problem that gave me the most trouble was one about the permutation group on 4 letters. It defined 2 elements of H a subgroup of the permutation group: (124) and (13) and asked what could be the order of H. In the end the idea wasn’t that difficult, but the lack of information was kind of scary when I was taking the test. The basic trick was the H-orbit of 3 had cardinality of 4 and identity, (124), and (124)(124) all fixed 3. So the order of the 3 stabilizer of H was at least 3, thus the order of H was at least 12 and from Lagrange the order of H must either be 12 or 24 since the order of the permutation group on 4 letters is 24.