Solutions to MATH 120b; Homework Set 4
February 3, 2010
1. Let M be an abelian group of order 1,728. As 1, 728 = 33 26 , M = A B with
|A| = 27 and |B | = 64, and M is determined up to isomorphism by t
Solutions to MATH 120b; Homework Set 3
January 27, 2010
1. This is almost 5.8, but not quite. We prove the rst isomorphism exists; the
proof that the second exists is essentially the same. Let N = M3
Solutions to MATH 120b; Homework Set 2
January 20, 2010
1. Assume Q is projective. Then by Theorem 3.4, Q is a submodule of a free module
M . Thus M = i Mi with Mi Z. Let x Q# ; then x = i xi with xi
Solutions to MATH 120b; Homework Set 1
January 13, 2010
1. We prove the result by induction on m; for m = n or n + 1, the lemma holds by
hypothesis, so the induction is anchored. Assume for m n +1 and