Let P denote the direct product of countably many copies of the integers Z.

E.g., the set {f: Z+ --> Z | f is a function} under addition of functions.

Puzzle: Is P a free abelian group? Prove your answer.

P is not free abelian. i learned this many years ago, if memory serves, in a class of t. y. lam. i have been trying to remember the proof, which is much more likely than finding my notes. now i think i have it. i'm not sure this was the original proof that i saw, but it's probably close. any subgroup of a free abelian group is also free abelian, so it suffices to find a subgroup of P that is not free abelian. let A = { (a_n) in P | v(a_n) --> infinity as n --> infinity } where v(k) is the largest exponent of a power of 2 that divides k , v(0) = infinity . (here i've written elements of P as sequences, rather than functions.) we claim that A is not free abelian. if it were, then for any prime p , the group A/pA would be a vector space over the field Z/pZ , with the image of a basis of A being a basis of A/pA over Z/pZ . in particular, its dimension would be independent of p . first consider p = 2 . note that 2A = { (a_n) in P | v(a_n) --> infinity and each a_n is even } so there is an obvious countable basis for A/2A over Z/2Z , namely { e_i } , where e_i is 0 in all coordinates, except for a single 1 in the i-th coordinate. thus the dimension of A/2A over Z/2Z is countably infinite. now consider m = 3 . in this case, A/3A is uncountable: for any sequence (b_n) of 0's, 1's and 2's , we have the element (4^n b_n) in A , and all have different classes in A/3A . therefore the dimension of A/3A over Z/3Z is uncountable. this shows that A , and therefore, P is not free abelian. mike