# Q (the rationals) not free abelian yet iso to Z^2?

TopCat
So I just proved that Q (the rationals) is not free abelian since it must have an empty basis. But Q is isomorphic to Z^2 = ƩZ which is equivalent to Q having a nonempty basis. I must be wrong about the isomorphism but I don't see why.

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## Answers and Replies

Science Advisor
Homework Helper
there is a map ZxZ*-->Q but it isn't injective.

TopCat
D'oh! Thanks, mathwonk.