What is the exact relationship between Lie groups and Lie algebras? I know it's not bijective because all commutative Lie groups have isomorphic Lie algebras.

1$\begingroup$ ... if they have the same dimension. $\endgroup$ – Torsten Schoeneberg Jan 20 '14 at 11:48
Up to isomorphism, there is one simply connected Lie group for every Lie algebra. Indeed, there is also a homomorphism of simply connected Lie groups for every homomorphism of the corresponding Lie algebras so one gets an equivalence of categories this way.
This pans out nicely in yr commutative example: the simply connected abelian groups are just the Lie algebras under addition. All other abelian Lie groups have a nonsimply connected torus component and look like $T^k\times\mathbb{R}^{nk}$.
To get the remaining Lie groups into the picture, recall that the universal cover of any Lie group is a simply connected Lie group.

3$\begingroup$ Warning: infinitedimensional Lie algebras may have no associated groups. $\endgroup$ – Allen Knutson Feb 5 '10 at 21:40
I'd suggest taking a look at "Lecture 23" of FultonHarris. In particular, given a complex Lie algebra, there's a unique connected, simply connected complex Lie group G with that Lie algebra. The other connected Lie groups with the same Lie algebra are quotients of G by discrete subgroups of its centre Z(G) (so G is the universal cover of these other groups). The representations of the Lie algebra are in correspondence with the representations of G. To see which representations of the Lie algebra are representations of one of the other Lie groups with the same Lie algebra, you just need to check which ones factor through the discrete subgroup of the centre that defines the other Lie group. If you're also interested in the real case, check out their "Lecture 26". Here the situation is more complicated.