Lie algebra of abelian Lie group is compact?

Question

Is it true that the Lie algebra of and abelian group is compact? For a compact Lie algebra I mean a Lie algebra of some compact Lie group.

Answer 1

Connected abelian Lie groups can be classified in that all of them are $\mathbb{R}^a \times (S^1)^b$ for some numbers $a$ and $b$. (If you want to prove this on your own, it helps to investigate the exponential map from the Lie algebra to the Lie group more closely.) Now, do you know the Lie algebra of this Lie group? Do you know of a compact Lie group with that same Lie algebra?

Edit: As Eric Wofsey pointed out my answer is more roundabout than is really needed. If $G$ is an abelian group, then the Lie algebra $\mathfrak{g}$ is also abelian in the sense that the commutator bracket is trivial. So your $\mathfrak{g}$ is just some $\mathbb{R}^N$ with trivial bracket and now the question reduces to finding a compact (abelian) Lie group with that particular Lie algebra. Do you see what it must be?