Is the Lie Algebra of a connected abelian group abelian? Is the Lie Algebra of a connected abelian group abelian? I guess that this should be true, but how do you prove it?
 A: Yes, and connectedness is not necessary. I know three proofs:
Proof 1
When $G$ is abelian, the inverse map
$$i:G\to G,\quad g\mapsto g^{-1}$$
is a group homomorphism. Hence, its differential at $1\in G$
$$di_1:\mathfrak{g}\to\mathfrak{g},\quad X\mapsto -X$$
is a Lie algebra homomorphism. But then
$$-[X,Y]=di_1([X,Y])=[di_1(X),di_1(Y)]=[-X,-Y]=[X,Y],$$
so $[X,Y]=0$.
Proof 2
For any Lie group $G$, the differential at $1$ of the map $\mathrm{Ad}:G\to GL(\mathfrak{g})$ is $\mathrm{ad}:\mathfrak{g}\to\mathrm{End}(\mathfrak{g})$ where $\mathrm{ad}(X)(Y)=[X,Y]$. But when $G$ is abelian, $\mathrm{Ad}$ is the constant map to the identity (since $\mathrm{Ad}(g)$ is the differential of the map $G\to G,a\mapsto gag^{-1}$ which is the identity when $G$ is abelian), so $\mathrm{ad}=0$.
Proof 3
For any Lie group $G$ we have that for $X,Y\in\mathfrak{g}$, 
$$\exp(sX)\exp(tY)=\exp(tY)\exp(sX),\forall s,t\in\mathbb{R}\quad\iff[X,Y]=0.$$
If $G$ is abelian, the left-hand side always hold, so $[X,Y]=0$ for all $X,Y\in\mathfrak{g}$.
Remark about the converse
The last proof can be used to prove the converse when $G$ is connected. This is because $\exp$ restricts to a diffeomorphism from a neighborhood of $0$ in $\mathfrak{g}$ to a neighborhood of the identity in $G$ and a connected group is generated by any neighborhood of the identity.
However, connectedness is necessary for the converse. For example, if $T$ is any abelian connected Lie group and $H$ is any non-abelian finite group, then $G=T\times H$ is a non-abelian Lie group with abelian Lie algebra.
