# Connected Lie group, Homomorphism of exponential

G Lie connected group, show that exp: Lie(G) $\rightarrow$ G is an Homomorphism of group iff G is abelian.

I think I've made the $\Leftarrow$:

If G is abelian $\Rightarrow$ Lie(G) is abelian $\Rightarrow$ $[X,Y]=0$ $\Rightarrow$ thanks to the Backer-Hausdorf-Campbell formula $exp(X+Y)=exp(X)exp(Y)$ $\Rightarrow$ exp is an Homomorphism (since G is connected and therefore exp is surjective).

How can I see the $\Rightarrow$ implication?

Assume $\exp : \text{Lie}(G) \to G$ is a morphism. Then it's clear that elements in the image of $\exp$ commute, since $\exp(X) \exp(Y) = \exp(X + Y) = \exp(Y + X) = \exp(Y) \exp(X)$. Since $\exp$ is a local diffeo at $0$, it follows the identity $e \in G$ has an abelian neighborhood $U$ (i.e. one in which every two elements commute). It's well-known that, since $G$ is connected, any element of $G$ is a product of elements in $U$, and from this it follows that $G$ is abelian.