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If $G$ is a (edit: simply connected)Lie group, when does a direct sum decomposition of its Lie algebra (into a direct sum of subalgebras) correspond to a (semi)direct product decomposition of $G$? Does it suffice for one of the summands of the Lie algebra to be the center of the algebra? I (think I) have verified that if $H$ and $N$ are the subgroups corresponding to the summands and $N$ is normal then $H$ and $N$ generate $G$ (by multiplying the images of the two summands under the exponential map, then applying the inverse function theorem and the fact that $G$ is generated by any neighborhood of the identity). But I have trouble verifying that $H \cap N = \{e\}$. Clearly more conditions are required since the two torus furnishes easy examples of such $H$ and $N$ with nontrivial intersection. As I mentioned I am particularly interested in the case when one of the subalgebras is the center of the Lie algebra of $G$.

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The Lie algebra does not see the topology of the group at large, so I doubt you can say much in general—if the Lie group were to be simply connected, maybe... – Mariano Suárez-Alvarez Jan 26 '12 at 3:47
@Mariano Suarez-Alvarez: Thanks! let's say it's simply connected. – zzz Jan 26 '12 at 4:02
I see. If $G$ is simply connected and $\mathfrak{g}= \mathfrak{h}\oplus \mathfrak{n}$ then this isomorphism induces a homomorphism $\phi : G \rightarrow H\times N$. then composing with multiplication $H\times N \rightarrow G$ gives a left inverse for $\phi$ (again because $G$ is simply connected and the derivative of the composition is the identity on $\mathfrak{g}$) But how to show it's also right inverse... – zzz Jan 26 '12 at 7:12
oh $\phi$ is surjective because its image contains a neighborhood of the identity by the inverse function theorem. So we're done... I think. – zzz Jan 26 '12 at 7:27

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