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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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    oh $\phi$ is surjective because its image contains a neighborhood of the identity by the inverse function theorem. So we're done... I think.2012-01-26

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