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I want to prove the following. Let $G$ be a compact connected Lie group with Lie algebra $\mathfrak g$ and center $Z_G.$ It is not hard to prove that $\mathfrak g$ is reductive. Therefore, we can decompose $\mathfrak g=Z_{\mathfrak g}\oplus[\mathfrak g,\mathfrak g]$ as direct sum of ideals where $[\mathfrak g,\mathfrak g]$ is semisimple. Let $G_{\mathrm{ss}}$ be the analytic subgroup of $G$ with Lie algebra $[\mathfrak g,\mathfrak g].$ Then we have $G=(Z_G)_0G_{\mathrm{ss}}$ where $(Z_G)_0$ is the identity component of $Z_G.$

I am reading "Lie Groups Beyond Introduction" by Knapp where the following argument is given. Choose covering groups $\widetilde{G_{\mathrm{ss}}}$ of $G_{\mathrm{ss}}$ and $\widetilde{(Z_G)_0}$ of $(Z_G)_0$ respectively. Then $\widetilde{G_{\mathrm{ss}}}\times\widetilde{(Z_G)_0}$ is a covering group of $G_{\mathrm{ss}}\times Z_G$ with Lie algebra $Z_{\mathfrak g}\times [\mathfrak g,\mathfrak g]$ which is isomorphic to $\mathfrak g.$ Therefore, we must have $\widetilde{G_{\mathrm{ss}}}\times\widetilde{(Z_G)_0}$ is a covering group of $G.$ I understand till this point. But it seems that from this one can prove at once what I have proposed. How is it so?

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    $\begingroup$ You might argue along the lines that $\widetilde{G_{ss}}\times \widetilde{(Z_G)_0}$ is a covering group of $(Z_G)_0 G_{ss}$, that $(Z_G)_0 G_{ss}\subset G$, and $G$ cannot be any bigger than $(Z_G)_0 G_{ss}$ and still have $\widetilde{G_{ss}}\times \widetilde{(Z_G)_0}$ as a covering group $\endgroup$
    – Ryan Hendricks
    Commented Jun 8, 2020 at 0:56
  • $\begingroup$ @fierydemon. I think much simpler argument avoiding covering groups exists. Consider the map $(g,h)\mapsto gh$ from $(Z_G)_0\times G_{ss}$ to $(Z_G)_0G_{ss}.$ The derivative of this map is clearly onto and hence an isomorphism as the corresponding Lie algebras have the same dimension. Thus the map is local diffeomorphism. As the exponential map is a local diffeomorphism and $G$ is connected. The map $(g,h)\mapsto gh$ is surjective. $\endgroup$
    – A beginner mathmatician
    Commented Jun 11, 2020 at 7:39

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