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Let M is a Riemannian manifold and $\tilde{M}$ is its universal covering and G is connected subgroup of the isometries of M . I know that there is a covering group $\tilde{G}$ such that acts on $\tilde{M}$ . My question is " is $\tilde{G}$ simply connected?

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Not necessarily, consider $M=\mathbb{R}^n$ endowed with the Euclidean metric. Its group of isometries is $O(n)$ which is not simply connected, but $\mathbb{R}^n$ is simply connected.

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