Let $G$ be a closed and connected semisimple subgroup of the Euclidean group $E(n)$ (the group of isometries of $\mathbb R^n$).
Can we prove that $G$ is conjugate to a subgroup of $O(n)$?
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$\begingroup$ I think you can do it by noting that we have the quotient homomorphism $q: E(n)\to\mathrm{O}(n)$, and homomorphic images of semisimple Lie algebras are semisimple. $\endgroup$– Robin GoodfellowCommented Aug 12, 2020 at 17:32
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2$\begingroup$ More directly, use the fact that $G$ is semi-simple. The restriction of the natural (linear)representation of $E(n)$ on $\mathbb{R}^{n+1}$ preserves the subrepresentation $\mathbb{R}^n$ and hence, by Weyl's theorem, it must be the direct sum of $\mathbb{R}^n$ with a $1$-dimensional complimentary subrepresentation $L\subset\mathbb{R}^{n+1}$. This subrepresentation determines a fixed point of $G$ acting on $\mathbb{R}^n$. Hence $G$ is conjugate to a subgroup of $\mathrm{O}(n)$. $\endgroup$– Robert BryantCommented Aug 12, 2020 at 19:47
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