Consider the wreath product $G=\mathbb{Z}_2\wr O_n(\mathbb{R}),$ where $O_n(\mathbb{R})$ is the set of orthogonal groups over reals. Can we show $G$ embeds in a nice enough group (for example, some subgroup of $\mathbb{GL}_{2n}(\mathbb{R})$ perhaps ?).
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9$\begingroup$ A theorem of Burnside shows that a subgroup of finite exponent in $\mathrm{GL}(k,\mathbb{R})$ must be finite. Even the lamplighter group $\mathbb{Z}_2 \wr \mathbb{Z}$ is not a subgroup of $\mathrm{GL}(k,\mathbb{R})$. $\endgroup$– AGenevoisCommented Aug 27, 2020 at 5:26
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3$\begingroup$ AGenevois' comment answers the question in any reasonable interpretation, but I'm still not sure exactly what is being asked. What representation of $O_n(\mathbb{R})$ is used to define the wreath product? Why refer to the 'set' of orthogonal groups? $\endgroup$– Mark WildonCommented Aug 27, 2020 at 12:11
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