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Sorry to inundate the feed with a question quite similar to my last, but again I've been drawing pictures for quite a while with little success. Does anyone have any idea how to represent the product group $\textrm{SO}(2)\times \mathbb{Z}_2$ by a solid (corresponding to rotations only)?

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    @p.s. Problem is that you can still rotate that about an orthogonal plane, thus screwing up commutativity again...2011-12-02

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As the comment by p.s. mentioned, your first concern is to find the subgroup isomorphic to $\mathrm{SO}_2\times\mathbb{Z}/2$, and there aren't any inside $\mathrm{SO}_3$: any subgroup of $\mathrm{SO}_3$ isomorphic to $\mathrm{SO}_2$ is the set of rotations around a fixed axis, and there are no other rotations that commute with all of them. You can find an appropriate subgroup in $\mathrm{O}_3$ (add the reflection in the plane perpendicular to the axis), but I don't think it is the subgroup of $\mathrm{O}_3$ defined by any solid (because a solid fixed by all rotations will automatically also be symmetric with respect to reflection in a plane containing the axis). This argument suggests that you won't even have much luck either if your solid is situated in a higher-dimensional space (in which case the $\mathrm{O}_3$ of the $3$-space containing you solid becomes the image by restriction of a subgroup of $\mathrm{SO}_n$).