A symmetry is an isometry of $\mathbb{R}^2$. Given a subset $S$ of $\mathbb{R}^2$, the symmetry group $Sym(S)$ is the group of all symmetries that map $S$ to $S$. My question is, is there an example of a finite group which is not isomorphic to any symmetry group? Or, is every finite group isomorphic to some symmetry group?
A finite group that is not a symmetry group
group-theorysymmetry
Best Answer
In the comments a link was posted giving an $n$-dimensional construction, but in your question you ask for the group to be the group of symmetries of a 2-dimensional set. In this case, the claim is false, essentially because the group of isometries of $\mathbb{R}^2$ is too small.
Specifically, which elements of the isometry group have order 3? One can check these are all of the form 'rotation by $2\pi/3$ or $4\pi/3$ about some point,' from the classification of isometries of $\mathbb{R}^2$ (they're all translations, rotations, reflections, or glide reflections; one can easily check that none of those can have order 3 besides the rotations just mentioned).
Take $G = \mathbb{Z}/3\mathbb{Z} + \mathbb{Z}/3\mathbb{Z}.$ It has 9 elements, 8 of which have order 3. All elements commute. If we tried representing it as a subgroup of the group of isometries of $\mathbb{R}^2,$ we'd run into the following problem: Rotations about two different points don't commute!