A finite group that is not a symmetry group 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: Being a finite subgroup of $\textrm{Isom}(\mathbb{R}^2)$ is a very special property. First I will give as simple example of a finite group which is not such a subgroup, and then bellow describe all finite groups of "symmetries" as you have defined them.
Consider $\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2$ with generators $a$, $b$, $c$. Each generator must act as an isometry of order 2, but there is a very explicit classification of isometries of $\mathbb{R}^2$: they must be

*

*the identity (order 1)

*a reflection (order 2)

*rotation (order 2 only if a rotation by $\pi$)

*translation (infinite order)

*glide reflection (infinite order)

So $a$, $b$, and $c$ must be reflections, or rotations by $\pi$. We also know that these generators must commute. Two reflections commute if and only if the mirror lines intersect at right-angles. Two rotations by $\pi$ commute if and only if they are equal (otherwise their product is a translation along the line through their centres of rotation). Finally a reflection and a rotation by $\pi$ commute if and only if the mirror line passes through the centre of rotation (otherwise their product is a glide reflection along the line orthogonal to the mirror through the centre of rotation).
This doesn't leave many options. We could let $a$ and $b$ be reflections in intersecting orthogonal mirrors, and $c$ be the rotation by $\pi$  centred on this intersection point, but then $c=ab$, and $\langle a,b,c\rangle\cong\mathbb{Z_2}\times\mathbb{Z_2}$, which is not the group we were after. It follows no group of isometries of $\mathbb{R}^2$ is isomorphic to $\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2$.

The general picture
If $G$ is finite, then it globally fixes a point in the plane. To see this, let $x\in\mathbb{R}^2$ be any point, and let $y$ be the "centre of mass" of the orbit $Gx$ which is compact since $G$ is finite. Then $G$ fixes $y$.
Groups of isometries which fix a point contain only rotations about that point, and reflections in mirrors through that point. It follows easily that the only finite groups of isometries of the plane are

*

*the trivial group, which is the group of symmetries of, for example, the shape "F" drawn in the plane

*the cyclic group $\mathbb{Z}_n$, which is the symmetry group of a ratchet wheel with $n$ teeth

*the dihedral group $\mathbb{D_n}$, which is the symmetry group of a regular $n$-gon.

Thus, these are the only finite groups of symmetries of $\mathbb{R}^2$, and every finite subgroup of $\textrm{Isom}(\mathbb{R}^2)$ is the group of symmetries of a set $S\subset\mathbb{R}^2$.
A: 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!
