Can all groups be thought of as the symmetries of a geometrical object?

Question

It is often said that we can think of groups as the symmetries of some mathematical object. Usual examples involve geometrical objects, for instance we can think of $\mathbb{S}_3$ as the collection of all reflections and rotation symmetries of an equilateral triangle, similarly we can think of $D_8$ as the symmetry group of a square.

Cayley's Theorem along with the fact that the symmetry group of a regular $n$-simplex is isomorphic to $\mathbb{S}_{n+1}$ allows us to think of any finite group as a subset of the symmetry group of some geometrical object. Which brings me to the following questions:

1. Can every finite group be represented as the collection of all symmetries of a geometrical object? That is, are all finite groups isomorphic to some Symmetry group?

2. Can such a result (the representation of groups as distance-preserving transformations of some geometrical object) be extended to infinite groups? If so, how?

Thanks in advance (:

Answer 1

Let $G$ be a finite group of order $n>1$.

In $\Bbb R^n$ with standard basis $e_1,\ldots, e_n$, we construct a geometric object with trivial symmetry group: Let $X=\{\frac 1ke_k|1\le k\le n\}\cup \{0\}$. Then $0\in X$ is the only point with distance $\le 1$ to all other points, hence must remain fixed by any symmetry movement. After that, $\frac 1ke_k$ is the only point in $X$ at distance $\frac 1k$ to $0$, hence must also remain fixed.

By considering the action on itself by left multiplication, a finite group $G$ of order $n$ can be viewed as a subgroup of $\Bbb S_n$, and this acts on $\Bbb R^n$ by permuting coordinates, which is an orthogonal linear transformation, hence "geometric".

The point $p=(1,2,3,\ldots, n)$ is left fix only by the identity, hence its orbit $Gp$ is a geometric object on which $G$ acts freely. However, we rather consider the orbit $Y:=G(3p+X)$.

Let $\alpha$ be a symmetry movement of $Y$. The points $G\cdot 3p$ are distinguished by the fact that they have $n$ points (namely "their" copy of $X$) in distance $\le 1$; this is because any other point in $G\cdot 3p$ differs in at least two coordinates by at least $3$, hence is at distance $\ge 3\sqrt 2$ and hence the various copies of $X$ are well enough separated. Hence we find $g\in G$ with $\alpha(3p)=g(3p)$. Then $g^{-1}\circ \alpha$ leaves $3p$ fixed and also must respect the copy of $X$ belonging to $3p$, hence must be the identity. We conclude that the symmetry group of $Y$ is isomorphic to $G$.

Answer 2

Yes. To any group $G$ (and choice of generating set $S$) you can associate its Cayley graph, which has a vertex for each group element $g$, and an edge between the vertices corresponding to $g$ and $gs$ for each $s$ in $S$. The left action of $G$ on itself corresponds to rigid motions of the graph. This graph is finite if and only if $G$ is a finite group.

If you know a little more topology, a corollary of Van Kampen's theorem is that every group $G$ is the fundamental group of a 2-dimensional CW complex $X$, so in particular the group $G$ acts by deck transformations on the universal cover $\tilde X$. It even turns out that every finitely presented group $G$ is the fundamental group of a 4-dimensional topological manifold. In the same vein, Eilenberg and Mac Lane gave a "functorial" construction of a (typically huge) geometric object $BG$, an example of what they term a $K(G,1)$—a space whose topology is in some sense completely determined by $G$, its fundamental group. This allows one to use methods from algebraic topology on even finite groups.

ETA: The representation of infinite, discrete groups as distance-preserving transformations of geometric objects is a central concern of Geometric Group Theory! Meier's Groups, Graphs and Trees or Clay and Margalit's Office Hours With a Geometric Group Theorist make excellent introductions to this field.

Answer 3

Often, motivation for studying groups is given by symmetries of polytopes, e.g., regular polygons, regular polyhedra and higher-dimensional anlagogues. And in fact, every finite group is the symmetry group of a polytope, which I would say is as geometric as you can get.

• Every group is the symmetry group of a polytope (as constructed in this answer).

• Almost every group is the symmetry group of a vertex-transitive polytope (orbit polytope).

  • Babai, László. "Symmetry groups of vertex-transitive polytopes." Geometriae Dedicata 6.3 (1977): 331-337.
  • Friese, Erik, and Frieder Ladisch. "Affine symmetries of orbit polytopes." Advances in Mathematics 288 (2016): 386-425.9

• I also remember to have read that every group is the symmetry group of a lattice polytope, but I cannot find the source right now.

For me, a general idea here is to look at Frucht's theorem from graph theory: every group is the symmetry group of a graph. Graphs are not really geometric objects $-$ they are combinatorial objects. However, there are tools to construct polytopes from these graphs that do reflect the symmetries of the graph (e.g. eigenpolytopes).

This is especially evident in the case of vertex-transitive graphs/polytopes: the groups that can be represented as symmetry groups of vertex-transitive graphs and as symmetry groups of vertex-transitive polytopes are exactly the same.