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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 (:

Slender Threads
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    See also this question. There are $49,487,365,422$ distinct groups of order $1024$. Probably not all are isomorphic to some nice symmetry group. The following MO-question is a duplicate. – Dietrich Burde Jun 07 '19 at 18:08
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    You might want to read about the Cayley graph of a group – Maxime Ramzi Jun 07 '19 at 18:18
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    Do you consider finite, undirected graphs "geometric"? Frucht's Theorem says that every finite group $G$ is isomorphic to the automorphism (i.e., symmetry) group $\operatorname{Aut}(\Gamma)$ of some finite, undirected graph $\Gamma$. A strong form of the theorem says that in fact for every $G$ there are infinitely many nonisomorphic finite, undirected graphs $\Gamma$ for which $\operatorname{Aut}(\Gamma) \cong G$. – Travis Willse Jun 07 '19 at 19:38
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    A theorem of Babai says that we can find graphs that aren't too large, relatively speaking: For any $G \not\cong \Bbb Z_3, \Bbb Z_4, \Bbb Z_5$ we can find a graph $\Gamma$ such that $|\Gamma| \leq 2 |G|$. – Travis Willse Jun 07 '19 at 19:38
  • By "infinite groups" do you mean, say, finitely generated infinite groups, Lie groups, or something else? – Travis Willse Jun 07 '19 at 19:39
  • Related: https://math.stackexchange.com/questions/51578/what-property-of-groups-is-related-to-the-symmetries/ – user1729 Jun 08 '19 at 07:27

3 Answers3

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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.

Rylee Lyman
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    (+1) for the excellent book recommendations. – Santana Afton Jun 07 '19 at 19:30
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    But the representation of $G$ on a Cayley graph of $G$, while faithful, is not necessarily full—for some groups, every Cayley graph has some "extra" symmetries that are not elements of the group representation. – K B Dave Jun 07 '19 at 19:34
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    @KBDave You are right, but I must confess that this defect does not bother me! I frequently give up control of the full symmetry/isometry group of an object $X$ in favor of having an action of my group $G$ on $X$ whose properties I like. – Rylee Lyman Jun 07 '19 at 19:38
  • @KBDave : it becomes full if we allow labels on the edges – Maxime Ramzi Jun 07 '19 at 19:57
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    @Max you can also replace each edge with a graph which looks like a directed edge. (But you have to be careful to not add in the symmetry of order two which "rotates" this edge along its length, so keeping the end points fixed). – user1729 Jun 08 '19 at 07:32
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    (easier: in the original graph add two new vertices to each edge, and then add a "spur" (single edge) on to one of the new vertices to comply with the directed edge.) – user1729 Jun 08 '19 at 07:56
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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 $G$.

Hagen von Eitzen
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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.

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.

M. Winter
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