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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 $mathbbS_3$ 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 $mathbbS_n+1$ 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 (:

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 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.

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 $X=frac 1ke_kcup 0$.
Then $0in X$ 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 $Gcdot 3p$ 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 $Gcdot 3p$ differs in at least two coordinates by at least $3$, hence is at distance $ge 3sqrt 2$ and hence the various copies of $X$ are well enough separated.
Hence we find $gin G$ with $alpha(3p)=g(3p)$. Then $g^-1circ alpha$ 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 $G$.