Does every group have an object of symmetry?

Question

I'm aware of Cayley's theorem, which says that every group is isomorphic to some subgroup of a symmetric group.

But it's not clear to me whether symmetric groups themselves (apart from their name) capture the notion of geometric symmetry that "objects of symmetry" have (and by geometric symmetry I mean the type of symmetry expressed when we talk about the rotations and flips of a square ($D_4$), or the symmetries of a cube ($S_4$))

Some stackexchange posts answer the question, but I can't tell if the first one is talking about symmetry (as in symmetric group) or symmetry (as in symmetry of a square), and the second answer is a bit too technical for me....

Furthermore, group explorer doesn't have objects of Symmetry for $Q_4$ and $Z_2 \times Z_4$. Is that for lack of imagination, an incomplete database, or because there is no object of symmetry for these groups (and the many others there)?

Thanks

Answer 1

If what you allow as a “geometric object” is sufficiently broad to match the kinds of groups you allow, the answer is positive. I’ll first restrict to the finite case, which from your examples seems to be the case you’re mainly interested in, and then discuss the infinite case.

For a finite group $G$, by Frucht’s theorem (linked to in a comment under the first answer you linked to), every group is isomorphic to the automorphism group of a finite undirected graph. Embed the graph $(V,E)$ in $\mathbb R^{|V|}$ by bijectively mapping the vertices to the canonical basis vectors and the edges to line segments between the vertices they are incident upon. The isometry group of the resulting geometric object is isomorphic to $G$.

The isometries of a Euclidean space are linear transformations, so specifying the images of all basis vectors under an isometry specifies the isometry. Since an automorphism of the graph specifies the images of all basis vectors, it uniquely defines an isometry; the object is invariant under this isometry; and the composition law of these isometries is the composition law of the automorphisms. Conversely, every isometry of the object corresponds to an automorphism of the graph. Hence the group of isometries is isomorphic to the group of automorphisms, which is isomorphic to $G$.

This doesn’t work in the infinite case, since there are groups of arbitrarily large cardinality (e.g. the free group over a set of arbitrarily large cardinality) and the Euclidean group only has the cardinality of the continuum. However, Frucht’s theorem was extended to infinite groups and graphs (see this section of the Wikipedia article, with references), so if we allow “geometric objects” in arbitrary powers of $\mathbb R$, we can embed an infinite graph $(E,V)$ whose automorphism group is isomorphic to $G$ in the subspace of $\mathbb R^V$ with finitely many non-zero components by again mapping the vertices to canonical basis vectors and the edges to line segments connecting them. Then a linear transformation is again uniquely determined by the images of all basis vectors (this is where we need the restriction to finitely many non-zero components), and it follows that the group of linear transformations of the resulting “geometric object” is isomorphic to $G$.

Answer 2

As I understand it the objects you are looking for are subsets $S$ of a $n$-dimensional Euclidian space where you consider a map $f: S \to S$ to be an symmetry of $S$ if the map $f$ preserves distances and angles, i.e. is a rigid motion.

Now your question is: for every group $G$, is there such an object $S$ such that $G$ is the group of all symmetries of $S$? It is also somewhat implicit in the question that you assume $G$ to be finite.

We can split it into two questions:

1) For every $G$ is there an object $S$ such that $G$ appears as a subgroup of the symmetries of $S$?

2) If the answer to question 1 is yes and we are looking at such an object, can paint different colors on it, or draw smiley faces, or carve holes or attach handles to the object in order to get rid of some of the symmetries and only end up with those in $G$?

Question 2 is quite interesting. Take for instance the group $A_5$ of all rotation symmetries of the dodecahedron. Clearly it is a subgroup of the group of all symmetries of the dodecahedron, wich also contains relflections. Can we mutilate the dodecahedron in such a way that only rotational symmetries remain? The answer is yes, but not very easy to find (although I'm sure Wikipedia has a picture).

However I will only say something about question 1 here. We note a couple of things:

If you picture examples of symmetric objects (cubes, spheres etc) you notice that they often have some sort of center point that is preserved by all the symmetries. We make a bold move forward and narrow the question down to:

1': For every finite group $G$, is there an object $S$ an a point $O$ in the Euclidian space where $S$ lives such that every symmetry of $S$ leaves $O$ in place and such that $G$ is a subgroup of the group of all symmetries of $S$?

Let's fist approach the situation from the other side and imagine that we do have such an object $S$ sitting inside an $n$-dimensional Euclidean space, which also contains a point $O$ with the special property that every symmetry of $S$ leaves $O$ in its place.

The reason for introducing the point $O$ is that we can bring in some linear algebra. Given the 'special' point $O$ we can think of the surrounding Euclidean space as the space $\mathbb{R}^n$ where $O$ is the origin. Notions such as 'span' suddenly make sense so we'll restrict our attention to the subspace of $\mathbb{R}^n$ spanned by $S$. Since we didn't say what $n$ was we may well assume that this subspace is all of $\mathbb{R}^n$.

Now the crux is that every symmetry of $S$ extends to a map from all of $\mathbb{R}^n$ to itself, and since the symmetry preserves angles and distances, the parallelogram law tells us that these maps are linear!

Conversely you perhaps remember from linear algebra that in order to be able to talk about distances and angles you need to have an inner product $\langle . , . \rangle$ on your vector space. The condition of preserving angles and distances says then that a symmetry $f: \mathbb{R}^n \to \mathbb{R}^n$ is not only linear but also satisfies $\langle f(x), f(y) \rangle = \langle x, y \rangle$ for all vectors $x, y$; in other words it preserves the inner product or is an orthogonal linear transformation.

Hence we can reformulate question 1 in more linear algebraic terms:

1'': for every group $G$ is there a number $n$ and a group of orthogonal linear transformations of $\mathbb{R}^n$ that is isomorphic to $G$?

The answer is yes. A simple way is to embed $G$ into $S_n$ and then let $S_n$ act on $\mathbb{R}^n$ by permuting the basis vectors.

Now that we have realized $G$ as a subgroup of the symmetries of all of $\mathbb{R}^n$, we'd like to prettify things by realizing it as a subgroup of the group of symmetries of a smaller subset $S$ of $\mathbb{R}^n$. This can be done as follows. Take a generic point $x$. Look at the set of $|G|$ points $g_1(x), g_2(x), ...$ where $g_1, g_2, ...$ are the elements of $G$, realized as linear transformations.

You end up with nicely symmetric set of points. Finally take there convex hull to get a more tangible solid object $S$.

Answer 3

Here is an exercise:

Let $G$ be a subgroup of the symmetric group $\mathfrak{S}_X$ of permutations of a finite set $X$. Show that there exists $k$ and a relation $R\subset X^k$ such that $G$ is equal to the automorphism group of $(X,R)$.

Where $\mathrm{Aut}(X,R)$ is by definition $\{g\in \mathfrak{S}_X:\;gR=R\}$, the group $\mathfrak{S}_X$ acting on $X^k$ by $g(x_1,\dots x_k)=(gx_1,\dots,gx_k)$.

Hence, every group as group of permutations of some given finite set $X$ can be viewed a group of symmetry of some "relational" structure on $X$ itself.

A hint to the exercise: one can choose $k=|X|$. Requiring $k=1$ is very restrictive (one only gets stabilizers of subsets), and even $k=2$ (realizing $G$ as stabilizer of some directed graph structure on $X$) is certainly too restrictive although I don't have an example in mind at the moment.