Does every group have an object of symmetry

abstract-algebrageometrygroup-theorysymmetry

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

Best Answer

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 $G$.

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