Every group as full symmetry group of points in $\mathbb R^d$

Question

Does every finite group $G$ have the property that it is isomorphic to a full symmetry group of some set of points in $\mathbb R^n$ for some $n$

Answer 1

Yes. The construction is not delicate; you just have to make something which is as generic as possible while still having $G$-symmetry. It could easily be possible to do something cleverer and shorter than what I'm about to do, but I want to emphasize that all I'm about to do below is to enforce genericity in what seems to me like the most obvious way to get the argument to work at each step.

Start with any faithful orthogonal linear representation of $G$ on a real inner product space $V = \mathbb{R}^n$ containing no trivial subrepresentation. A good example to visualize is the action of the cyclic group $C_k, k \ge 3$ of order $k$ on $\mathbb{R}^2$ by rotation because the most obvious version of this construction will not work in this case; if you just pick a vector and act on it you'll produce dihedral symmetry, and all the work I'm about to do is geared towards avoiding this sort of thing. In general you can take the action of $G$ on the set of functions

$$\left\{ f : G \to \mathbb{R} \mid \sum_{g \in G} f(g) = 0 \right\}.$$

Pick a collection of points $R$ in $V$ whose properties we'll specify later. We'll take our set of points to have the form

$$GR = \bigcup_{g \in G} gR.$$

By construction, the isometry group certainly contains $G$. The game is to figure out what we need to ask of $R$ for the isometry group of $GR$ to be exactly $G$.

So, suppose $\varphi : V \to V$ is some isometry of $GR$. For starters, we'd like to guarantee $\varphi(0) = 0$. This is guaranteed if the "center of mass" of $GR$ is $0$, which is in turn guaranteed if $\sum_{g \in G} g$ acts by $0$ on $V$. But this is equivalent to the condition that $V$ has no trivial subrepresentations.

Now pick a point $r_0 \in R$. We'd like to guarantee $\varphi(r_0) = g r_0$ for some $g \in G$. We can do this by requiring

Property 1: Every point in $R$ has a unique distance from the origin.

Then the only points in $GR$ the same distance away from the origin as $r_0$ are the points $g r_0, r \in G$, so $\varphi(r_0)$ must be such a point.

Pick some other point $r_1 \in R$. We'd now like to guarantee $\varphi(r_1) = g r_1$ for the same $g \in G$ as above. To do this we'll require

Property 2: There is some constant $D$ such that any two points in $R$ are less than $D$ apart, but any point in $R$ is greater than $D$ away from any point in $gR, g \neq e$.

Now, since $d(\varphi(r_1), \varphi(r_0)) = d(r_1, r_0) < D$, it follows that $\varphi(r_1)$ must lie in the same copy $gR$ of $R$ as $r_0$ does. To force $\varphi(r_1) = g r_1$, we'll require

Property 3: Every point in $R$ has a unique distance from every other point in $R$.

This uniquely forces $\varphi(r_1) = g r_1$ as desired. Since $r_1$ was arbitrary, we've shown that $\varphi(r) = gr$ for every $r \in R$. We'd like to conclude from this that $\varphi = g$. Since $\varphi$ is an isometry fixing the origin, it is linear, and so to guarantee this it suffices to require

Property 4: The vectors in $R$ span $V$.

Every property we've asked for is guaranteed by the following construction. Say that a vector $v \in V$ is generic if its stabilizer under the action of $G$ is the identity. Since the action of $G$ on $V$ is faithful, the subset of non-generic vectors in $V$ is a finite union of subspaces of strictly lower dimension, and hence a vector is generic with probability $1$.

If $v$ is a generic vector, let $2D$ be the minimum distance between $v$ and any $gv, g \neq e$, and take $R$ to be a random finite set of at least $\dim V$ points all of which are strictly less than $\frac{D}{2}$ away from $v$. This satisfies all of the properties above with probability $1$, so we're done.

Example. Take $G = C_k, k \ge 3$ acting on $V = \mathbb{R}^2$ by rotation. If we pick $R$ to be a single nonzero vector we get something with dihedral symmetry (Property 4 is not satisfied). But we can fix this by picking $R$ to be a pair of vectors very close together with different distances from the origin.