Can every group be realized as an isometry group?

Question

The question is actually from Löh's Geometric Group Theory: an introduction on page 40:

Exercise 2.E.10: Is there for every group $G$, an $n\in \mathbb{N}$ and a subset $X\subset \mathbb{R}^n$ such that the isometry group of $X$ is isomorphic to $G$?

I am aware of the paper Realizing Finite Groups in Euclidean Space by Albertson and Boutin, which says the answer is yes if $|G|<\infty$. But for $|G|$ infinite, say countably infinite, I do not have a quick answer yet.

Most likely, the answer is no. If so, can anyone help give an example of such $G$ with $G$ countable?

Answer 1

No. There are a great many countably infinite groups which are not even subgroups of $\mathrm{GL}_n(\mathbf R)$ for any $n \in \mathbf N$ (they are "not linear").

For example let $G$ be the group of all finitely supported permutations of $\mathbf Z$. This group $G$ contains an isomorphic copy of every finite group, so it cannot embed in $\mathrm{GL}_n(\mathbf R)$ for any $n$. This follows for example from Jordan's theorem on finite linear groups. To get a finitely generated example you can take $H$ to be the subgroup of $\mathrm{Sym}(\mathbf Z)$ generated by the transposition $(0,1)$ and the infinite cycle $n \mapsto n+1$. Note that $G \le H$.

For a finitely presented example one can take Higman's group $$G = \langle a, b, c, d \mid a^b = a^2, b^c = b^2, c^d = c^2, d^a = d^2\rangle,$$ which has no nontrivial finite quotient. By contrast, finitely generated linear groups are residually finite (a theorem of Mal'cev).

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Edit: In the comments it was pointed out that the OP could have in mind the looser definition of isometry group as the group of all permutations $\pi : X \to X$ preserving Euclidean distance. In this case some more explanation is required.

Let $X \subset \mathbf R^n$ and let $G$ be the isometry group of $X$ in the above sense. By translating $X$ we may assume that $0 \in X$, and by reducing the dimension if necessary we may assume that $X$ spans $\mathbf R^n$.

We claim that every isometry of $X$ extends uniquely to an (affine linear) isometry of $\mathbf R^n$. Let $B \subset X$ be a basis. Let $\pi$ be an isometry of $X$. It follows from the law of cosines that $$\langle \pi(x) - \pi(0), \pi(y) - \pi(0)\rangle = \langle x, y \rangle \qquad (*)$$ for all $x, y \in X$. Applied to $x, y \in B$ and considering a Gram determinant, it follows that $B' = \pi(B) - \pi(0)$ is also a basis. Let $f$ be the unique (affine linear) isometry of $\mathbf R^n$ such that $f(0) = \pi(0)$ and $f(b) = \pi(b)$ for all $b \in B$. Now applying $(*)$ with $y \in B$ and the same identity with $f$ in place of $\pi$ shows that $$\langle \pi(x) - \pi(0), b'\rangle = \langle f(x) - f(0), b' \rangle$$ for all $x \in X$ and $b' \in B'$, which implies that $f(x) = \pi(x)$ for all $x \in X$.

Thus $G$ is isomorphic to a subgroup of $\mathrm{Isom}(\mathbf R^n) \le \mathrm{AGL}_n(\mathbf R) \le \mathrm{GL}_{n+1}(\mathbf R)$.

Answer 2

As you are reading Löh's book on GGT, it is sensible to give a standard, important, and famous example of a non-linear group, and which is one of a family of groups which appears throughout her book, and throughout GGT.

In particular, consider the Baumslag-Solitar group $$\operatorname{BS}(2, 3)=\langle a, b\mid ba^2b^{-1}=a^3\rangle.$$

A group is Hopfian if every surjective endomorphism is an automorphism, and $\operatorname{BS}(2, 3)$ is the standard example of a finitely presented non-Hopfian group (this is Löh's Exercises 2.E.21).

Finitely generated residually finite groups are Hopfian (this is Löh's Exercise 4.E.28), and so $\operatorname{BS}(2, 3)$ is non-residually finite. Then apply @SeanEberhard's answer/Mal'cev's theorem to get that the group is non-linear (see here for a sketch proof).