Is every infinite group isomorphic to some isometry (sub)group? A while ago I managed to convince myself that every finite group is isomorphic to a subgroup of a finite dimensional Euclidean space. But my little simplex construction doesn't extend to infinite groups. Is it the case that every infinite group is isomorphic to an isometry subgroup of a finite dimensional space?
I don't mind if groups are only subgroups of an isometry group, and I don't mind if the space is non-Euclidean.
My naive thought is that non-Euclidean spaces are necessary to deal with groups like $GL({\mathbb R}^n)$ that explicitly include scale transformations.
 A: Start by taking a finitely generated infinite simple group $G$ (see this MSE question for references), say, $G$ is a Higman group. Then $G$ (despite being countable) does not embed as a subgroup in any matrix group $GL(n, F)$ where $F$ is any field. This follows from a theorem of A.Malcev, which states that every finitely generated subgroup $G$ of  $GL(n, F)$ is residually finite, i.e., the intersection of all finite index subgroups of $G$ is trivial. Malcev's theorem implies that $G$ (if infinite) contains proper finite index subgroups, hence, proper normal finite index subgroups, hence, cannot be simple.  See a proof of Malcev's theorem here in the case $F={\mathbb C}$. 
On the other hand, if you do not mind infinite-dimensional vector spaces, then every group $G$ embeds into the isometry group of some Banach space (a complete normed vector space). Namely, let $V$ denote the space of bounded functions on $G$ with sup-norm. Then $G$ acts isometrically, linearly (and faithfully) on $V$ by the rule:
$$
g\cdot f(x)= f(xg) 
$$
where $f: G\to {\mathbb R}$ is in $V$. 
