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Nov 30, 2016 at 7:09 comment added Izaak Meckler What about a screw along an axis? That's an isometry acting without fixed points which isn't a translation.
Jul 13, 2011 at 12:38 comment added Theo Johnson-Freyd The isomorphism here is as manifolds, not as groups, unless I am very mistaken. There are many nonabelian Lie groups (the solvable connected simply-connected ones) that are isomorphic as manifolds to $\mathbb R^n$. Since the tangent bundle to any Lie group is trivializable by right translations, every Lie group admits a left-invariant metric (pick any metric on the underlying vector space of the Lie algebra). In particular, Alain below gives a noncommutative group with left-invariant flat metric, which is as a manifold just $\mathbb R^3$, but not as a group.
Jul 10, 2011 at 20:40 history answered Igor Belegradek CC BY-SA 3.0