 Finite groups cannot act freely on Euclidean space

If a nontrivial finite group acted freely on ${\mathbb{R}}^n$, a prime-order cyclic subgroup $C_p$ would too. The quotient is an $n$-manifold with contractible universal cover. The singular version of cellular free resolution from a contractible universal cover identifies its cohomology with group cohomology of $C_p$. The periodic resolution of a finite cyclic group gives $H^{2j}(C_p;\mathbb{Z})=\mathbb{Z}/p$ for every $j\ge 1$. But Poincaré duality for noncompact manifolds makes the quotient cohomology zero above dimension $n$. This contradiction excludes the action, for arbitrary homeomorphisms, not only Euclidean isometries.