I am interested in the classifying space $BG$ of a finite group $G$.


A real representation $V$ of $G$ of dimension $r$ defines a real vector bundle over $BG$ of rank $r$. If the determinant of this representation is trivial, then this bundle is orientable and a choice of orientation determines an Euler class $e(V) \in H^r(BG,\mathbb{Z})$.


Do these classes generate $H^\*(BG,\mathbb{Z})$ as a ring?


This is easy to show when $G$ is abelian. In that case we need only consider $G = \mathbb{Z}\_n$ and the cohomology ring is generated in degree 2 by the Euler class of the $2\pi/n$-rotation representation.


More generally, for any $G$, the map $H^1(BG,SO(2)) \to H^2(BG,\mathbb{Z})$ is an isomorphism since the latter is torsion and this is a Bockstein-type operation. We can consider the element of $H^1(BG,SO(2))$ as a homomorphism $G \to SO(2)$ giving us a 2d representation and it is easy to see from the definition of the Bockstein that the above map is the Euler class of this representation.



Is the cohomology ring $H^*(BG,\mathbb{Z})$ generated by Euler classes?
Let $G = \mathbb{Z}/p \times \mathbb{Z}/p$ for $p$ odd. Then $H^3(BG; \mathbb{Z}) \cong \mathbb{Z}/p$ is not in the subring generated by Euler classes, since the non-trivial irreducible representations are of rank $2$.


EDIT: The problem of determining the subring of $H^\*(BG; \mathbb{Z})$ generated by all Chern classes has been much studied. You might start with [Atiyah's 1961 paper](http://www.numdam.org/article/PMIHES_1961__9__23_0.pdf).


