 Cup square of a Thom class

For an oriented rank-$r$ vector bundle, let $U\in H^r(D(E),S(E);\mathbb{Z})$ be its Thom class. Forgetting relative supports sends $U$ to ${\pi }^{\ast }e(E)$, where $e(E)$ is the Euler class. Compatibility of relative and mixed cup products givesFor odd $r$, graded commutativity of the cup product makes $2U^2=0$. The Thom isomorphism theorem is injective, so $2e(E)=0$. This explains why an odd-rank Euler class can have only two-torsion.