If $M$ is a compact oriented parallelizable 4 dimensional manifold, then it admits a flat metric

Question

In order to prove that the first Pontryagin class of the tangent space of a compact oriented parallelizable 4 dimensional manifold is zero, my Professor suggested me to prove that such hypothesis imply that $M$ admits a flat metric and then $p_1(TM)$ will be zero since it is equal to $c_2(TM\otimes \mathbb{C})$ (second Chern class). I couldn't managed to prove this statement, buy already proved that $p_1(TM) = 0$ since in this case $c_2(TM\otimes \mathbb{C})$ will be de Euler class of the tangent space and $e(TM)$ is zero because $M$ is parallelizable. I would like to know how to prove that $M$ admits a flat metric, so I hope someone can help me. I would also be grateful if someone gives me a concrete example of a manifold in which this happens.

Answer 1

Of course not. The simplest example is to take any closed oriented 3-manifold $N$ (for instance, $S^3$) and multiply it by the circle. The resulting manifold $M$ is parallelizable (since every orientable 3-manifold is) but admits a flat metric if and only if $N$ admits a flat metric.