I'm trying to find generating manifolds for the cobordism group $\mathit{MO}\_5(K(\mathbb Z/2, 2))\cong (\mathbb Z/2)^4$, which can be represented as the cobordism group of closed 5-manifolds $M$ together with a class $B\in H^2(M;\mathbb Z/2)$. I've found three of the four generators; the fourth should be a 5-manifold $M$ such that $w\_1(M)w\_2(M)\in H^3(M;\mathbb Z/2)$ is nonzero, and here I've gotten stuck.


Such an $M$ cannot be a product of lower-dimensional manifolds, nor can it be the total space of a fiber bundle over the circle.


I'm happy to hear general approaches or ideas as well, such as ways of modifying a manifold to change its cohomology or Stiefel-Whitney classes in useful ways.



Is there a closed 5-manifold $M$ with $w_1(M)w_2(M)\ne 0$?
Note that the third Wu class is $\nu\_3 = w\_1w\_2$, so on a closed connected smooth $n$-manifold $M$, $\operatorname{Sq}^3 : H^{n-3}(M; \mathbb{Z}\_2) \to H^n(M; \mathbb{Z}\_2)$ is given by $\operatorname{Sq}^3(x) = w\_1(M)w\_2(M)x$. 


As $\operatorname{Sq}^i(x) = 0$ if $i > \deg x$, when $n = 5$ we see that $\operatorname{Sq}^3 : H^2(M; \mathbb{Z}\_2) \to H^5(M; \mathbb{Z}\_2)$ is the zero map. By Poincaré duality, it follows that $w\_1(M)w\_2(M) = 0$.


