Fundamental groups of non-orientable closed four-manifolds The fundamental group of a closed orientable manifold is finitely presented, and every finitely presented group arises as the fundamental group of a closed orientable four-manifold; see this question. 
Not every finitely presented group arises as the fundamental group of a closed non-orientable four-manifold. One necessary condition is that it contain an index two subgroup which corresponds to the orientation double cover. Is this the only restriction?

Let $G$ be a finitely presented group with an index two subgroup. Is there a closed non-orientable four-manifold $M$ with $\pi_1(M) \cong G$?

 A: Let $G$ be a finitely presented group
$$ G = \langle g_1, g_2, \cdots, g_n | R_1, R_2, \cdots, R_m \rangle$$
One standard way to realize this as the fundamental group of a compact $4$-manifold is to construct a $5$-dimensional manifold. 
1) Start with $D^5$ and attach a $1$-handle for every generator $g_i$.  Here the attaching map you use does not matter all that much but to get the process moving let's choose the attaching maps so the resulting handlebody will be orientable.  We will want to change this to answer your question. . . but this works for now. 
2) You then attached a $2$-handle for every relator $R_j$.  The $2$-handles are attached via a circle mapping to the handlebody constructed in step (1).  A map from a circle to the previous handlebody is determined (up to homotopy) by a conjugacy-class in the fundamental group of the handlebody, i.e. a word in the free group, which is exactly what the relator $R_j$ is.  To make this a handle attachment you need to ensure the circle is in the boundary of the handlebody, and it is embedded, and has a trivial normal bundle.  This can all be achieved due to our choice of making the handlebody orientable in step (1) and a little transversality. 
So this is a 5-manifold with the appropriate fundamental group.  But the manifold has boundary, so we check to see if the boundary (which is a closed manifold) also has the same fundamental group.  Here we use the fact that our 5-manifold deformation-retracts to a $2$-skeleton.  So by a little transversality, this $5$-manifold does have the same fundamental group as its boundary.  This is where we see we could have made this argument for $4$-manifolds, $5$-manifolds, etc, but this argument falls flat for $3$-manifold fundamental groups. 
In the case you have an epi-morphism $G \to \mathbb Z_2$, you modify step (1) so that your $1$-handle attachment is orientation-reversing provided the loop maps to something non-trivial via the homomorphism $G \to \mathbb Z_2$.  Step (2) is unchanged.  The boundary $4$-manifold has the same fundamental group as the $5$-manifold constructed, again by dimension counting (transversality). 
