$4$-manifold with fundamental group $\Bbb Z/4\Bbb Z$ Before I write my question, I want to write some thoughts.
Let $M$ be a connected topological manifold such that $\pi_1(M)=\Bbb Z/3\Bbb Z$. Then, considering its orientation $2$-fold cover, which is connected, I can say $M$ is orientable. Now, an example of such a closed $3$-manifold is $L(3,1)$.
Now, this type of argument can not be given if I consider $\pi_1(M)=\Bbb Z/4\Bbb Z$ to conclude $M$ is orientable. But Euler characteristic of an odd-dimensional closed manifold is always zero, so we cannot say $\Bbb Z/4\Bbb Z$ is the fundamental group of any closed connected non-orientable $3$-manifold, as  $H_1(M,\Bbb Z)$ is infinite when $M$ is closed non-orientable connected $3$-manifold.
Again this logic can not be given for $4$-dimensional closed connected manifold. So, I am wondering if the following fact. I assume closed means compact without boundary.

Does there exist closed connected $4$-manifolds both orientable and
non-orientable type having fundamental group $\Bbb Z/4\Bbb Z$?

Any help will be appreciated.
 A: As pointed out, every finitely presented group arises as the fundamental group of a closed smooth orientable four-manifold.
The same is not true of non-orientable manifolds as $\mathbb{Z}/3\mathbb{Z}$ illustrates. A necessary condition is that the group must have an index two subgroup (i.e. the fundamental group of the orientable double cover). This turns out to be sufficient. That is, a finitely presented group is the fundamental group of a closed smooth non-orientable four-manifold if and only if it has an index two subgroup; see this question.
A: A non-orientable example: consider the automorphism $f : S^2 \times S^2$ given by $(x, y) \mapsto (y, -x)$ where $-$ denotes the antipode map. This map has order $4$ and gives a free action of $\mathbb{Z}/4$ on $S^2 \times S^2$, so its quotient is a closed $4$-manifold $X$ with $\pi_1(X) \cong \mathbb{Z}/4$. Since $\chi(S^2 \times S^2) = 4$ we have $\chi(X) = 1$ so $X$ is non-orientable; alternatively, we can check that $f$ acts by $-1$ on $H^4(S^2 \times S^2)$.
