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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.

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    $\begingroup$ Every finitely presented group is the fundamental group of a compact $4$- dimensional manifold: take the standard 2-dimensional CW complex with that fundamental group, immerse it into $\Bbb R^4$ and take a regular neighborhood. $\endgroup$
    – markvs
    Sep 15 '20 at 19:50
  • $\begingroup$ Does this argument gives both orientable and non-orientable, can you elaborate more. I will really appreciate if you write this as an answer. $\endgroup$
    – User
    Sep 15 '20 at 19:57
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    $\begingroup$ @JCAA: that construction seems to me to produce either an open manifold or a manifold with boundary. I think you want to immerse into $\mathbb{R}^5$ and take the boundary of a tubular neighborhood, as Somnath Basu does here: mathoverflow.net/a/15421/290 (I don't know whether the result is orientable or not though.) $\endgroup$ Sep 15 '20 at 20:01
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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.

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  • $\begingroup$ My bad! I should check this website before asking this question. Thanks. +1 for your help. $\endgroup$
    – User
    Sep 16 '20 at 6:22
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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 $H^4(S^2 \times S^2)$.

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  • $\begingroup$ But, here we can not take orientation two-cover to get another orientable manifold. Am I right? Thanks for your answer. $\endgroup$
    – User
    Sep 15 '20 at 19:55
  • $\begingroup$ @Math: why not? The orientation double cover is the quotient of $S^2 \times S^2$ by $f^2$, which sends $(x, y)$ to $(-x, -y)$. It turns out to be the real Grassmannian $\text{Gr}_2(\mathbb{R}^4)$. $\endgroup$ Sep 15 '20 at 19:57
  • $\begingroup$ Sorry, I was thinking that $\pi$ induced an index $2$-subgroup of $\Bbb Z/4\Bbb Z$ as covering induced map is injective. $\endgroup$
    – User
    Sep 15 '20 at 20:03

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