Manifolds with a finite but not trivial fundamental group

Question

I came across this nice result:

Theorem: If $M$ is a connected smooth manifold with finite fundamental group, then its first de Rham cohomology is trivial: $$H^1_{dR}(M)=0.$$

However, I don't know any example of manifold with a finite but not trivial fundamental group. What are the nice examples of such manifolds?

Answer 1

Real projective space $\mathbb{RP}^n$ has fundamental group $\mathbb Z/2\mathbb Z$ for $n\geq 2$. This the quotient of the sphere $S^n$ by the antipodal action $x\sim -x$. In fact $S^n$ is a $2$-sheeted universal cover, which implies by covering space theory that its fundamental group is of order $2$.

Answer 2

The special orthogonal group $SO(n)$ for $n\ge 3$ has fundamental group $\mathbb Z/2\mathbb Z$.

Answer 3

The Poincare homology sphere has fundamental group of order $120$, the binary icosahedral group. Any homology sphere must have a perfect fundamental group, but it need not be trivial.

Answer 4

In analogy to the answer of Robert Israel: The special linear group $\operatorname{SL}_n(\mathbb{R})$ (For a proof it is a smooth manifold see here.) has fundamental group $\mathbb{Z}_2$ for $n\geq 3$.

I am not totally sure, whether the projective orthogonal groups are smooth manifolds (somebody would need to verify that), but we have (See here.) for $n\geq 1$: $$\pi_0\operatorname{PO}(2n) \cong\mathbb{Z}_2$$ $$\pi_0\operatorname{PO}(2n+1) \cong 1$$ $$\pi_1\operatorname{PO}(4n) \cong\mathbb{Z}_2^2$$ $$\pi_1\operatorname{PO}(4n+2) \cong\mathbb{Z}_4$$ $$\pi_1\operatorname{PO}(2n+1) \cong\mathbb{Z}_2.$$