Topological space with fundamental group $\mathbb{Z}/3\mathbb{Z}$. In my course "Introduction To Algebraic Topology" I had following test problem:

Exemplify a topological space with fundamental group $\mathbb{Z}/3\mathbb{Z}$.

I was supposed to use this theorem:

Let $Y$ be a simply connected topological space. If a group $G$ (finite or countable) acts on $Y$ freely and properly discontinuously, then fundamental group of the quotient space $Y/G$ is naturally isomorphic to $G$.

So problem is I just wasn't able to come up with simply connected space such that $\mathbb{Z}/3\mathbb{Z}$ acts on it freely and properly discontinuously. 
Any ideas? Thanks!
 A: There is an explicit construction:
Consider the CW complex given by three $0$ cells, three $1$ cells and a two cell pasted in (this is a filled in triangle.)
Now, considering the circle with a base point, and pasting in this $CW$-complex by identifying all three vertices with the base point (in a three fold way.)
The fundamental group of the resulting space is exactly $\mathbb Z_3$ by construction, since the two cell gives the relation $a^3=1$ on the fundamental group.
This is called the presentation complex by the way.

If you want to construct the Cayley complex (the universal cover of the presentation complex), you will need to take three disks and paste them to the $1$ skeleton of the first construction. $\mathbb Z_3$ will act by cyclic permutation of the $2$-cells, and taking the quotient will give you exactly the presentation complex. Note that this is a simply connected space and that $\mathbb Z_3$ acts freely on the two cells.
A: How about the unit sphere $S^3$? If $A$ is a $2$-by-$2$ rotation matrix
for angle $2\pi/n$, then the block matrix $\pmatrix{A&0\\0&A}$ acts on $S^3$ without fixed points.
