Space $X$ so that $\pi_{1}(X)=\mathbb{Z}_{3}$

Question

Given a group $G=\langle S|R \rangle$ we can construct a 2-dim complex $K$ s.t $\pi_{1}(K)=G$. I get that the main idea is to construct 2-cells that "read" the relators and glue them to the wedge that represents the free group on $S$. With that in mind I can see for example that if $G=\mathbb{Z}\times\mathbb{Z}$ we glue the rectangle $aba^{-1}b^{-1}$ in the graph $R_{2}$ (rose graph) that represents the free group on $a,b$ and after that glueing we get a torus. WIth the same procedure we can easily see that if $G=\mathbb{Z}_{2}$ we get the projective plane. But what about all the other cyclic groups? For example which space has fundamental group $\mathbb{Z}_{3}$? I am having trouble to see if this is a familiar space.

Answer 1

Here's a way to get adequate spaces, but not using the procedure you describe (MathsIsFun described what this procedure looks like in the case of a cyclic group).

Suppose you have a (discrete) group $G$ acting very nicely (there are, of course, precise hypotheses) on a simply-connected space $X$.

Then covering theory tells us that the quotient $X/G$ has $\pi_1\cong G$.

The case $G=\mathbb Z, X= \mathbb R$ with action by translation gives $S^1$ with the correct $\pi_1$, and the case $G=\mathbb Z/2, X= S^n$ with the antipodal action gives $\mathbb RP^n$, with the correct $\pi_1$ as well.

For cyclic groups $G$, note that $G$ can be seen as a subgroup of the multiplicative group of nonzero complex numbers. Unfortunately, $\mathbb C^*$ isn't simply connected, but fortunately for us, it embeds as a group into a simply-connected group : the unit group of the quaternions $\mathbb H$. If we look at the elements of norm $1$, we thus get a group embedding $G\to S^3$, and thus, an action by translation produces a quotient $S^3/G$ with $\pi_1\cong G$

(you may be interested in looking up "lens spaces")

In particular, every finite subgroup of $S^3$ (with the "quaternionic" multiplication) can be realized as the fundamental group of a quotient of $S^3$.

Answer 2

The real projective plane has fundamental group $\mathbb Z_2$, not $\mathbb Z_3$. To obtain a space with fundamental group $\mathbb Z_n$, $n\geq 2$, you can form the quotient of the 2-disk $D^2$ identifying each point $x\in \partial D^2 = S^1$ with its counterclockwise rotation by an angle of $\frac{2\pi}{n}$.

The projective plane is the case $n=2$ in this construction.