I have a past paper question for a first course in algebraic topology, which asks one to calculate the first three homology and homotopy groups for the space $L_n$, defined as follows:
Let $G=\{z\in\mathbb C|z^n=1\}\cong\mathbb Z_n$ act on $S^3=\{(z_1,z_2)| |z_1|^2+|z_2|^2=1\}$ by $z(z_1,z_2)=(zz_1,zz_2)$ and define $L_n$ to be the quotient space. (The question claims that the action of $G$ is properly discontinuous.)
This space $L_n$ looks suspiciously like a lens space $L(p,q)$ with $p=q=n$, with the exception that $p,q$ are not coprime, as usually required in the definition.
I think I can manage to calculate homology and homotopy groups for the usual lens space, but I am a bit lost as to what $L_n$ is.
My question is, why are $p,q$ in the definition of a lens space $L(p,q)$ required to be coprime?
Recall that $L(p,q)$ is the quotient of $S^3$ by the action of $\mathbb{Z}/p\mathbb{Z}$ (thought of as the $p$th roots of $1$) given by $z(z_1,z_2) = (zz_1, z^q z_2)$. Hence, the space $L_n$ is actually the space $L(n,1)$ in the $(p,q)$ notation (and note that $\gcd(n,1) = 1$, as is required for the $(p,q)$ notation.)
As Qiaochu (and Wikipedia) point out, requiring $\gcd(p,q)=1$ is equivalent to asking that the action be free. Free actions are nice for two reasons. First, as you pointed out, freeness guarantees the quotient is a manifold. Second, it also guarantees the quotient map $S^3\rightarrow L(p,q)$ is a covering, which reduces much (but not all) of the algebraic topology considerations of $L(p,q)$ to those of $S^3$.