When to attach 2-cells in Cayley complexes? In Hatcher's Algebraic Topology section 1.3, Cayley complexes are explained. The book states that we get a Cayley complex out of a Cayley graph by attaching a 2-cell to each loop. There is an example showing the Cayley complex for $\mathbb{Z}\times\mathbb{Z}$ (the fundamental group of the torus). We attach one 2-cell to each loop and we get $\mathbb{R}^{2}$ with vertical and horizontal tiling. I understand this.
The book then says (example 1.47) that the Cayley complex of a cyclic group of order $ n $ is $n$ disks with boundaries identified. I can't for the life of me figure out where the $n$ disks come from. In the Cayley graph, we have one loop $e \to x \to x^2 \to \cdots \to x^n = e$. I guess the relation $x^n = e$ somehow generates $n$ loops, but I don't understand why.
The next example is for $\mathbb{Z}_2*\mathbb{Z}_2$ in which two 2-cells are attached to each loop. I also don't understand why two.
I'm looking for a canonical description of the algorithm to build Cayley complexes, and the application of the algorithm to build Cayley complexes for finite cyclic groups and $\mathbb{ℤ}_2*\mathbb{ℤ}_2$.
Thank you.
 A: This is essentially the same answer as user32240 but I will try to explain it differently.  Hatcher's description is a bit sloppy.  The correct thing to say is that if $R$ is the set of defining relators for $G$, then each element $r\in R$ labels a loop based at every vertex of the Cayley graph. To each of these based loops, you add a 2-cell. 
The reason for this is you want the group $G$ to act freely on the Cayley complex.  Now if you have a relator of the form $r=s^n$ where $s$ is not a proper power, then each loop labeled by $s^n$ in the Cayley graph can be read from $n$ different starting vertices and so you need a 2-cell for each one.
So, for example, if $G=\mathbb Z_2$ with presentation $\langle a\mid a^2=1\rangle$, then you want to have 2 2-cells and have $\mathbb Z_2$ permute them so that you have a free action. The two 2-cells come from the loop labelled $a^2$ at 1 and the looped labeled by $a^2$ at $a$. You can think of the 2 2-cells as the northern and southern hemisphere of a sphere.
If you attached only one $2$-cell, you would get a disk.  The group $G$ would fix the center of this disk and so the action is not free.  Although the projective plane is a disk with antipodal points identified, the quotient map is not a covering map.  By attaching 2 disks you get a covering map.
Incidentally, this issue is not handled properly in the book of Lyndon and Schupp if memory recalls.  Cohen makes a big point of this in his book and on the necessity of using $n$ disks for relators $r^n$.
A: The Cayley complex of $\mathbb{Z}/n\mathbb{Z}$ contains $n$ vertices and $n$ edges connected in a circle. So if you want to attach a disk, you have $n$ choices because $(123),(231),(312)$ gives three different attaching disks. For example in the $\mathbb{R}\mathbb{P}^{2}$ case, you glue a disk along the relation $(12)$, and another one along the relation $(21)$. The resulting complex is just the universal cover - the sphere. 
This is hard to visualize for $n$ large, and as Hatcher commented the resulting complex is not a surface in general. For a picture and more elementary explanation you may check Page 52, where Hatcher constructed $X_{G}$ explicitly. That's why Hatcher say $\widetilde{X}_{G}$ consists of $n$-disks with their boundary circles identified. 
The $\mathbb{Z}_{2}*\mathbb{Z}_{2}$ case is similar, but here I think Hatcher is only trying to kill off enough relations so that the resulting space is simply connected. He observed that all the relations (or circles) near a point must pass through either of the two boundary circles, so it is enough to "fill in" four disks to "kill" the 2 circles. If one did this to every vertex then one got his complex. 
