problem 14 of section 1.2 from Hatcher

Question

Consider the quotient of a cube $I^3$ obtained by identifying each square face with the opposite square face via the right-handed screw motion consisting of a translation by one unit in the direction perpendicular to the face combined with a one-quarter twist of the face about its center point. Show this quotient space X is a cell complex with two 0-cells, four 1-cells, three 2-cells, and one 3-cell. Using this structure, show that $\pi_1(X)$ is the quaternion group of order eight.

I find that $\pi_1(X^1)$, the 1-skeleton, is $\mathbb{Z}*\mathbb{Z}*\mathbb{Z}$. The attaching 2-cells give three relations, and they simplify $\pi_1(X^1)/N$ into $$\pi_1(X,A)=\{\alpha, \beta\mid \alpha \beta\bar \alpha \bar \beta=0, \ \beta \bar \alpha \beta \bar \alpha=0. \}$$ It is not the quaternion group since we can see that $\alpha^n$ cannot be simplified, thus the group is not finite. I think something is wrong about my
computation. Can someone kindly helps?

Answer 1

Let's first draw a picture of the cube with its identifications:

$\hskip1.75in$

Edges with the same label are identified according to their direction, vertices with the same symbol are identified, and opposite faces are also identified.

First, note that $\pi_1(X^1) = \mathbf{Z} * \mathbf{Z} * \mathbf{Z}$, with generators $ab^{-1}, ac^{-1}, ad^{-1}$, say. To calculate the fundamental group, then, we consider what relations attaching 2-cells gives to our generators: $$\pi_1(X) = \frac{\langle ab^{-1},ac^{-1},ad^{-1} \rangle}{\langle bc^{-1}ad^{-1},ba^{-1}dc^{-1},ac^{-1}db^{-1}\rangle}$$ where the relations come from attaching the top, front, and right faces, respectively. Now let $i = ab^{-1}$, $j = ac^{-1}$, $k = ad^{-1}$; we then have a presentation $$\pi_1(X) = \frac{\langle i,j,k \rangle}{\langle i^{-1}jk,i^{-1}k^{-1}j,jk^{-1}i\rangle}$$ which we can rewrite as $$\pi_1(X) = \langle i,j,k \mid i = jk,\ j=ki,\ ij = k\rangle$$ We claim this is a presentation for the quaternion group. Multiplying $i = jk$ by $i$ and $ij = k$ by $k$, we get $i^2 = k^2 = ijk$. The middle relation says $j = ki = kjk$, and multiplying this by $j$ we obtain $j^2 = jkjk = ijk$. Since this process is reversible, our presentation is equivalent to $$\pi_1(X) = \langle i,j,k \mid i^2 = k^2 = j^2 = ijk \rangle$$ which is the presentation for the quaternion group listed on the Group Properties Wiki, for example.