Determine the covering space of $S^1 \vee S^1$ corresponding to subgroup of $\pi_1(S^1 \vee S^1)$ generated by cubes of all elements.

Question

This is exercise 1.3.13 in page 80 from Allen Hatcher's book Algebraic topology.

Determine the covering space of $S^1 \vee S^1$ corresponding to subgroup of $\pi_1(S^1 \vee S^1)$ generated by cubes of all elements.

Answer 1

Let $G$ be the subgroup of $\pi_1(S^1\vee S^1)$ generated by cubes. First, note that $G$ is a characteristic subgroup, and in particular a normal subgroup. The quotient $\pi_1(S^1\vee S^1)/G$ is thus the Burnside group $B(2,3)$: the group with two generators modulo the relations that the cube of every element is $1$. Thus the covering space associated to $G$ can be identified with the Cayley graph of $B(2,3)$ with respect to its two canonical generators.

To be more explicit about the group $B(2,3)$, according to this source, $B(2,3)$ has $27$ elements and is isomorphic to the group of matrices of the form $\begin{pmatrix} 1 & x & y \\ 0 & 1 & z \\ 0 & 0 & 1 \end{pmatrix}$ for $x,y,z\in\mathbb{F}_3$. I would assume the two generators can be taken to be $\begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$ and $\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}$, but I have not checked the details.