This is a question related to the exercise 2218 from the book "Problems and Solutions in Mathematics" by Ta-Tsien, $2^{nd}$ Ed.
Let $Z$ denote the figure 8 space, $Z = X \vee Y$, $X$ and $Y$ circles. Les $x,y \in \pi_1(Z,*)$ be the elements in $Z$ defined by $X$, $Y$, where $*$ denotes the vertex.
Part of the first question is:
Let $h:\pi_1(Z,*) \rightarrow \mathbb{Z}/6\mathbb{Z}$ be the homomorphism satisfying $h(x)=2$ and $h(y)=3$, and let $p:\tilde{Z}\rightarrow Z$ denote the covering space corresponding to the kernel of h. ($p_*(\pi_1(\tilde{Z},*)))=ker(h).$)
And, in the solution, the following statement is made:
Since $h(x)=2$ and $h(y)=3$ and $\pi_1(Z,*)$ is generated by $x$ and $y$, it follows that the homomorphism $h$ is surjective. Thus $\pi_1(Z,*)/ ker(h)$ is isomorphic to the group $\mathbb{Z}/6\mathbb{Z} = \mathbb{Z}_6$.
Followed by the statement I do not understand:
Hence the covering space $p:\tilde{Z}\rightarrow Z$ is a 6-fold cover.
How do you deduce this last statement from the previous one ?
For every regular covering map $p$ with the covering group $G$, by the definition of $G$ we have that $p$ is n-to-one map where $n$ is the order of $G$.