I'm looking for a topological space $X$ and a covering space $(\tilde X,p)$ such that the deck transformation group Aut$_{X}(\tilde X)$ is isomorphic to $\mathbb{Z}/\mathbb{6Z}$.
My idea is to find out a topological space which fundamental group is isomorphic to $\mathbb{Z}$. Then I could seek a regular covering space $(\tilde X,p)$ such that $p_*(\pi_1(\tilde X)) \cong \mathbb{6Z}$. Finally, I know that if $(\tilde X,p)$ is a regular covering space then Aut$_{X}(\tilde X) \cong \pi_1(X)/p_*(\pi_1(\tilde X))$.
For example I could take $X= \mathbb{S}^1$ and $\tilde X=\mathbb{R}/\mathbb{6Z}$. Indeed $\mathbb{S}^1$ admits as unversal covering space $(\mathbb{R},q)$ with $q(t)=e^{2\pi i t}$. Now, the other covering spaces can be built as quotient between $\mathbb{R}$ and $G$ with $G$ subgroup of $\mathbb{Z}$, so $G=\mathbb{nZ}$ and $(\mathbb{R},r)$ is a covering space of $\mathbb{R}/G$. Thus, $r$ and $q$ induce a covering map $p$ between $\mathbb{R}/G$ and $G$ such that $p_*(\pi_1(\mathbb{R}/G))=G$ (this last part comes out from the proof of existence theorem for covering spaces). Therefore, as $Z$ is abelian and so $G$ is normal, Aut$_{X}(\tilde X) \cong \mathbb{Z}/G$ and we take $G=6\mathbb{Z}$ as desired.
Is this proof correct?
Is it possible to find other $X$ and $(\tilde X,p)$?
Best Answer
Your proof is correct and you see that for your example $X = S^1$ you get $\tilde{X} = S^1$ and $p(z) = z^6$.
However, there are infinitely many such examples. Using the universal covering, Lee Mosher has given a concrete example. What the universal covering space of $D^2 \cup_{z^6} S^1$ looks like was described in the answers to What is the universal cover of $S^1 \cup_f D^2$? (you have to adapt it to $f(z) = z^6$). For $X$ you can also take any other path-connected, locally path-connected and semilocally $1$-connected space with fundamental group $\mathbb{Z}_6$.
More generally the following is known (here we only consider coverings $p : \tilde{X} \to X$ such that $X$ is connected and locally path-connected and $\tilde{X}$ is connected):
The group $Deck(p)$ of deck transformations is isomorphic to the quotient group $\pi_1(X, x_0)/p_\ast(\pi_1(\tilde{X}, \tilde{x}_0))$ provided $p_\ast(\pi_1(\tilde{X}, \tilde{x}_0))$ is a normal subgroup of $\pi_1(X, x_0)$, where $\tilde{x}_0 \in \tilde{X}$ and $x_0 = p(\tilde{x}_0)$.
You can use this to construct a great variety of examples, because if $X$ has a universal covering (e.g. if $X$ is semilocally $1$-connected), then for any subgroup $G$ of $\pi_1(x,x_0)$ you find a covering $p : \tilde{X} \to X$ such that $G = p_\ast(\pi_1(\tilde{X}, \tilde{x}_0))$.