Automorphism groups of Graphs What are the automorphism groups of the following regular covering spaces?

I think the first picture is an 8-degree cover of the figure 8, whereas the second one is an infinite degree cover. I tried to use the theorem about regular regular covering maps which asserts that if $p: Y \rightarrow X$ is regular, then Aut($Y \rightarrow X$) $\cong \pi_{1}(Y,y_{0}) / p_{\#}\pi_{1}(X,x_{0})$
Thanks
 A: It appears that you misquoted the theorem: the conclusion should be $\operatorname{Aut}(Y \to X) \cong \pi_1(X,x_0) / p_* \pi_1(Y,y_0)$.  
But in fact, you can "eyeball" the automorphisms for these covers, using these two facts:


*

*A covering transformation is uniquely determined by where it sends a single point.  

*If $p: Y \to X$ is regular, then for each $x \in X$ and each pair of lifts $y, y' \in Y$ of $x$, there is a covering transformation taking $y$ to $y'$.  


So in your case the automorphism group is in one-to-one correspondence with the lifts of the basepoint in $X = S^1 \vee S^1$.  Let me denote the paths with a single arrowhead by $a$, and the paths with a double arrowhead by $b$.  
In (a), we have 8 vertices (= lifts of the basepoint in $X$), so the automorphism group has order eight.  Moreover, the automorphism group is generated by $a$ and $b$, where I now think of the paths as giving me a covering transformation: I can go from the basepoint to any other point using $a$ and $b$.  To figure out what the group is, I need to know: (i) what are the orders of $a$ and $b$, and (ii) how do $a$ and $b$ commute.  From the graph, I see that $a^2$ sends each vertex to itself, so corresponds to the identity transformation.  Similarly, $b^4$ is also the identity.  Finally, I also see that $abab$ is the identity, so $ab = b^3 a$.  Thus the group of automorphisms for the cover in (a) is $\langle a, b \mid a^2, b^4, (ab)^2 \rangle$.  This is the dihedral group of order 8.  
Similarly, I'll leave you to verify that in (b), the group of covering transformations is $\langle a, b \mid a^2, aba^{-1}b^{-1} \rangle \cong \mathbb{Z} \oplus \mathbb{Z}/(2)$.  (To see that there are no more relations, observe that every word in this group can be written uniquely as $a^\epsilon b^n$ where $\epsilon \in \{0,1\}$ and $n \in \mathbb{Z}$, and the bijection between this group and the lifts of the basepoints should become clear.)  
Having given this somewhat intuitive explanation, let me show how the theorem can be used.  For example, in (a), what are the (based) loops in the covering space $Y$, i.e., what is $\pi_1(Y)$?  Well, we know that $a^2$, $b^4$, and $(ab)^2$ are some of them.  It turns out that quotienting out these three relations, will quotient out all the other elements in $\pi_1(Y)$ as well.  So since $\pi(X) \cong \mathbb{Z} * \mathbb{Z} = \langle a, b \rangle$, we have $\operatorname{Aut}(Y \to X) \cong \langle a, b \rangle / \langle a^2, b^4, (ab)^2 \rangle$, which is what we found above.  
