Describe all path-connected $3$ fold covers of $X=S^1 \vee \Bbb{R}P^2$ (please justify why your list is exhaustive). Which are regular (i.e. normal) and why?
I get that $X$ has a universal cover, hence the covers would be in one-to-one correspondence with subgroups of $\pi_1(X)$. Now, $\pi_1(X)=\Bbb{Z} * \Bbb{Z}_2$. So, I basically need to find the subgroups of $\Bbb{Z} * \Bbb{Z}_2$ of index $3$ and the normal covers would correspond to normal subgroups. But how do I find the normal subgroups of $\Bbb{Z} * \Bbb{Z}_2$? What is the general structure of subgroups of $\Bbb{Z} * \Bbb{Z}_2$ anyway?
Let $p : E\rightarrow B$ be a 3 fold cover, where $B = S^1 \vee \mathbb{R}P^2$. Let $x\in B$ be the basepoint. As $p$ is a 3-fold cover, $|p^{-1}(x)| = 3$ and $\pi_1(B,x)$ induces a right action on the set $p^{-1}(x)$. If this action is transitive, then the covering space $E$ is connected.
So the question becomes, in how many ways can $\mathbb{Z}*\mathbb{Z}_2$ act transitively on the set $\{1,2,3\}$. We may use the following presentation for our group: $$\langle a,b | b^2 = e \rangle $$
Now we enumerate the possiblities of where the elements can map too.
$\textbf{Case 1:}$ $1*b = 1$, $2*b = 2$.
As $b$ has order 2, this forces the action of $b$ to be trivial. Now if $a$ was also trivial, the action would not be transitive, and the cover wouldn't be connected. Hence $1*a = 2$ or $1*a = 3$. Making this choice forces you to choose $a^2$ and $a^3$ as well, with $a^3 = id$. Both of these actions determine homeomorphic covering spaces; the circle with three vertice, and at each vertex, a copy of $\mathbb{R}P^2$.
$\textbf{Case 2:}$ $1*b = 2$, $2*b = 1$. Again, if $a$ acts trivially, the action can't be transitive. Now suppose $ 1*a = 1$, then $2*a = 3$ and $3*a = 2$. This corresponds (up to homeomorphism) to the covering space given by a 2-sphere with two vertices (say at the north and south pole). At the north pole, there is a single circle. at the south pole, two arcs, with opposite end points for both arcs residing in a copy of $\mathbb{R}P^2$.
$\textbf{Case 3:}$ Now if $a*1 = 2$ instead, then $a*2 =3$. If it didn't, then both $a$ and $b$ would leave the point $3$ fixed, and hence the action wouldn't be transitive. So, This covering space is given by a two sphere with two vertices, and a copy of $\mathbb{R}P^2$ with a single vertex. The vertex on the $\mathbb{R}P^2$ corresponds to the point $3$ in our set. There are two arcs emminateing from this vertex and each arc terminates at a different pole of the 2-sphere. Finally, there an an arc connecting the two vertices of the two sphere.
Up to homeomorphism, these are the three different connected 3-fold covers of $S^1 \vee \mathbb{R}P^2$. To answer part two, try and draw the covers, then look for the ones which are symmetric.