Free group of rank-$2$ has no non-trivial abelian normal subgroup

Question

I am doing this problem and write a solution. Can someone help me by giving some feedback? Thanks in advance.

Let $p:X\to \Bbb S^1\lor\Bbb S^1$ be a covering with $X$ path-connected and $\pi_1(X)$ is non-trivial abelian. Then, $p:X\to \Bbb S^1\lor\Bbb S^1$ is irregular.

$X$ is a graph, as it is a covering of a graph. The fundamental group of $X$ is a free group with some number of generators and this number is the same as the number of edges outside a maximal tree of $X$.

Also, $\pi_1(X)$ is non-trivial abelian implies there is exactly one edge outside a maximal tree of $X$, i.e. $\pi_1(X)=\Bbb Z$.

So, $X$ has the following form: Consider a circle and choose $n$-points on this circle, and add exactly two infinite-tree at each point. For $n=1,2,3$ some coverings are given.

Now, in each of these images consider two vertices, bold black, and bold green. There is no deck-transformation that interchange bold black and bold green vertices: after removing the green one there is a component with loop, on the other hand removing the black one each component is contractible.

Edit After Lee Mosher's Answer

Answer 1

The examples that you have worked out exhibit a pattern, of which you have stated some special cases. Here's the general pattern:

Proposition: If $X$ has an abelian fundamental group $\pi_1(X)$, then $X$ is homeomorphic to the quotient of a circle $C$ and a finite collection of trees $T_1,...,T_k$, by choosing one vertex $x_i \in T_i$ on each tree and a collection of distinct vertices $y_i \in C$, and identifying $x_i \sim y_i$ for each $i=1,...,k$.

And although in your application the graph $X$ is a connected covering space of $Y$, in fact this proposition holds in general for any connected graph $X$. The proof is not hard, and uses only facts about connected graphs and their fundamental groups.

Furthermore, once you've proved that proposition, then you can easily prove the additional fact that any graph automorphism of $X$ takes $C$ to $C$ (Hint: For each open edge $e \subset X$, $X-e$ is connected if and only if $e \subset C$). Thus, knowing in your special case that the set of trees $T_i$ is not trivial, you can conclude that the group of graph automorphisms of $X$ does not act transitively on the vertex set of $X$. It follows that the deck transformation group of the covering map $X \mapsto Y$ does not act transitively on the vertex set of $X$.