Successive quotienting of deck transformations Let $\widetilde{X}\to X$ be a covering space, and let $G=D(\widetilde{X}\to X)$ denote the group of deck transformations. In this question is is shown that the space $\widetilde{X}/G$ is also a covering space of $X$, so that there is a tower of covering spaces
$$\widetilde{X}\to\widetilde{X}/G\to X\,.$$
My question: does this process terminate? That is, given a covering space $\widetilde{X}\to X$, can I find a tower of covering spaces
$$\widetilde{X}=X_n\to X_{n-1}\to\cdots\to X_2\to X_1=X$$
such that
$$X_{n}=X_{n+1}/D(X_{n+1}\to X_n)$$
for all $n$?
For simplicity, I don't mind assuming that $X$ obeys various "niceness" conditions.
 A: Your question translates into a pure group theory question, like this.
Since we're starting from the top and working down, let me reverse the direction of your indexing, and ask whether there exists a tower of covering spaces
$$\widetilde X = X_1 \to X_2 \to \cdots \to X_{n-1} \to X_n = X
$$
such that $X_{n+1} = X_n / D(X_n \to X_{n+1})$.
Let me assume all the usual hypotheses needed to apply covering space theory: all spaces are path connected and locally path connected. Let me also assume that base points in $X$ and $\widetilde X$ are consistently chosen, and that as we work down the tower they continue to be consistently chosen, so I can leave them out of the notation for fundamental groups.
It follows that we have a tower of subgroup inclusions
$$G = \pi_1(X) = \underbrace{\pi_1(X_n)}_{H_n} > \underbrace{\pi_1(X_{n-1})}_{H_{n-1}} > ... > \underbrace{\pi_1(X_2)}_{H_2} > \underbrace{\pi_1(X_1)}_{H_1} = \pi_1(\widetilde X)
$$
having the property that $H_{i+1} = N_G(H_i)$ and $D(X_{i+1} \to X_i) = N_G(H_i) / H_{i} = H_{i+1} / H_i$. The notation $N_G(H)$ means the normalizer of $H$ in $G$, i.e. the largest subgroup of $G$ in which $H$ is normal, and so $N_G(H) / H$ is just the quotient group.
So, here's how to get a strong counterexample, namely an example such that the tower never even gets going. For this, we simply need an example of a group $G$ and a subgroup $H=H_1$ such that $N_G(H)=H$, and the tower never gets going because $D(\widetilde X \to X) = N_G(H)/H = \text{Id}$.
For such an example, take $G$ to be the rank $2$ free group and $H$ to be a rank $1$ free factor; it's well known that $H$ is its own normalizer in $G$.
We can thus take $X$ to be a rank $2$ rose and $\widetilde X$ to be the covering space corresponding to one of the two loops of the rose. This $\widetilde X$ looks like a copy of that loop with a copy of "half" of the universal covering tree of $X$ attached to a point of that loop.
