I'm reading the book "Trees" by Serre and I'm having doubts about my understanding of the following statement.
In page $25$ it says the following "A tree of representatives of $X$ mod $G$ is any subtree $T$ of $X$ which is the lift of a maximal tree in $G/X$; because prop. 11 of no 2.3 , every orbit of $G$ in $\text{vert}(X)$ contains exactly one vertex of $\text{vert}(X)$."
Here prop. 11 says that a maximal subtree in a connected graph that contains every vertex of such graph and $G$ acts without inversion.
I want to check if what I understood is correct. I understood that because the tree is maximal in $G/x$ it contains every representative of the orbits (vertexs in $G/X$), then a tree of representatives must contain exactly one vertex of each orbit. Really I don't understand the last statement about the orbits of $G$ in $\text{vert}(X)$, so I just interpreted it the way it would make sense to me.
Also does "$X$ mod $G$" stand the same as $G/X$?
If my interpretation is somewhat correct I would love any correction and if it's not right I'd really want to get your understanding and intuition of the statement.
First, in my copy of Trees (Springer Monographs in Mathematics, corrected second printing 2003), Proposition 11 has nothing about actions. This is what it says (in page 21):
Proposition 11 is in Chapter 1, Section 2, which does not have any material about actions of groups. Acting without inversions is introduced at the beginning of Chapter 1, Section 3, Subsection 3.1, on page 25. So your paraphrase of Proposition 11 is incorrect, by including extraneous material that has nothing to do with that proposition ($G$ acting without inversion). That isn't going to help with your understanding.
(As a minor note, the notation that Serre uses for the quotient graph is "$G\backslash X$", not $G/X$ as you do...)
The statement in question is:
Proposition 14, before this statement, tells us that if $X$ is a connected graph and $G$ acts without inversion, then every subtree of $G\backslash X$ lifts to a subtree of $X$.
Now recall that the vertex set of $G\backslash X$ are the orbits of the action of $G$ on $X$ (the equivalence classes of vertices under the equivalence relation $x\sim y\iff \exists g\in G(gx=y)$) and the edges of $G\backslash X$ are the orbits of edges under the action of $G$.
Given a vertex $\mathbf{x}$ of $G\backslash X$, a "preimage" of $\mathbf{x}$ is then a single vertex $x\in X$ that lies in the orbit $\mathbf{x}$.
That means that if $G$ acts without inversions on a connected graph $X$, then we can take a maximal tree of $G\backslash X$ (because every graph has maximal trees), and lift it to a tree $T$ of $X$. By definition, each orbit corresponds to a single vertex in $G\backslash X$, so the tree $T$ contains one vertex from each orbit.
That is, it contains one representative from each and every orbit, not "contains every representative of the orbits", which would mean that it contains the entire orbit (that could be a problem with how you are expressing yourself in English rather than a misunderstanding of the issue; like saying "Soy una estación the tren" (I am a train station) instead of "Estoy en una estación de tren" (I am at a train station) not reflecting a mistaken belief that you are a train station, but rather reflecting an error in translation of "I am at" as "I am"). So, yes, a tree of representatives would necessarily contain one and only one vertex from each orbit of the action.
And, no, "$X$ mod $G$" is not the same as $G\backslash X$. Note that $G\backslash X$ is a specific graph. Wheras here we are talking about "representatives of $X$ mod $G$", which is a collection of vertices that represent element of $X$ up to equivalence modulo $G$. If you said "representatives of $G\backslash X$", you would be saying "representatives of the quotient graph", whereas here we are talking about "representatives of the graph up to equivalence modulo $G$". You are talking about different kinds of objects. Here, "$X$ mod $G$" is an identifier, not an object, whereas $G\backslash X$ is an object.