Fixed points of a group-operation on a tree, Serre's book "Trees" 6.3.4. and Prop 27 Hello! I have a problem with the following Lemma, which is mentioned in Serre's book "Trees" on page 60. In the book it is the Example 6.3.4.:
Lemma: Let $G$ be a group acting (without inversion) on a tree $X$. Let $X^G$ be the set of fixed points of $G$ in $X$ ($X^G$ is a subgraph of $X$). Let $G'$ be a subgroup of finite index in $G$ with $X^{G'}\neq\emptyset$. Then $X^G\neq\emptyset$.
Proof: Let $H$ be a normal subgroup of finite index in $G$ contained in $G'$ (for example, the intersection of the conjugates of $G'$). We have $X^H\neq\emptyset$ and $G/H$ acts on the tree $X^H$.


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*since $G/H$ is finite, it has a fixed point, whence $X^G\neq\emptyset$.


Question 1: Why is the index of $H$ in $G$ finite? Couldn't it happen, that the intersection of all conjugates of $G'$ equals the trivail group in $G$?
Question 2: If $G/H$ is finite, why it is clear that the action of $G/H$ has a fixed point in $X^H$?
Question 3:(Proof of Prop. 27, page 65)  If we look at the situation where $G$ is a fin. generated nilpotent group, we can choose $H$ such that $G/H$ is cyclic (not necessary finite, i think). Now let $X^H\neq\emptyset$. Then in the book Serre concludes, that $G/H$ has a fixed point and whence $X^G\neq\emptyset$.
Question 2 and 3 are on the same conclusion, i think. It seems like he use the same argument. But which one is it?
Thanks for thinking about it and help.
 A: The first two questions have been answered. The third question is also easy. If $G/H$ is cyclic and we assume that $H$ has fixed points, let $T$ be the subtree of fixed points of $H$. Then $G/H$ acts on that tree. Since $G/H$ is cyclic and the action is without inversions, it either has a fixed point, whence $G$ has a fixed point or it has a stable line $l$ (prove it!). The first option leads to a fixed point for the whole $G$ and the second option leads to a stable line for the whole $G$ on which $G$ acts as $G/H$ (since $H$ fixes the line pointwise). 
I guess the confusion came from not reading the whole statement of Serre. Serre does not claim that $G/H$ always fixes a point (that would be silly since $\mathbb Z$ acts on the tree $\mathbb R$ by translations). He only says that either it fixes a point or stabilizes a line in $X^H$.  
A: I guess question 2 has been answered in detail (though the proof given is for cyclic finite groups ?), but I think Serre also proves it in his book explicitly on Page 36, prop 19 (If a group G acts on a tree X without inversion, then X having an invariant vertex under G action, is equivalent to saying the set G.P is bounded for some vertex P). If G is finite, clearly G.P is a finite set and hence of finite diameter[I think we define two vertices to be connected, if they are joined by a finite path]
