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 $X^H$.

- 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.