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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.

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Your question 1 is a known exercise in group theory. See artofproblemsolving.com/Forum/viewtopic.php?t=19724 and artofproblemsolving.com/Forum/viewtopic.php?t=22796 . –  darij grinberg Sep 26 '11 at 16:27
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For questions 2 and 3, it might be useful to remind us of how Serre defines a tree (there seem to be multiple definitions of infinite trees) and what "without inversion" means (I hope it means something that implies that the order of $G/H$ is odd, since otherwise there can be counterexamples). –  darij grinberg Sep 26 '11 at 16:33
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OK, I have looked up in Serre what "tree" and "without inversion" mean. (For the record: A tree is a connected graph - possibly infinite, possibly even locally infinite - without cycles. An automorphism of a tree is said to be "without inversion" if no two vertices which it maps to each other are connected by an edge.) If $X^H$ is a finite subtree, then question 2 is answered by artofproblemsolving.com/Forum/viewtopic.php?f=42&t=277243 . –  darij grinberg Sep 26 '11 at 16:40
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Even if $X^H$ is not a finite subtree, we can reduce question 2 to the case of a finite subtree. In fact, let me show that any finite-order automorphism of a tree either has a fixed point or an inversion. To prove this, consider any vertex $x$ of the tree, and denote the automorphism by $f$. Let $F$ be the set of images of $x$ under $f^i$ for $i\in\mathbb N$. Since $f$ is finite-order, this set $F$ is finite. Also, $F$ is clearly an $f$-invariant set. Add to the set $F$ all intermediate vertices on the shortest paths connecting vertices of $F$ (pairwise). The resulting set is still ... –  darij grinberg Sep 26 '11 at 16:46
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... $f$-invariant, because the image of an intermediate vertex on the shortest path connecting two vertices $x$ and $y$ under $f$ must itself be an intermediate vertex on the shortest path connecting the vertices $f\left(x\right)$ and $f\left(y\right)$. Also, the resulting set is connected, as can be easily seen. Thus, we have obtained a connected finite $f$-invariant subset of the vertex set of our tree. In other words, we have obtained a finite $f$-invariant subtree. Now apply the theorem from artofproblemsolving.com/Forum/viewtopic.php?f=42&t=277243 (note that I also ... –  darij grinberg Sep 26 '11 at 16:48

2 Answers 2

up vote 2 down vote accepted

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 $X^H$.

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Nice. Yes. In my conclusion, i always thought, that in our counter example we only have option to get this fixed point. But you right, we still have in the case of $G/H$ the second option, which is immediately clear, since if $G/H$ is infinite it is isomorphic to $\mathbb Z$. Thank you very much! It was an error in reosoning of me. –  Eric Sep 27 '11 at 8:05
    
The book said, you should apply Porp. 26. but you have to use Prop 25. then your argumentation is immedietaly clear. –  Eric Sep 27 '11 at 8:11

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]

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