Action of a finite group on a tree

Question

The following is my homework question:

Let $G$ be a finite group and $A$ be a tree. Let $G$ acts on $A$ by graph automorphisms. Show that there exists a vertex or an edge that is fixed by all the elements of $G$.

I am considering the Cayley graph of $\mathbb{Z}$ with respect to the generating set $\{1\}$. It is a tree obviously. If $G$ acts on this Cayley graph with translations, there will be no fixed vertex or an edge.

What is wrong with my counter example?

Also any hints or ideas about the question would help a lot.

Thanks in advance.

Answer 1

Your example does not work because the group $G=\mathbb{Z}$ is infinite.

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To prove the result you are after, start with the following. Concluding from here is relatively straight forward; you need to use the fact that $G$ is finite to obtain a finite subtree $T$ of $A$ which $G$ acts on.

Lemma. If a group $G$ acts on a finite tree $T$, then $G$ fixes an edge or vertex of $T$.

Proof. As $T$ is finite, there are $3$ possibilities for the vertex structure of $T$:

1. $T$ contains vertices both of degree $>1$ and of degree $=1$.
2. $T$ only contains vertices of degree $1$ (i.e. two vertices connected by an edge).
3. $T$ is a single vertex.

Suppose $T$ contains vertices of degree $>1$, i.e. we are in the situation of (1). Then $G$ permutes the vertices of degree $>1$. That is, writing $T'$ for the minimal subtree of $T$ containing all vertices of degree $>1$ of $T$, then $G$ acts on $T'$.

Therefore, if we are in case (1) we can always obtain a smaller tree $T'$ on which $G$ acts. We can repeat this procedure on $T'$ and so on, so we get a chain $T\supset T'\supset T''\supset\cdots$. As $T$ is finite, this chain terminates at a tree $\hat{T}\subset T$ which $G$ acts on and which is either a two vertices connected by an edge (case (2)) or a single vertex (case (1)). Hence, we see that $G$ fixes an edge (case (2)) or a vertex (case (1)) of $T$. QED