2
$\begingroup$

Let $T$ denote the unique (countably infinite) tree in which every vertex has degree $3$ and let $G=Aut(T)$ be the group of graph automorphisms of $T$.

I'm interested in the properties of this group, especially in the normal subgroups of $G$. Is there a known classification of all normal subgroups? If not, what is known and what is conjectured about them?

Improve this question
$\endgroup$
3
  • $\begingroup$ Is your tree planar or abstract? $\endgroup$
    – Alex Degtyarev
    Commented Feb 21, 2014 at 16:55
  • $\begingroup$ Abstract. In particular $G$ is the full automorphism group of $T$. $\endgroup$
    – Dominik
    Commented Feb 21, 2014 at 16:57
  • $\begingroup$ You might want to look up branch groups; see for example the book by Bartholdi, Grigorchuk and Sunik arxiv.org/pdf/math/0510294.pdf. $\endgroup$
    – shane.orourke
    Commented Feb 21, 2014 at 19:21

2 Answers 2

Reset to default
11
$\begingroup$

Let $G_+$ be the index 2 subgroup of $G$, consisting of automorphisms which act on $T$ without inversions, which means that if such an automorphism preserves an edge, it fixes this edge pointwise. Equivalently, this is the subgroup of $G$ generated by (pointwise) edge-stabilizers. It follows from a paper by J.Tits in 1970 ("Sur le groupe des automorphismes d'un arbre") that the subgroup $G_+$ is simple. Therefore, this is the only proper normal nontrivial subgroup of $G$.

Edit. More precisely, $G_+$ is the maximal subgroup of $G$ whose every element acts on $T$ without inversions. This subgroup necessarily has index 2 in $G$. Equivalently, if we regard $T$ as a bipartite graph, then $G_+$ is the index 2 subgroup in $G$ preserving the parts of the vertex set.

Improve this answer
$\endgroup$
6
  • 1
    $\begingroup$ One could also have a look at Serre's book "trees": springer.com/mathematics/algebra/book/978-3-540-44237-0 $\endgroup$
    – Ian Agol
    Commented Feb 21, 2014 at 23:20
  • $\begingroup$ Thanks for the answer. But wouldn't it be easier to say that, given the unique 2-coloring of $T$, $G_+$ consists simply of the color-preserving automorphisms of $T$? $\endgroup$
    – Dominik
    Commented Feb 24, 2014 at 20:58
  • $\begingroup$ @Dominik: Yes, it is equivalent, just the matter of taste. $\endgroup$
    – Misha
    Commented Feb 24, 2014 at 22:07
  • $\begingroup$ The first given definition of $G_+$ is not correct. One correct definition of $G_+$ is that it is the subgroup of those $g$ such that for all $x$ we have $d(x,gx)\in 2\mathbf{Z}$. One other is that it is the set of $g$ preserving a 2-coloring (a map from vertices to $\{0,1\}$ such that adjacent vertices have different colors) - of course there are 2 2-colorings, not just 1, and this is why this subgroup has index 2. The coset $G-G^+$ contains loxodromic elements, which have no inversion. $\endgroup$
    – YCor
    Commented Jan 16, 2018 at 1:23
  • 1
    $\begingroup$ To conclude, the famous notion of groups acting on a tree without inversion is not a natural notion. Essentially, it has only been coined so as to ensure that if there's a bounded orbit then there's a fixed vertex. The natural notion is the notion of bipartite-preserving action, i.e. an action through the group of index $\le 2$ of automorphisms preserving some bi-coloring. $\endgroup$
    – YCor
    Commented Jan 20, 2018 at 3:08
1
$\begingroup$

Back in 1983, N. Gupta and S. Sidki (Math Z.) were constructing finitely generated infinite p-groups as groups of automorphisms of p-regular trees. There is a later treatment of their work by Gilbert Baumslag, but I don't have an immediate reference for his exposition. I think that Baumslag mostly looks at the special case where p=3. This work might be irrelevant to Dominik, but it might be of interest.

All trees are planar.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .