Free product as automorphism group of graph

Question

Let $A$ and $B$ be two groups. We define following graph $X$. The set of vertices is the left cosets $gA$ and $gB$ where $g\in A*B$ (By $A*B$, I mean the free product of $A$ and $B$). The edges of the graph X correspond to the elements of $A*B$, and we use $e_g$ to denote the edge associated to $g\in A*B$. This does have the unfortunate consequence that the edge associated to the identity is $e_e$. Each edge is associated to an unordered pair of vertices: $\operatorname{Ends}(e_g) = \{gA,gB\}$. I want to show that $\operatorname{Aut}(X)=A*B$. I can show that $A*B$ is a subset of $\operatorname{Aut}(X)=A*B$, but I do not any idea for the converse.

Answer 1

This is not true, $\operatorname{Aut}(X)\neq A\ast B$. First, note that the tree you construct is a biregular tree, where $A$ acts by fixing a vertex and permuting its edges (and the trees extending from these edges). The group $B$ acts similarly, and this extends to a faithful action of $A\ast B$.

Theorem: $\operatorname{Aut}(X)\neq A\ast B$

To see that $\operatorname{Aut}(X)\neq A\ast B$, note that in general the tree which you construct contains $S_{|A|}\ast S_{|B|}$ in its automorphism group, because vertices have valency $|A|$ and $|B|$ respectively (and the adjacent edges can be be permuted). By a relatively simple argument, $S_{|A|}\ast S_{|B|}$ does not embed into $A\ast B$ (if either of $A$ or $B$ is infinite, just look at cardinality!). Hence, the isomorphism does not hold.

A slightly different, but fundamentally identical, way of seeing that the isomorphism cannot hold is as follows: Take two non-isomorphic groups $A_1$ and $A_2$ of the same order. Then $A_1\ast B\not\cong A_2\ast B$, but the graph $X$ you obtain is in each case the same. For example, $P=C_4\ast C_3$ and $Q=(C_2\times C_2)\ast C_3$ both give the same tree, and these actions extend to a natural, faithful action of $S_4\ast C_3$ on the same tree (where $S_4$ permutes the four adjacent edges of some valency $4$ vertex). Derek Holt pointed this out in the comments, below (which lead to the above proof).

Lee Mosher has pointed out in the comments that the following result also holds. This implies that above theorem.

Theorem: If one of $A$ or $B$ has order greater than two then $\operatorname{Aut}(X)$ has cardinality strictly greater than that of $A\ast B$.

Note that if both $A$ and $B$ have order precisely two then you are looking at a combinatorial line, which has automorphism group $A\ast B\cong D_{\infty}$.

One way of proving this result is as follows.

If either of $A$ or $B$ is infinite (suppose, without loss of generality, $A$ is infinite with $|A|\geq|B|$) then this holds because the above argument proves that $S_{A}$ embeds into $\operatorname{Aut}(X)$, and it is well-known that $|S_{A}|>|A|\geq|A\ast B|$.

Suppose both $A$ and $B$ are finite, with $|A|>2$. We wish to prove that $\operatorname{Aut}(X)$ is uncountable, so we shall find an uncountable subgroup of $\operatorname{Aut}(X)$. Begin by fixing a vertex $v$ which is associated to a coset of $A$, and enumerate all but one of its child vertices, labelling them $v_1, v_2, \ldots, v_{|A|-1}$. Note that these labelled vertices are $B$-vertices. A permutation of these children gives an element of $\operatorname{Aut}(X)$. We can do the same labelling with the $A$-descendants (and their children) of each of the $v_i$. Then, enumerate the $A$-vertices apart from in the subtree tree we are ignoring. Then the elements of the subgroup of $\operatorname{Aut}(X)$ which fixes the vertex $v$ and which fixes the subtree we are ignoring can thus be represented by an element of the Cartesian product $S_{|A|-1}\times S_{|A|-1}\times\cdots$. This Cartesian product is uncountable, hence $\operatorname{Aut}(X)$ is also uncountable.

Note that because the elements of $A\ast B$ are only of finite length, in the above argument elements of $A\ast B$ have finite support in the Cartesian product (and hence we do not violate the fact that there are only countably many elements of $A\ast B$).

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The comments below the question imply that the OP is interested in Theorem 3.28 (p81) of John Meier's book Groups, Graphs and Trees. The exact theorem is as follows. A biregular tree is an infinite bipartite tree where all vertices of a specific class have the same valancy.

Theorem Every free product of groups $A\ast B$ can be realized as a group of symmetries of a biregular tree $\mathcal{T}$, and a fundamental domain for this action consists of a single edge and its two vertices. Further, if $A$ and $B$ are both finite, this tree is $\mathcal{T}_{|A|, |B|}$.

However, this theorem is just saying $A\ast B\hookrightarrow \operatorname{Aut}(X)$. It says nothing about an isomorphism.