The graph of free product group.

Question

For the free product $A*B$=G, where $A$ and $B$ are groups, there is a graph defined by: the edge set E(G)$\backsimeq$G and the vertex set V(G)$\backsimeq$G $G/A \bigsqcup G/B$, and to g=$a_1b_1...a_kb_k$, the edge $e_g$ is defined by $(V_{gA},V_{gb})$. Also this graph is a tree. I am confused by these notations, do you know anything about this graph?

Answer 1

What you are looking for is the Bass-Serre tree of the free product decomposition. One can define it algebraically as is done in Serre's book "Trees", or topologically as is done by Scott and Wall.

From the algebraic perspective, the vertex set partitions into two subsets, one subset being in one-to-one correspondence with the collection of cosets $G/A$, the other in one-to-one correspondence with the collection of cosets $G/B$. It looks like the notation has been corrupted or miscopied in what you quote. It looks like the vertex corresponding to $gA \in G/A$ should be denoted $V_{gA}$, and the one corresponding to $gB \in G/B$ should be denoted $V_{gB}$.

Also, the edge set is in one-to-one correpondence with $G$ itself, the edge corresponding to $g \in G$ being labelled $E_g$. Finally, the endpoints of $E_g$ are $V_{gA}$ and $V_{gB}$.

Of course there is some work to do, in order to prove that this graph is a tree. For the proof, see Serre's book or the Scott-Wall paper.

Answer 2

If $G$ is a group and $S\subseteq G$ is a subset of $G$ which generates $G$ and such that $\{s^{-1}:s\in S\}=S$ (we say in this case that $S$ is symmetric) then the Cayley graph of $G$ with respect to $S$ is the graph $\mathscr C $ which has

• as set of vertices the set $G$

• for each $g\in G$ and each $s\in S$ an arrow $g\to sg$, which we usually label with the element $s$.

You seem to want to construct the Cayley graph of a free product $G=A*B$ of two groups, but as I mentioned in a comment above for this to make sense you need to specify, along with the group $G$, a subset of $G$ which generates it. As your question stands, therefore, it does not make sense.