# free products and group actions

###### Theorem 1.

(See Lang, Exercise 54 p. 81) Suppose $G_{1},\ldots G_{n}$ are subgroups of $G$ that generate $G$. Suppose further that $G$ acts on a set $S$ and that there are subsets $S_{1},S_{2},\ldots S_{n}\subset S$, and some $s\in S-\cup S_{i}$ such that for each $1\leq i\leq n$, the following holds for each $g\in G_{i},g\neq e$:

• $g(S_{j})\subset S_{i}$ if $j\neq i$, and

• $g(s)\in S_{i}$.

Then $G=G_{1}\star\ldots\star G_{n}$ (where $\star$ denotes the free product).

Proof: Any $g\in G$ can be written $g=g_{1}g_{2}\ldots g_{k}$ with $g_{i}\in G_{j_{i}},j_{i}\neq j_{i+1},g_{i}\neq e$, since the $G_{i}$ generate $G$. Thus there is a surjective homomorphism $\phi:\coprod G_{i}\twoheadrightarrow G$ (since $\coprod G_{i}$, as the coproduct, has this universal property). We must show $\ker\phi$ is trivial. Choose $g_{1}g_{2}\ldots g_{k}$ as above. Then $g_{k}(s)\in S_{j_{k}}$, $g_{k-1}(g_{k}(s))\in S_{j_{k-1}}$, and so forth, so that $g_{1}(g_{2}(\ldots(g_{k}(s)\ldots)\in S_{j_{1}}$. But $e(s)=s\notin S_{j_{1}}$. Thus $\phi(g_{1}g_{2}\ldots g_{k})\neq e$, and $\phi$ is injective.

Title free products and group actions FreeProductsAndGroupActions 2013-03-22 17:34:56 2013-03-22 17:34:56 rm50 (10146) rm50 (10146) 5 rm50 (10146) Theorem msc 20E06