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 $g\in G_{i},g\neq e$ :

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$g(S_{j})\subset S_{i}$ if $j\neq i$ , and
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$g(s)\in S_{i}$ .

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
Canonical name	FreeProductsAndGroupActions
Date of creation	2013-03-22 17:34:56
Last modified on	2013-03-22 17:34:56
Owner	rm50 (10146)
Last modified by	rm50 (10146)
Numerical id	5
Author	rm50 (10146)
Entry type	Theorem
Classification	msc 20E06