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

•
$g(S_{j})\\subset S_{i}$ if $j\eq i$ , 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 $g_{i}\\in G_{j_{i}},j_{i}\eq j_{i+1},g_{i}\eq e$ , 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 $e(s)=s\otin S_{j_{1}}$ . Thus $\\phi(g_{1}g_{2}\\ldots g_{k})\eq e$ , and $\\phi$ is injective.

单词	FreeProductsAndGroupActions
释义	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 $g\\in G_{i},g\eq e$ : • $g(S_{j})\\subset S_{i}$ if $j\eq i$ , 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 $g_{i}\\in G_{j_{i}},j_{i}\eq j_{i+1},g_{i}\eq e$ , 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 $e(s)=s\otin S_{j_{1}}$ . Thus $\\phi(g_{1}g_{2}\\ldots g_{k})\eq e$ , and $\\phi$ is injective.
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