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.
随便看	preservation of uniform integrability PresheafOfATopologicalBasis presheaf of a topological basis PrimalElement primal element Primality primality PrimalityCertificate primality certificate PrimalityTesting primality testing PrimaryDecomposition primary decomposition PrimaryDecompositionTheorem primary decomposition theorem primary decomposition theorem PrimaryDecompositionTheorem1 PrimaryIdeal primary ideal PrimaryPseudoperfectNumber primary pseudoperfect number Prime prime PrimeConstant prime constant