free products and group actions

Theorem 1.

(See Lang, Exercise 54 p. 81) Suppose G1,Gn are subgroupsMathworldPlanetmathPlanetmath of G that generate G. Suppose further that G acts on a set S and that there are subsets S1,S2,SnS, and some sS-Si such that for each 1in, the following holds for each gGi,ge:

  • g(Sj)Si if ji, and

  • g(s)Si.

Then G=G1Gn (where denotes the free productMathworldPlanetmath).

Proof: Any gG can be written g=g1g2gk with giGji,jiji+1,gie, since the Gi generate G. Thus there is a surjectivePlanetmathPlanetmath homomorphismMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath ϕ:GiG (since Gi, as the coproductMathworldPlanetmath, has this universal propertyMathworldPlanetmath). We must show kerϕ is trivial. Choose g1g2gk as above. Then gk(s)Sjk, gk-1(gk(s))Sjk-1, and so forth, so that g1(g2((gk(s))Sj1. But e(s)=sSj1. Thus ϕ(g1g2gk)e, and ϕ is injectivePlanetmathPlanetmath.

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