free products and group actions
Theorem 1.
(See Lang, Exercise 54 p. 81) Suppose ${G}_{\mathrm{1}}\mathrm{,}\mathrm{\dots}\mathit{}{G}_{n}$ are subgroups^{} of $G$ that generate $G$. Suppose further that $G$ acts on a set $S$ and that there are subsets ${S}_{\mathrm{1}}\mathrm{,}{S}_{\mathrm{2}}\mathrm{,}\mathrm{\dots}\mathit{}{S}_{n}\mathrm{\subset}S$, and some $s\mathrm{\in}S\mathrm{}\mathrm{\cup}{S}_{i}$ such that for each $\mathrm{1}\mathrm{\le}i\mathrm{\le}n$, the following holds for each $g\mathrm{\in}{G}_{i}\mathrm{,}g\mathrm{\ne}e$:

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$g({S}_{j})\subset {S}_{i}$ if $j\ne i$, and

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$g(s)\in {S}_{i}$.
Then $G\mathrm{=}{G}_{\mathrm{1}}\mathrm{\star}\mathrm{\dots}\mathrm{\star}{G}_{n}$ (where $\mathrm{\star}$ denotes the free product^{}).
Proof: Any $g\in G$ can be written $g={g}_{1}{g}_{2}\mathrm{\dots}{g}_{k}$ with ${g}_{i}\in {G}_{{j}_{i}},{j}_{i}\ne {j}_{i+1},{g}_{i}\ne e$, since the ${G}_{i}$ generate $G$. Thus there is a surjective^{} homomorphism^{} $\varphi :\coprod {G}_{i}\twoheadrightarrow G$ (since $\coprod {G}_{i}$, as the coproduct^{}, has this universal property^{}). We must show $\mathrm{ker}\varphi $ is trivial. Choose ${g}_{1}{g}_{2}\mathrm{\dots}{g}_{k}$ as above. Then ${g}_{k}(s)\in {S}_{{j}_{k}}$, ${g}_{k1}({g}_{k}(s))\in {S}_{{j}_{k1}}$, and so forth, so that ${g}_{1}({g}_{2}(\mathrm{\dots}({g}_{k}(s)\mathrm{\dots})\in {S}_{{j}_{1}}$. But $e(s)=s\notin {S}_{{j}_{1}}$. Thus $\varphi ({g}_{1}{g}_{2}\mathrm{\dots}{g}_{k})\ne e$, and $\varphi $ is injective^{}.
Title  free products and group actions 

Canonical name  FreeProductsAndGroupActions 
Date of creation  20130322 17:34:56 
Last modified on  20130322 17:34:56 
Owner  rm50 (10146) 
Last modified by  rm50 (10146) 
Numerical id  5 
Author  rm50 (10146) 
Entry type  Theorem 
Classification  msc 20E06 