simply transitive

Let $G$ be a group acting on a set $X$ . The action is said to be simply transitive if it is transitive and $\\forall x,y\\in X$ there is a unique $g\\in G$ such that $g.x=y$ .

Theorem.

A group action is simply transitive if and only if it is free and transitive

Proof.

Necessity follows since $g.x=x$ implies that $g=1_{G}$ because $1_{G}.x=x$ also. Now assume the action is free and transitive and we have elements $g_{1},g_{2}\\in G$ and $x,y\\in X$ such that $g_{1}.x=y$ and $g_{2}.x=y$ . Then $g_{1}.x=g_{2}.x\\implies g_{2}^{-1}.g_{1}.x=(g_{2}^{-1}g_{1}).x=x$ hence $g_{2}^{-1}g_{1}=1_{G}$ because the action is free. Thus $g_{1}=g_{2}$ and so the action is simply transitive.∎

单词	SimplyTransitive
释义	simply transitive Let $G$ be a group acting on a set $X$ . The action is said to be simply transitive if it is transitive and $\\forall x,y\\in X$ there is a unique $g\\in G$ such that $g.x=y$ . Theorem. A group action is simply transitive if and only if it is free and transitive Proof. Necessity follows since $g.x=x$ implies that $g=1_{G}$ because $1_{G}.x=x$ also. Now assume the action is free and transitive and we have elements $g_{1},g_{2}\\in G$ and $x,y\\in X$ such that $g_{1}.x=y$ and $g_{2}.x=y$ . Then $g_{1}.x=g_{2}.x\\implies g_{2}^{-1}.g_{1}.x=(g_{2}^{-1}g_{1}).x=x$ hence $g_{2}^{-1}g_{1}=1_{G}$ because the action is free. Thus $g_{1}=g_{2}$ and so the action is simply transitive.∎
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