Assume that the group $G$ acts on $X$ . Then define $\forall x,y \in G \forall g \in G: g(x,y) := (g(x),g(y))$, so that $G$ acts on $X^2$ by this action. I have to prove that $G$ has at least two orbits on $X^2$. How can i do that ?
Personally i think that this is strange because assume $X = \{1\}$. Then $\{e\}$ acts on $X$ by $\forall x \in X\forall g\in G: g(x) := 1$. Then i just can imagine the orbit $G(1,1)$ which is $X^2$. What is wrong here or am i right ?!
Action on a doubly transitive group - induced action
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group-theory
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0The question should say that $X$ has at least two elements. – 2012-10-01
1 Answers
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Assume that $X$ has at least two elements and that $G$ acts non-trivially. Observe that the orbit of the point $(x,x)\in X^2$ and the orbit of $(y,z)\in X^2$ (for any $y\neq z$) will be distinct. Hence you have at least two orbits.
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0Yes, your'e right, but then the question is not very interesting ;-) – 2012-10-02