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, as in the example you wrote yourself - if $|X|=1$ then $G$ could act transitively on $X^2$ – 2012-10-01
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0I saw this, too. But in the question i am not given this assumption. Probably an incorrect question then. But thanks ! – 2012-10-01
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0And it should also be given that the action is non-trivial. So at least two mistakes in the phrasing of the question. – 2012-10-01
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0True. Thanks for the corrections. I am new on this topic so i am glad to learn from you guys. – 2012-10-01
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1I agree that $X$ needs at least two elements, but I don't think a trivial $G$ action is a problem. When the action is trivial, *every* point in $X\times X$ is its own orbit, so there are at least 2 of them. – 2012-10-01
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0Yes, your'e right, but then the question is not very interesting ;-) – 2012-10-02