# Action on a doubly transitive group - induced action

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 ?!

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The question should say that $X$ has at least two elements. – Chris Eagle Oct 1 '12 at 22:34

## 1 Answer

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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Yes, as in the example you wrote yourself - if $|X|=1$ then $G$ could act transitively on $X^2$ – Dennis Gulko Oct 1 '12 at 22:41
I saw this, too. But in the question i am not given this assumption. Probably an incorrect question then. But thanks ! – Andre Oct 1 '12 at 22:42
And it should also be given that the action is non-trivial. So at least two mistakes in the phrasing of the question. – Dennis Gulko Oct 1 '12 at 22:44
True. Thanks for the corrections. I am new on this topic so i am glad to learn from you guys. – Andre Oct 1 '12 at 22:47
I 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. – Jason DeVito Oct 1 '12 at 22:52