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 September 20th, 2017, 10:52 AM #11 Senior Member   Joined: Jan 2015 From: usa Posts: 104 Thanks: 1 So the full answer is the following?: If $\{[x_i]\}$ is a complete set of orbits of $X$, then $\{[(x_i, x_j)]\}$ is surely a distinct set of $m^2$ orbits of $X \times X$. So $m^2 \leq M$. If the action is trivial then each orbit consists of exactly one element: $[x] = \{x\}$. Then $X$ has $|X|$ distinct orbits and $X \times X$ has $|X|^2$. Then $m^2=M$. Now let us prove that $m^2=M$ implies that G acts trivialy on X: Let $\displaystyle x_i\in X$. $\displaystyle x_i$ belongs to one and only one orbit. And let $\displaystyle m^2=M$. Let $\displaystyle \mathcal{O}_{ij}=\{ [(x_r,\ x_s)]\ | \ x_r\in[x_i], \ x_s\in [x_j]\}$. The number of all $\displaystyle \mathcal{O}_{ij}$ is $\displaystyle m^2$. And every $\displaystyle \mathcal{O}_{ij}$ contains at least one orbit $\displaystyle [(x_i,\ x_j)]$. So, $\displaystyle \mathcal{O}_{ij}$ contains unique orbit $\displaystyle [(x_i,\ x_j)]$, and the orbit has to contain all the pairs $\displaystyle (x_r,\ x_s)$ where $\displaystyle x_r\in[x_i], \ x_s\in [x_j]$: $\displaystyle \mathcal{O}_{ij}=\left\{ \{(x_r,\ x_s) \ | \ x_r\in[x_i], \ x_s\in [x_j] \} \right\}.$ Suppose, there are $\displaystyle x_i, x_k\in [x_i]$ such that $\displaystyle x_i\neq x_k$. Then there has to be such $\displaystyle g\in G$ such that $\displaystyle g\left((x_i,\ x_i)\right)=(x_i,\ x_k)$. But it is impossible because the map $\displaystyle x\mapsto gx, \ \ x\in X, \ \ g\in G$ is bijective. So, there is a contradiction, and one has $\displaystyle x_i, x_k\in [x_i]$ implies $\displaystyle x_i=x_k$. Last edited by mona123; September 20th, 2017 at 10:58 AM.
September 20th, 2017, 10:56 AM   #12
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Quote:
 Originally Posted by mona123 So the full answer is the following?: If $\{[x_i]\}$ is a complete set of orbits of $X$, then $\{[(x_i, x_j)]\}$ is surely a distinct set of $m^2$ orbits of $X \times X$. So $m^2 \leq M$ for sure.
Copying my handwaving isn't a proof. I omitted the proof of this assertion, which you need to supply. Proof by "for sure."

Quote:
 Originally Posted by mona123 If the action is trivial then each orbit consists of exactly one element: $[x] = \{x\}$. Then $X$ has $|X|$ distinct orbits and $X \times X$ has $|X|^2$.
Can you prove that? It's obvious, yet your copy/paste leaves me wondering if you understand why it's true. Forgive me if I'm being demanding. At this level of math you need to force yourself to nail things down else you're doomed going forward.

Last edited by Maschke; September 20th, 2017 at 11:00 AM.

 September 20th, 2017, 10:59 AM #13 Senior Member   Joined: Jan 2015 From: usa Posts: 104 Thanks: 1 Sorry i just wanted to write a fast reply when you are still connected
September 20th, 2017, 11:01 AM   #14
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Quote:
 Originally Posted by mona123 Sorry i just wanted to write a fast reply when you are still connected
No prob. None of my business really.

 September 20th, 2017, 11:03 AM #15 Senior Member   Joined: Jan 2015 From: usa Posts: 104 Thanks: 1 in that case the orbits of $X\times X$ are $\left\{[x,x]\right\}$
September 20th, 2017, 11:19 AM   #16
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Quote:
 Originally Posted by mona123 in that case the orbits of $X\times X$ are $\left\{[x,x]\right\}$
No, there are only $|X|$ of those. Where are the other $|X|^2 - |X|$?

Like I say none of my business since AB Victor has already provided the solution and you didn't ask for advice in how to succeed group theory class. The only virtue of my comments is that if you can nail down this proof it will help you going forward. Start by writing down exactly what a trivial action is.

Just out of curiosity, what class is this? Group actions are covered in basic abstract algebra, but I never saw things like the dicyclic groups at that level.

Leaving for the afteroon, back later. Don't worry about when I'm here, I'm here most of the time.

Last edited by Maschke; September 20th, 2017 at 11:22 AM.

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