
Abstract Algebra Abstract Algebra Math Forum 
 LinkBack  Thread Tools  Display Modes 
September 20th, 2017, 10:52 AM  #11 
Senior Member Joined: Jan 2015 From: usa Posts: 101 Thanks: 0 
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  
Senior Member Joined: Aug 2012 Posts: 1,956 Thanks: 547  Quote:
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: 101 Thanks: 0 
Sorry i just wanted to write a fast reply when you are still connected 
September 20th, 2017, 11:01 AM  #14 
Senior Member Joined: Aug 2012 Posts: 1,956 Thanks: 547  
September 20th, 2017, 11:03 AM  #15 
Senior Member Joined: Jan 2015 From: usa Posts: 101 Thanks: 0 
in that case the orbits of $X\times X$ are $\left\{[x,x]\right\}$

September 20th, 2017, 11:19 AM  #16  
Senior Member Joined: Aug 2012 Posts: 1,956 Thanks: 547  Quote:
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.  

Tags 
action, finite, group, set 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
example of a finite group....  rayman  Abstract Algebra  7  March 4th, 2012 05:41 AM 
Prove that this finite set is a group  nata  Abstract Algebra  2  October 30th, 2011 05:29 PM 
Let G be a finite group and h:G–>G an isomorphism, such that  johnmath  Abstract Algebra  2  November 27th, 2010 10:52 PM 
finite group  stf123  Abstract Algebra  1  September 28th, 2007 12:08 PM 
T'group finite (helpme please thanks)  sastra81  Abstract Algebra  1  January 9th, 2007 07:05 AM 