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March 8th, 2010, 04:08 PM  #1 
Senior Member Joined: Nov 2009 Posts: 129 Thanks: 0  group action on set
Let X be a Gset. Show that G acts faithfully on X if and only if no two distinct elements of G have the same action on each element of X. thanks. 
March 9th, 2010, 07:17 AM  #2 
Senior Member Joined: Apr 2008 Posts: 435 Thanks: 0  Re: group action on set
I've seen many of your topics before, and my first response is always the same: What have you got so far? I only say this because these look much like homework problems, and I don't want to do others' homework. 
March 9th, 2010, 11:33 AM  #3 
Senior Member Joined: Nov 2009 Posts: 129 Thanks: 0  Re: group action on set
The book says If N= {e}, then the identity element of g is the only element that leaves every x in X fixed, we then says that G acts faithfully on X. This is what I try: Let g_1,g_2 in G and x in X, => (g_1e)x=x for each x in X and (g_2e)x=x for each x in X. then g_1=g_2 since they are not distinct elements in G. <= If no two distinct elements g_1 and g_2 of G, (g_1e)x=(g2e)x = (g_1e)xx^1 = (g_2e)xx^1 = g_1e=g_2e = e . 
March 10th, 2010, 11:05 PM  #4 
Senior Member Joined: Nov 2009 Posts: 129 Thanks: 0  Re: group action on set
the proof i did before was wrong. This is my new proof: we define ker = { g?Ggx=x for all x?X} = {e} => let g_1, g_2 ?G and x?X for any g_1=e and g_2=e there exist an x in X such that g_1x=x and g_2x=x. Therefore g_1x=g_2x <= if for no any two distinct g_1,g_2?G there exist an x in X such that g_1x=x and g_2x=x. 
March 11th, 2010, 05:54 AM  #5  
Senior Member Joined: Oct 2007 From: Chicago Posts: 1,701 Thanks: 3  Re: group action on set Quote:
I assume you are starting with the assumption that G is acting faithfully. You are trying to show that ifg_1x=g_2x, then g_1=g_2. Does this show that? Quote:
What are you trying to prove?  

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