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March 8th, 2010, 04:08 PM   #1
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group action on set

Let X be a G-set. 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.
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March 9th, 2010, 07:17 AM   #2
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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.
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March 9th, 2010, 11:33 AM   #3
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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=(g-2e)x = (g_1e)xx^-1 = (g_2e)xx^-1 = g_1e=g_2e = e .
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March 10th, 2010, 11:05 PM   #4
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Re: group action on set

the proof i did before was wrong. This is my new proof:
we define ker = { g?G|gx=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.
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March 11th, 2010, 05:54 AM   #5
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Re: group action on set

Quote:
Originally Posted by tinynerdi
=> 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
For any g_1=e? That just means "for e".
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:
<= 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.
This is not a statement. What is the assumption in the backwards direction?
What are you trying to prove?
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