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Abelian group

  • Thread starter POtment
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  • #1
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Homework Statement


Let G be an abelian group. Show that the set of all elements of G of order 2 forms a subgroup of G. Find all elements of order 2 in Z6.

The Attempt at a Solution


The elements of Z6 are 1,4,5. I'm not sure how to find the set of all elements of order 2. Can someone help with that? I think I can prove from there.
 

Answers and Replies

  • #2
Tom Mattson
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The Attempt at a Solution


The elements of Z6 are 1,4,5.
Nope. How did you come up with that? And are you talking about the additive group [itex]\mathbb{Z}_6[/itex], or the multiplicative group [itex]\mathbb{Z}_6[/itex]? It does make a difference.
 
  • #3
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We are talking about the additive group of Z6. Can you nudge me in the right direction to find the correct elements?
 
  • #4
Dick
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If it's additive then an element g has order 2 if g+g=0 and g is not zero. It should be pretty easy to find them. But why do you say the elements are 1,4,5???
 
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  • #5
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OK. I understand what you are saying and know how to construct the table to find the elements of order 2. However, I still don't understand what the elements of Z6 are. Is it as obvious as 0,1,2,3,4,5?
 
  • #6
Dick
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It's as obvious as that, yes. So for which ones is g+g=0 mod 6?
 
  • #7
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(0,0), (3,3)
 
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  • #8
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This one is solved. Thanks for nudging me in the right direction!
 
  • #9
Dick
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I didn't ask for which pairs is (a,b) is a+b=0. Why would I?? Look up the definition of 'order 2'. Now write it on a blackboard ten times. Now step back and look at it. Now tell me why most of the pairs in your 'guess' aren't interesting.
 
  • #10
Dick
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(0,0), (3,3)
Better. So the subgroup is made of the elements 0 and 3.
 

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