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Torsion group, torsion subgroup

  • Thread starter hkhk
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  • #1
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hkhk

if G= Z4 x Z what would be the torsion group T(G)?

and what is the factor group of G/ T(G) ?
 
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  • #2
quasar987
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What have you tried?
 
  • #3
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the cyclic group <(1,0)> is the torsion group
( (0,0) (1,0) (2,0) (3,0))
?
 
  • #4
quasar987
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That makes no sense. From the top.. the definition of the torsion T(G) of a group G is by definition the set [itex]T(G)=\{ g\in G : g^n=e\ \mbox{for some n\in\mathbb{N}} \}[/itex] where e denotes the identity element in G. That is to say, it is simply the set of elements that have finite order!

Do you know any group in which every element has finite order? If so, that will be a torsion group.

Do you know any group in which no element other than the identity has finite order? If so, that will be a torsion free group.
 
  • #5
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1. then Q is a torsion free abelian group . true?

2. i was trying to find a set of elements of G= Z4 x Z that has finite order
 
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  • #6
quasar987
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What do these i) and ii) refer to?
 
  • #7
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i edited the question
 
  • #8
quasar987
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My post was in reference to question 1. If you can't think of a group in which every element has finite order (respectively one in which no element other than the identity has finite order), think of the groups you know and find T(G) for them.
 
  • #9
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Z4 is a group of finite order
and Z is not
so Z4 is a torsion group, while Z is torsion free, am i on the right track

sorry i do not have a very strong background in this topic and i am teaching myself this so that i can understand more advanced math
thanks a lot for your help
 
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  • #10
quasar987
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Well you're absolutely right: Z is torsion free since as is well known of anyone, mn=0 (m>0) cannot happen for n other than 0. And Z4 is torsion since every nonzero element in Z4 has order 4.

As for the second question, what have you tried and where are you stuck?
 

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