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Infinite groups with elements of finite order

  1. Oct 4, 2011 #1
    Can anyone think of an example of an infinite group that has elements with a finite order?
     
  2. jcsd
  3. Oct 4, 2011 #2

    Ben Niehoff

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    "Infinite", as in having an infinite number of elements? The circle group U(1) comes to mind, having all of the cyclic groups Z_n as subgroups.
     
  4. Oct 4, 2011 #3
    How about (0,1) , in Z(+)Z/2 ?
     
  5. Oct 5, 2011 #4
    Yes, as in having an infinite number of elements. What if we wanted a group with order infinity that had a relatively small number of elements of finite order?
     
  6. Oct 5, 2011 #5
    Ben:
    I imagine eddy wants the elements to have finite order under the operation of the ambient group, not under the operation of the subgroups.
     
  7. Oct 5, 2011 #6

    Ben Niehoff

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    But in my example they are the same. For example, [itex]e^{i \pi /4}[/itex] has finite order in U(1), while [itex]e^{i \pi / \sqrt{2}}[/itex] has infinite order.
     
  8. Oct 5, 2011 #7
    How does the element 1 have finite order in the group you described Bacle?
     
  9. Oct 5, 2011 #8

    Ben Niehoff

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    Yet another construction is to take any finite group G and construct the group [itex]H = G \times \mathbb{Z}[/itex]. Then H has elements of finite order given by (g, 0), and all elements of the form (g, z) for [itex]z \neq 0[/itex] have infinite order.
     
  10. Oct 5, 2011 #9
    Can you explain this further Ben? The way I see it, since G is finite, when you add two elements of H the result might not be in H.
     
  11. Oct 5, 2011 #10

    Ben Niehoff

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  12. Oct 5, 2011 #11
    Eddyski3:

    The group Z(+)Z/2 is the direct sum of groups, which carries the operation of the first group in the first component, and the operation of the second group in the second component, so that:

    (a,b)+(a',b'):= (a+'Za', b+'Z/2b') , so you add the first

    components as if you were in Z, and the other components as if you were in Z/2,

    so (0,1)+(0,1)=(0+0,1+1)=(0,0).

    Ben:
    You're right, I missed your point; any rational multiple of pi will have finite order.
     
    Last edited: Oct 5, 2011
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