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Is this ok to prove?

  1. Dec 4, 2004 #1
    so I need to show that in any group, it's elements and inverse has the same order.

    so can I say that since it is a group, we know that there exists a unique inverse for each element. So each element would have 1 inverse. And then, that means we have the same number of elements as number of inverses?

    does that work? or am I missing something?
     
  2. jcsd
  3. Dec 4, 2004 #2

    shmoe

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    Can you state the definition of the order of an element in a group?
     
  4. Dec 4, 2004 #3

    so the order is the number of elements in a group....

    is this suppose to be a hint? :)
     
  5. Dec 4, 2004 #4
    semidevil...

    The order of an element g in a group G is n such that g^n = identity where n is the smallest positive integer >= 1. To prove an element and its inverse have the same order you can say:

    g^n = identity
    (g^-1) ^ n = (g^n)^-1 = (identity)^-1 = identity
     
  6. Dec 4, 2004 #5

    shmoe

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    That's the order of the group, not an element. I was hoping to get you to check the definition carefully;).
     
  7. Dec 4, 2004 #6
    Heh, sorry shmoe...didn't mean to steal your thunder.
     
  8. Dec 4, 2004 #7

    shmoe

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    No thunder to be stolen, we're all here to help (or be helped) :smile:
     
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