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Prove every even ordered group has an element of order 2

  1. Nov 16, 2005 #1
    Can someone help me on where to begin? What do I know about even ordered groups that could help?


    My first idea was to incorporate the fact that for an element to be of order 2, it must be it's own inverse. (This made me think of the identity element- I don't know if that's what the proof is about). Secondly, I thought of using a cayley table for arbitrary operation, and for arbitrary amounts of elements. The only problem there is I didn't know how to fill out the table properly, as there was no operation, and I had no idea how to incorporate the fact that G is even ordered into the table.
     
  2. jcsd
  3. Nov 16, 2005 #2

    shmoe

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    The inverse idea is the right track. What would happen if you had no element of order two?
     
  4. Nov 16, 2005 #3
    Well... if there were no element of order 2, then there would be no element that was its own inverse. Sorry, but I don't know where to go from here... what i think up next seems awkward to me.

    I think that you're trying to get me to show that G would be odd ordered, but I'm still unsure of how. What is a good example of an odd ordered group?
     
  5. Nov 16, 2005 #4

    shmoe

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    The integers modulo any odd number give an odd group (under addition), [tex]\bbmath{Z}_n[/tex].

    Just think about the elements that aren't their own inverse for the moment. How many are there? Specifically are there an even number or an odd number of them?
     
  6. Nov 17, 2005 #5
    i guess you can't use cauchy's theorem...?
     
  7. Nov 17, 2005 #6

    shmoe

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    This is probably from an earlier chapter. Cauchy isn't necessary here (though of course works).
     
  8. Nov 17, 2005 #7
    i thought so, it seemed a little too easy
     
  9. Nov 17, 2005 #8

    matt grime

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    Suppose I tell you in a set S there is a way to pair up some of the elements of S exactly. What can you say about the number of elements paired up? Can you relate this to the inverses? What elements don't pair with something different? (And yes it's good you've spotted the identity element as being special)
     
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