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Homework Help: Abstract algebra question concerning center of a group

  1. Nov 30, 2011 #1

    xcr

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    1. The problem statement, all variables and given/known data

    If a is the only element of order 2 in a group G, prove that a is an element of Z(G).
    [Z(G) is the notation used by the book for center of group G]


    2. Relevant equations

    Z(G)={a is an element of G: ag=ga for every g that is an element of G}


    3. The attempt at a solution

    I know that if a has order 2 (|a|=2) then a ≠ the identity of the group, say e, and a=a^-1.
    I just don't see where I would go from here in showing the center of a group.
     
  2. jcsd
  3. Nov 30, 2011 #2
    What is the order of [itex]gag^{-1}[/itex]??
     
  4. Nov 30, 2011 #3

    xcr

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    I would say two but I don't really have any reasoning for saying that...
     
  5. Nov 30, 2011 #4

    xcr

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    Actually, after looking at it, I would say that the order of gag^-1 is 2 because if the order of a is 2, then (a^2)=e. So (gag^-1)^2=(g^2)(a^2)(g^-2)=(g^2)(e)(g^-2)=(g^2)(g^-2)=e
     
  6. Dec 1, 2011 #5
    Indeed. Now use that there is only one element of order 2...
     
  7. Dec 1, 2011 #6

    xcr

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    Still don't see where you are going with it
     
  8. Dec 1, 2011 #7
    There is only one element of order 2. What can you conclude??
     
  9. Dec 1, 2011 #8

    xcr

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    That the element is not the identity and it is also its inverse.
     
  10. Dec 1, 2011 #9
    You have found that both a and [itex]g^{-1}ag[/itex] are elements of order 2.

    But the question states that there is ONLY ONE element of order 2. So what can you conclude??
     
  11. Dec 1, 2011 #10

    xcr

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    Then a=gag^-1. So multiplying on the right by g would give me ag=ga, ta-da
     
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