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Math Help - Find order of this element

  1. #1
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    Find order of this element

    Let G be a finite group that has the properties that it contains 2 elements a and b such that:

    1. a^5=e
    2. aba^- ^1    =b^2
    3. a ≠ e
    4. b ≠ e

    Find O(b).
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  2. #2
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    Note that aba^{-1}=b^2 \Rightarrow (aba^{-1})^2 = ab^2a^{-1}=b^4 \Rightarrow a(aba^{-1})a^{-1} = a^2ba^{-2}=b^4

    Can you continue?
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  3. #3
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    I'll give it another shot and report back
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  4. #4
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    Quote Originally Posted by Defunkt View Post
    Note that aba^{-1}=b^2 \Rightarrow (aba^{-1})^2 = ab^2a^{-1}=b^4 \Rightarrow a(aba^{-1})a^{-1} = a^2ba^{-2}=b^4

    Can you continue?
    I understand what you did here, but I'm not really seeing how I'll be able to get rid of all the b's just so it equals e...
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  5. #5
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    Quote Originally Posted by dynas7y View Post
    I understand what you did here, but I'm not really seeing how I'll be able to get rid of all the b's just so it equals e...
    Try applying the same 'process' I did to a^2ba^{-2}
    Last edited by Defunkt; October 4th 2010 at 09:15 AM.
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  6. #6
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    Ok, I've tried applying the process for others such as b^1^0, or really any n where b^n can be broken down into factors of b^2 and I think I'm just not seeing it. When attempting to find something such as b^5, if I broke that down into b and b^4, then the order would matter since the group is not necessarily abelian, right? I may just be missing something or it's not catching my eye but I don't see right now how I'm going to get b^n=e, for some n. Any additional "hints" or points in the right direction are appreciated
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  7. #7
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    What I meant was that you do this: a^2ba^{-2}=b^4 \Rightarrow a^2b^2a^{-2}=b^8  \Rightarrow a^2(aba^{-1})a^{-2} = a^3ba^{-3} = b^8.

    Can you continue in this way until you reach a^5ba^{-5} ?


    You can also prove the more general theorem: if aba^{-1} = b^i then a^mba^{-m} = b^{(i^m)}
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  8. #8
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    Quote Originally Posted by Defunkt View Post
    What I meant was that you do this: a^2ba^{-2}=b^4 \Rightarrow a^2b^2a^{-2}=b^8  \Rightarrow a^2(aba^{-1})a^{-2} = a^3ba^{-3} = b^8.

    Can you continue in this way until you reach a^5ba^{-5} ?


    You can also prove the more general theorem: if aba^{-1} = b^i then a^mba^{-m} = b^{(i^m)}
    So continuing until I reach a^5ba^{-5} we obtain:

    a^4ba^{-4}=b^1^6 \Rightarrow a^4b^2a^{-4}=b^3^2  \Rightarrow a^4(aba^{-1})a^{-4} = a^5ba^{-5} = ba^{-5} = b^3^2.

    Now what?
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  9. #9
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    If a^5=e , what is a^{-5} ?
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  10. #10
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    Quote Originally Posted by Defunkt View Post
    If a^5=e , what is a^{-5} ?
    I'm not quite sure, but wouldn't a^{-5}=e as well?
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  11. #11
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    I think I get it in that if a^{-5}=e, then we have b=b^{32}, which gives us e=b^{31}. Thus, the order of b must be a divisor of 31, meaning it is either 1 or 31. However it can't be 1 since b is a non-identity element by the given, thus o(b)=31.

    However, can you please explain why a^5=e gives a^{-5}=e? Thanks.
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  12. #12
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    Since a^5a^{-5}=a^{-5}a^5=e and the inverse of the identity is the identity itself.
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