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Math Help - [SOLVED] Getting crazy with this excerice....

  1. #1
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    [SOLVED] Getting crazy with this excerice....

    Hello!!! thanks for reading!

    Here's the problem (regarding groups, but nothing too complex):
    I'm told that a and b are elements in G, which is a group.
    I'm also told that b<>e (where e is the unit element - thus e*a = a, etc...), and that o(a) = 5 (meaning: a^5=e, and 5 is the smallest natural number that satisfies that).
    I'm told that: aba^-1 = b^2
    I need to find o(b).

    This looks pretty innocent at first (and maybe it really is), but I've spent too much time on this question... tried every trick I know...
    I've proved, some excercises before, that o(aba^-1)=o(b), so I tried using that a lot, but it didn't really get me much.
    Of course I tried to do "^5" to both sides, and many many other things, playing with the inverses etc....
    Best I could get, with much effort, is that o(ba) = 5.
    I just don't know what to do :-\

    Thanks!
    Tomer.
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  2. #2
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    Quote Originally Posted by aurora View Post
    Hello!!! thanks for reading!

    Here's the problem (regarding groups, but nothing too complex):
    I'm told that a and b are elements in G, which is a group.
    I'm also told that b<>e (where e is the unit element - thus e*a = a, etc...), and that o(a) = 5 (meaning: a^5=e, and 5 is the smallest natural number that satisfies that).
    I'm told that: aba^-1 = b^2
    I need to find o(b).

    This looks pretty innocent at first (and maybe it really is), but I've spent too much time on this question... tried every trick I know...
    I've proved, some excercises before, that o(aba^-1)=o(b), so I tried using that a lot, but it didn't really get me much.
    Of course I tried to do "^5" to both sides, and many many other things, playing with the inverses etc....
    Best I could get, with much effort, is that o(ba) = 5.
    I just don't know what to do :-\

    Thanks!
    Tomer.
    that's a good question! i'll prove a more general result first:

    Lemma: let a,b be two elements of a group such that aba^{-1}=b^n. then a^{\ell}ba^{-\ell}=b^{n^{\ell}}, \ \ \forall \ell \in \mathbb{N}.

    Proof: by induction:

    for \ell = 1, the claim is just the hypothesis of the lemma. now suppose the claim is true for \ell. then:

    a^{\ell + 1}ba^{-(\ell + 1)}=a(a^{\ell}ba^{-\ell})a^{-1}=ab^{n^{\ell}}a^{-1}=(aba^{-1})^{n^{\ell}}=(b^n)^{n^{\ell}}=b^{n^{\ell + 1}}. \ \ \ \square
    .................................................. .................................................. ................................

    now it's easy to solve your problem: since we're given that aba^{-1}=b^2, we can apply the Lemma for

    n =2 and \ell = 5. thus: b=a^5ba^{-5}=b^{32}. hence b^{31}=e. since 31 is prime, we have o(b)=31. \ \ \ \square
    Last edited by NonCommAlg; June 23rd 2008 at 09:07 PM.
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  3. #3
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    Quote Originally Posted by NonCommAlg View Post
    a^{\ell + 1}ba^{-(\ell + 1)}=a(a^{\ell}ba^{-\ell})a^{-1}=ab^{n^{\ell}}a^{-\ell}=(aba^{-1})^{n^{\ell}}=(b^n)^{n^{\ell}}=b^{n^{\ell + 1}}. \ \ \ \square
    Do you mean ab^{n^\ell}a^{-1} instead?
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  4. #4
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    Quote Originally Posted by ThePerfectHacker View Post
    Do you mean ab^{n^\ell}a^{-1} instead?
    i meant exactly what i wrote! haha ... still, what part of that line are you refering to?
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    Quote Originally Posted by NonCommAlg View Post
    i meant exactly what i wrote! haha ... still, what part of that line are you refering to?
    Look at what I quoted in my first post. The stuff in between 2nd and 3rd "=" sign.
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  6. #6
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    Quote Originally Posted by ThePerfectHacker View Post
    Look at what I quoted in my first post. The stuff in between 2nd and 3rd "=" sign.
    Ah, of course! i fixed the typo. thank you!
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  7. #7
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    Wow, bless you!
    I guess it's solvable, though somehow sutdents are always fearful to prove more than they need .
    Thank you so much!
    Tomer.
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  8. #8
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    I been curious the key fact here was that 31 is a prime number. What happened if we had a^4=e then b = a^4ba^{-4} = b^{2^4} and so b^{15}=e. What can we conclude? It can still be that the order is 15.
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