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Math Help - Group Theory Help Please!!!

  1. #1
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    Group Theory Help Please!!!

    Hi all ,

    Just wanted to see if i was handling this Q correctly

    Q in a group G , a^5 = e, aba^-1 = b^2 for some a,b element of G.
    find the order of b?

    I said aba^-1 = b^2

    i) multiplied RHS and LHS by 'a'

    to get ab = b^2a

    ii) then i) multiplied RHS and LHS by b^-1 (b inverse)

    to get a = b^2ab^-1

    then put both side to the power of 5

    to get a^5 = e

    so e = (b^2ab^-1)^5

    e = b^10a^5b^-5

    e = b^10b^-5 (a^5 turns to e)
    = b^5

    so order of b = 5 ??????

    Not sure at all if this is correct any suggestions would be great.

    thanks

    chave
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  2. #2
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    so e = (b^2ab^-1)^5

    e = b^10a^5b^-5
    Take care, here you may have assumed that a and b commute.

    But, since aba^{-1}=b^2, we get that the commutator aba^{-1}b^{-1}=b, thus a and b don't commute!

    Edit: made a mistake in my development, ThePerfectHacker sent a proof just after your post. For commutativity, (xy)^5=x^5y^5 is wrong in general.
    Last edited by clic-clac; February 22nd 2009 at 02:14 PM. Reason: rofl
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  3. #3
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    I didn't assume they commute but am i allowed to put both sides
    to the power of 5????
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  4. #4
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    Quote Originally Posted by chave View Post
    Hi all ,

    Just wanted to see if i was handling this Q correctly

    Q in a group G , a^5 = e, aba^-1 = b^2 for some a,b element of G.
    find the order of b?
    If aba^{-1} = b^2 then a^k ba^{-k} = b^{2^k}. Prove this by induction!
    Now if k=5 then a^5 ba^{-5} = b^{32} \implies b^{31} = e.
    Therefore, b=e\text{ or }|b|=31 since 31 is a prime number.
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  5. #5
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    Sorry but I don't really understand perfect hacker what you mean here?? can you explain a bit more please?? thanks especially the induction bit and are you saying the order of b is 31????
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  6. #6
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    Quote Originally Posted by chave View Post
    Sorry but I don't really understand perfect hacker what you mean here?? can you explain a bit more please?? thanks especially the induction bit and are you saying the order of b is 31????
    Can you prove the induction?

    I said that b^{32} = b which means b^{31} = e. Therefore, either b is the idenity element i.e. b=e (which is trivial and probably is not the case in whatever problem you are working on) or the order of b is 31.
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  7. #7
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    ok i understand the second bit, no i cant get the induction proof??
    sorry again for annoying you.
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  8. #8
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    Quote Originally Posted by chave View Post
    ok i understand the second bit, no i cant get the induction proof??
    sorry again for annoying you.
    If k=1 then aba^{-1}=b^2, which is true by hypothesis of problem. Say that a^kba^{-k}=b^{2^k} then raise both sides to the power of 2 to get \left( a^k b a^{-k} \right)^2 = b^{2^{k+1}}. However, \left( a^k b a^{-k} \right)^2 = a^k b^2 a^{-k} = a^k (aba^{-1})a^{-k} = a^{k+1}ba^{-(k+1)}. This completes induction.
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  9. #9
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    thanks a lot mate
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