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Math Help - order of x (Abstract Algebra)

  1. #1
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    order of x (Abstract Algebra)

    Can someone help?

    Prove: If x and g are elements of the group G, prove that |x| = |g^(-1)xg|. Deduce that |ab| = |ba| for every a, b in G.

    |x| means the order of x. 1 is the identity element in this case.

    For G a group and x in G, the order of x to be the smallest positive integer n such that x^n = 1 (that is |x| = n).
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  2. #2
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    For the first part, let |x| = n and start taking products of gxg^{-1} until you see a pattern. Notice that (gxg^{-1})^2 = (gxg^{-1})(gxg^{-1}) = gx(g^{-1}g)xg^{-1} = gx^2g^{-1}. Similarly, (gxg^{-1})^n = gx^ng^{-1}. I think you can take it from here.

    For the second part, let |ab| = n. Then, if e is the identity,

    b = be

    \Longrightarrow b = b((ab)(ab)...(ab)) for n copies of ab

    \Longrightarrow b = ((ba)(ba)...(ba))b by associativity (notice that you have n copies of ba)

    \Longrightarrow e = (ba)^n by cancellation

    \Longrightarrow |ba| \le n

    If |ba| = k < n, then (ba)^k = e.

    So, (ba)^kb = b.

    By associativity, b(ab)^k = b.

    By cancellation, (ab)^k = e, which is a contradiction.

    Thus, |ba| = n.
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  3. #3
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    Quote Originally Posted by dori1123 View Post

    Deduce that |ab| = |ba| for every a, b in G.
    we have: ba=b(ab)b^{-1}. thus |ab|=|ba| by the first part of the problem.
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  4. #4
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    Thank you for helping me with the proof. But above only prove |x|=|g^(-1)xg| if the order of x is finite. What if the order of x is infinite? The definition I was given is that if the order of x is infinite, then we write |x| = 0. I don't know how to show |g^(-1)xg| = 0.
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  5. #5
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    Quote Originally Posted by dori1123 View Post
    Thank you for helping me with the proof. But above only prove |x|=|g^(-1)xg| if the order of x is finite. What if the order of x is infinite? The definition I was given is that if the order of x is infinite, then we write |x| = 0. I don't know how to show |g^(-1)xg| = 0.
    If g^{-1}xg has finite order then so does g(g^{-1}xg)g^{-1} = x by above.
    Thus, they are both finite or infinite.
    And if finite then they have the same order.
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