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Math Help - Abstract Algebra: Orders of Groups

  1. #1
    is up to his old tricks again! Jhevon's Avatar
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    Abstract Algebra: Orders of Groups

    Hello All,

    I'm stuck on part (b)

    Problem:

    (a) Prove that if a and b are elements of an Abelian group G, with \circ (a) = m and \circ (b) = n, then (ab)^{mn} = e ( e denotes the identity element in my text). Indicate where you use the condition that G is Abelian.

    (b) With G,~a and b as in part (a), prove that \circ (ab) divides \circ (a) \circ (b).

    (c) Give an example of an Abelian group G and elements a and b in G such that \circ (ab) \ne \circ (a) \circ (b). Compare part (b)



    Things that might come in Handy:

    Let G be a group and a \in G.

    We define the order of a, denoted \circ (a), to be the smallest, positive integer n such that a^n = e. If no such integer exists, we say a has infinite order.

    a^n of course means a operating on itself n times. that is, a^n = \underbrace{aaaa \cdots a}_{n \mbox{ times}} = a*a*a*a* \cdots *a

    Tell me if you guys need anymore information.




    What I have tried:

    (a) Proof: Let a,~b,~G be as in part (a). Since G is Abelian, (ab)^{mn} = a^{mn}b^{mn}

    Now, a^{mn}b^{mn} = a^{mn}b^{nm} = (a^m)^n(b^n)^m = e^ne^n = ee = e

    QED


    (b) This is where I'm stuck. Here's what I did so far.

    Let a,~b,~G be as in part (a). And let \circ (ab) = p.

    Then m,~n, \mbox{ and }p are all the smallest, positive integers such that: a^m = e, b^n = e and (ab)^p = a^pb^p = e. We want to show that p|mn, that is, mn = kp for some k \in \mathbb{Z}.

    Now, a^mb^n = ee = e = a^pb^p

    ...

    that's it, I'm not sure where to go from there. I've tried several things, but they all end up doing nothing.


    (c) My example was (\mathbb{Z}_3, \oplus) with a = [1] and b = [2]


    Thanks guys and gals
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  2. #2
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    Quote Originally Posted by Jhevon View Post
    (b) With G,~a and b as in part (a), prove that \circ (ab) divides \circ (a) \circ (b).
    Just use a theorem. Suppose x is a group (any group not just abelian) G and x has finite order k. If n is any positive integer so that x^n = e then k|n. Now apply this theorem here.

    (c) Give an example of an Abelian group G and elements a and b in G such that \circ (ab) \ne \circ (a) \circ (b). Compare part (b)
    Consider G=\mathbb{Z}_5 and a=[2],b=[3].
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  3. #3
    Senior Member JaneBennet's Avatar
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    (b)
    If \mathrm{o}(x)=p and x^k=e, write k=qp+r where 0\leq r<p. Hence e=x^k=(x^p)^qx^r=x^r and it follows that r=0 since p is the smallest positive integer such that x to that power is the identity.

    (c)
    Take a = b = any element of order 2. Then \mathrm{o}(a\circ b)=1 but \mathrm{o}(a)\cdot\mathrm{o}(b)=4.
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