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Math Help - Abstract Algebra

  1. #1
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    Abstract Algebra

    If G is an abelian group that contains elements of order 4 and 10 then what is the order of a*b ?

    I know that when the gcd(|a|,|b|)=1 then |ab|=|a||b|, but how do i find the order in this case?

    Help will be appreciated.
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  2. #2
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    Re: Abstract Algebra

    Hey mrmaaza123.

    With regards to Lagrange's theorem we know that a sub-group needs to be a factor of the total order of the group.

    Now in the case of gcd being 1, it means that these sub-groups are going to be disjoint (except for the identity element) which means that the order will be the product of the two sub-groups which will have |a|*|b|.

    But when you do not have a gcd of one, it means that you need to check the actual sub-groups themselves.

    The other alternative is that you factorize the groups into disjoint sub-groups and then apply the situation with gcd being equal to 1.
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  3. #3
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    Re: Abstract Algebra

    Thank you for the effort, but i do not know what the group G might be for this question. THe question simply says G.
    My book gives the answer as 20, and i am still having some problem in figuring out how that happened.
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  4. #4
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    Re: Abstract Algebra

    The factors of both are 2*2 and 2*5. You might want to find theorems that describe decompositions of groups with into ones with only prime factors.

    Also this is just an observation but 20 = 4*10/gcd(4,10) = 4*10/2 = 20, so maybe that can give you hint as what to look for.
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  5. #5
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    Re: Abstract Algebra

    Aha! i understood that. It seems to work for every question !
    I will look up for a proper result though.
    But,Thank you very much, i highly appreciate your help.
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  6. #6
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    Re: Abstract Algebra

    suppose that |a| = 4, and |b| = 10.

    since G is abelian: (ab)2 = (ab)(ab) = a(ba)b = a(ab)b = (aa)(bb) = a2b2.

    an induction proof shows (ab)n = anbn.

    thus (ab)20 = (a)20(b)20 = (a4)5(b10)2 = e5e2 = ee = e.

    thus |ab| divides 20. since |a| ≠ |b|, b ≠ a-1 (which has the same order as a), so |ab| ≠ 1.

    now (ab)4 = a4b4 = eb4 = b4 ≠ e, since |b| = 10. so |ab| ≠ 4.

    similarly, (ab)10 = a10 = a2 ≠ e, so |ab| ≠ 10.

    thus we are left with: |ab| = 2,5 or 20.

    if |ab| = 2, this means a2b2 = e, so that a4b2 = a2, that is:

    b2 = a2, and therefore b4 = a4 = e, contradicting |b| = 10.

    if |ab| = 5, then a5b5 = ab5 = e, whence a = b5, so that:

    a2 = b10 = e, contradicting |a| = 4.

    so the only possibility left is |ab| = 20, as no lower divisor of 20 can possibly be the order of |ab|.

    in general, for ABELIAN groups (only), if <a>∩<b> = {e}, then indeed, |ab| = lcm(|a|,|b|) in an abelian group (the proof in the general case is pretty much that same:

    clearly ablcm(|a|,|b|) = e, and if d|lcm(|a|,|b|), with (ab)d = e,

    then ad = b|b|-d, so that ad is in <a>∩<b> and thus equals {e}, so that |a| divides d,

    and similarly |b| divides d, so that lcm(|a|,|b|) also divides d, so that d = lcm(|a|,|b|).

    the important thing is that <a> and <b> only intersect in {e}, for example:

    if b = a-1, then |ab| = 1, even though lcm(|a|,|a-1|) = |a|).
    Last edited by Deveno; October 11th 2012 at 09:18 AM.
    Thanks from emakarov
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