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Math Help - Free abelian group

  1. #1
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    Free abelian group

    The theorem of free abelian group says,

    "If F is a free abelian group of finite rank n and G is a nonzero subgroup of F, then there exists a basis \{x_{1}, x_{2},...,x_{n}\} of F, an integer r(1 \leq r \leq n) and positive integers d_{1}, d_{2},...,d_{r} such that d_{1}|d_{2}|..|d_{r} and G is free abelian with basis \{d_{1}x_{1},....,d_{r}x_{r}\}."

    My question is
    If I choose a basis for G as \{x_{1}, x_{2},x_{3}\} and \{x_{1}, 2x_{2},4x_{3}\},respectively, G satisfies the above condition and I see that G is a subgroup of F.
    If I choose a basis for G as \{x_{1}, 3x_{2},5x_{3}\}, G does not satisfies the above theorem. But G looks still a subgroup of F to me.
    The members of G might be
    \{x_{1}, 2x_{1},,,5x_{3}, 10x_{3},,,,3x_{2}+5x_{3},,, -5x_{3},,,2x_{1}+3x_{2}-10x_{3},,,,\}.

    Please correct me if I am wrong on the concept of the above theorem.
    Thanks in advance.
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  2. #2
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    Quote Originally Posted by aliceinwonderland View Post
    The theorem of free abelian group says,

    "If F is a free abelian group of finite rank n and G is a nonzero subgroup of F, then there exists a basis \{x_{1}, x_{2},...,x_{n}\} of F, an integer r(1 \leq r \leq n) and positive integers d_{1}, d_{2},...,d_{r} such that d_{1}|d_{2}|..|d_{r} and G is free abelian with basis \{d_{1}x_{1},....,d_{r}x_{r}\}."

    My question is
    If I choose a basis for G as \{x_{1}, x_{2},x_{3}\} and \{x_{1}, 2x_{2},4x_{3}\},respectively, G satisfies the above condition and I see that G is a subgroup of F.
    If I choose a basis for G as \{x_{1}, 3x_{2},5x_{3}\}, G does not satisfies the above theorem. But G looks still a subgroup of F to me.
    The members of G might be
    \{x_{1}, 2x_{1},,,5x_{3}, 10x_{3},,,,3x_{2}+5x_{3},,, -5x_{3},,,2x_{1}+3x_{2}-10x_{3},,,,\}.

    Please correct me if I am wrong on the concept of the above theorem.
    Thanks in advance.
    The group you constructed does not violate the theorem. The theorem does not say any basis has this divisor property. The theorem says that that there is a basis that has this divisor property. Thus, even if you constructed a subgroup out of a basis not having the divisor then you can find a different basis that produces the same subgroup and does have this property.
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  3. #3
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    Thanks for your reply.

    The above G is generated by \{x_{1}, 3x_{2}, 5x_{3}\}.
    If I choose a basis for G that has the divisor property, lets say, \{x_{1}, x_{2}, 5x_{3}\}, I see that it does not generate the same G.
    I'll appreciate if you show me the equivalent basis (having a divisor property) that generates the same G.
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  4. #4
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    Quote Originally Posted by aliceinwonderland View Post
    The above G is generated by \{x_{1}, 3x_{2}, 5x_{3}\}.
    If I choose a basis for G that has the divisor property, lets say, \{x_{1}, x_{2}, 5x_{3}\}, I see that it does not generate the same G.
    I am working on a concrete example, but something went wrong in my procedure.
    My example is based on the description of my textbook [Hungerford, p73-74].

    Let F be a free abelian group (rank 3) and the basis is \{x_{1}, x_{2}, x_{3}\}. Let G be a subgroup of F (rank 3) and generated by \{2x_{1}, 3x_{2}, 5x_{3}\}. Let v be an element of G such that v=d_{1}y_{1} + k_{2}y_{2} + k_{3}y_{3} (y_{1}, y_{2},y_{3} are another bases of F), where v=d_{1}(y_{1} +q_{2}y_{2}+q_{3}y_{3}) + r_{2}y_{2} +r_{3}y_{3}.
    ......
    The minimality of d_{1} implies that r_{2}=r_{3}=0 so that bases of F can be replaced by \{y_{1} +q_{2}y_{2}+q_{3}y_{3}, y_{2}, y_{3}\}. I replaced each  y_{i} with x_{i} for i=1,2,3.

    Now, I pick v as 4x_{1}+9x_{2}+15x_{3} in G.
    4x_{1}+9x_{2}+15x_{3} = 4(x_{1}+2x_{2}+3x_{3})+ x_{2} +3x_{3}. According to the above description, x_{1}+2x_{2}+3x_{3} should belong to G, but it is not. Where are errors in my procedure?

    plz help me show the bases of the subgroup G generated by  \{2x_{1}, 3x_{2}, 5x_{3}\}, where the bases of G has the divisor property.
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