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 of F, an integer and positive integers such that and G is free abelian with basis ."
My question is
If I choose a basis for G as and ,respectively, G satisfies the above condition and I see that G is a subgroup of F.
If I choose a basis for G as , G does not satisfies the above theorem. But G looks still a subgroup of F to me.
The members of G might be
Please correct me if I am wrong on the concept of the above theorem.
Thanks in advance.
Thanks for your reply.
The above G is generated by .
If I choose a basis for G that has the divisor property, lets say, , 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.
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 . Let G be a subgroup of F (rank 3) and generated by . Let v be an element of G such that are another bases of F), where .
The minimality of implies that so that bases of F can be replaced by . I replaced each with for i=1,2,3.
Now, I pick v as in G.
. According to the above description, 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 , where the bases of G has the divisor property.