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**aliceinwonderland** 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 $\displaystyle \{x_{1}, x_{2},...,x_{n}\}$ of F, an integer $\displaystyle r(1 \leq r \leq n)$ and positive integers $\displaystyle d_{1}, d_{2},...,d_{r}$ such that $\displaystyle d_{1}|d_{2}|..|d_{r}$ and G is free abelian with basis $\displaystyle \{d_{1}x_{1},....,d_{r}x_{r}\}$."

My question is

If I choose a basis for G as $\displaystyle \{x_{1}, x_{2},x_{3}\}$ and $\displaystyle \{x_{1}, 2x_{2},4x_{3}\}$,respectively, G satisfies the above condition and I see that G is a subgroup of F.

If I choose a basis for G as $\displaystyle \{x_{1}, 3x_{2},5x_{3}\}$, G does not satisfies the above theorem. But G looks still a subgroup of F to me.

The members of G might be

$\displaystyle \{x_{1}, 2x_{1},,,5x_{3}, 10x_{3},,,,3x_{2}+5x_{3},,, -5x_{3},,,2x_{1}+3x_{2}-10x_{3},,,,\}$.

Please correct me if I am wrong on the concept of the above theorem.

Thanks in advance.