# Torsion free abelian group of rank 1

I find it hard to understand a part of a proof on torsion free abelian groups of rank 1. Let $A$ and $B$ be torsion free groups of rank 1 and of the same type. Let $a'$ and $b'$ arbitrary non zero elements from $A$ and $B$, respectively. Then the height of $a'$ and $b'$ $H(a')=(k_1,...,k_n,...)$ and $H(b')=(k'_1,...,k'_n,...)$ are equivalent. Let $n_1,...,n_s$ be those indices $n$ for which $k_n$ is different from $k'_n$ . Since $k_{n_i}$ and $k'_{n_i}$ are non negative integers, we can solve the equations $p_{n_1}^{k_{n_1}}... p_{n_s}^{k_{n_s}}x=a'$ and $p_{n_1}^{k'_{n_1}}... p_{n_s}^{k'_{n_s}}y=b'$ and denote their solutions by $a\in A$ and $b\in B$, respectively. Clearly $H(a)=H(b)$ since they are obtained from $H(a')$ and $H(b')$ by replacing the $k_n$ $(k'_n)$ of index $n_1,n_2,...,n_s$ by $0$. It follows that an equation $mx=ta$ $(t,m$ non zero integers) is solvable in $A$ if and only if the equation $my=tb$ admits a solution in $B$. Why is it true? I think it is related to the fact that in a torsion free abelian group of rank 1 two non zero elements are dependent but I think it is not enough. And why in torsion free groups may such equations have at most one solution? Someone can help me? Thanks

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Your nomenclature is a bit of a mess. Is $(k_n)_i$ supposed to be $k_{n_i}$? What is $((p_n)_1)$? Why the up-raised parentheses, ${}^{(}(k_n)_1$? – Arturo Magidin Jun 22 '11 at 18:03