Bases of Subgroups of Infinitely Generated Free Abelian Groups In the case when $F$ is a free abelian group of finite rank, it is known that for every subgroup $B$ of $F$ there are bases $\{b_1,\ldots,b_k\}$ and $\{f_1,\ldots,f_n\}$ of $B$ and $F,$ respectively such that $b_i=m_i f_i$ for all $i=1,\ldots,k.$ I can't find a similar result on the infinitely generated free abelian groups. Does it mean it is not true? If not true, are there any results on "partial proportionality"?
 A: If the analogous result for infinitely generated free abelian groups were true, then it would follow that, whenever $F$ is a free abelian group and $B$ a subgroup (automatically also free), the quotient is a direct sum of cyclic groups $\mathbb Z$ and $\mathbb Z/m$. Specifically, if $F$ has a free basis $\{f_i:i\in N\}$ and $B$ has a free basis $\{b_j:j\in K\}$, where $K\subseteq N$ and, for each $j\in K$, $b_j=m_jf_j$, then $F/B$ would be the direct sum of groups $\mathbb Z/m_j$ (generated by the coset of $f_j$) for $j\in K$ and groups $\mathbb Z$ (generated by the coset of $f_i$) for all $i\in N-K$. 
But every abelian group is a quotient of a free abelian group, and not every abelian group is a direct sum of cyclic groups. Probably the most familiar example is the additive group $\mathbb Q$ of rational numbers. 
So, if you choose some free abelian group $F$ that maps onto the rationals, say with generators $\{f_i:i\in\mathbb N-\{0\}\}$ and the homomorphism sending $f_i$ to $1/i\in\mathbb Q$, then the kernel $B$ of this homomorphism is free but not of the form you wanted.
