You cannot in general put a group structure on a set. There is a model of ZF with a set A that is "amorphous", ie it has no proper infinite, coinfinite subset and cannot be partitioned into finite sets; such a set has no group structure.
See e.g at http://groups.google.com/group/sci.math/msg/06eba700dfacb6ed
Sketch of proof that in standard Cohen model the set $A=\{a_n:n\in\omega\}$ of adjoined Cohen reals cannot be partition into finite sets:
Let $\mathbb{P}=Fn(\omega\times\omega,2)$ which is the poset we force with. The model is the symmetric submodel whose permutation group on $\mathbb{P}$ is all permutations of the form $\pi(p)(\pi(m),n)=p(m,n)$ where $\pi$ varies over all permutations of $\omega$, (that is we are extending each $\pi$ to a permutation of $\mathbb{P}$ which I also refer to as $\pi$) and the relevant filter is generated by all the finite support subgroups.
Suppose for contradiction that $p\Vdash\dot{f}:A\rightarrow \omega$ and also that $p$ forces that $f$ partitions into finite pieces; let $E$ (a finite set) be the support of $\dot{f}$. Take some $a_{i_0}\not\in E$ and extend $p$ to a $q$ such that $q\Vdash f^{-1}[m]=\{a_{i_0},\ldots a_{i_n}\}$. Then pick some $j$ which is not in $E$ nor the domain of $q$ nor equal to any of the $a_{i_0},\ldots a_{i_l}$. If $\pi$ is a permutation fixing $E$ and each of $a_{i_1},\ldots a_{i_n}$ and sending $a_{i_0}$ to $j$, it follows that $\pi(q)\Vdash f^{-1}[m]=\{a_j,a_{i_1},\ldots a_{i_n}\}$. But also $q$ and $\pi(q)$ are compatible and here we run into trouble, because $\pi[q]$ forces that $a_j$ is in $f^{-1}[m]$ and $q$ forces that it is not. Contradiction.