set theory - Does every non-empty set admit a group structure (in ZF)?

You cannot in general put a group structure on a set. There is a model of ZF with a set A that has no infinite countable 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

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Sketch of proof that in standard Cohen model the set $A={a_n:ninomega}$ of adjoined Cohen reals cannot be partitioned into finite sets:

Let $mathbb{P}=Fn(omegatimesomega,2)$ 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 $pVdash " bigcup_{iin I}dot{A_i}=A$ is a partition into finite pieces"; let $E$ (a finite set) be the support of this partition. Take some $a_{i_0}notin E$ and extend $p$ to a $q$ such that $qVdash ``{a_{i_0},ldots a_{i_n}}$ is the piece of the partition containing $a_{i_0}$". 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 $a_j$, it follows that $pi(q) Vdash " {a_j,a_{i_1},ldots a_{i_n}}$ is the piece of the partition containing a_j". But also $q$ and $pi(q)$ are compatible and here we run into trouble, because $q$ forces that $a_{i_0}$ and $a_{i_1}$ are in the same piece of the partition, and $pi(q)$ forces that this is not the case (and they are talking about the same partition we started with because $pi$ fixes $E$). Contradiction.

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