Usted no puede, en general, poner una estructura de grupo en un conjunto. No es un modelo de ZF con un conjunto que no tiene infinitos contables subconjunto y no puede ser dividida en conjuntos finitos; tal conjunto no tiene estructura de grupo.
Véase e.g a http://groups.google.com/group/sci.math/msg/06eba700dfacb6ed
Croquis de la prueba de que en la norma Cohen modelo el conjunto de $A=\{a_n:n\in\omega\}$ de comunica Cohen reales no se pueden crear particiones en finito de conjuntos:
Deje $\mathbb{P}=Fn(\omega\times\omega,2)$ 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 $\pi$) y el filtro relevante es generada por todos los finita apoyo de los subgrupos.
Supongamos por contradicción que $p\Vdash " \bigcup_{i\in I}\dot{A_i}=A$ is a partition into finite pieces"; let $E$ (a finite set) be the support of this partition. Take some $a_{i_0}\not\in E$ and extend $p$ to a $q$ such that $q\Vdash ``\{a_{i_0},\ldots a_{i_n}\}$ 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 $E$). Contradicción.
