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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 countable subset and cannot be partitioned into finite sets; such a set has no group structure.

Sketch of proof that in standard Cohen model the set $A=\{a_n:n\in\omega\}$ of adjoined Cohen reals cannot be partitioned 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 $p\Vdash " \bigcup_{i\in I}\dot{A_i}=A$ 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 $E$). Contradiction.

3 correcting an oversight in the proof; added 88 characters in body

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.

Sketch of proof that in standard Cohen model the set $A=\{a_n:n\in\omega\}$ of adjoined Cohen reals cannot be partition partitioned 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 $p\Vdash\dot{f}:A\rightarrow p\Vdash " \omega$ and also that $p$ forces that $f$ partitions bigcup_{i\in I}\dot{A_i}=A$is a partition into finite piecespieces"; let$E$(a finite set) be the support of$\dot{f}$. this partition. Take some$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_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}$to$j$, a_j$, it follows that $\pi(q)\Vdash f^{-1}[m]=\{a_j,a_{i_1},\ldots \pi(q) \Vdash " \{a_j,a_{i_1},\ldots a_{i_n}\}$ is the piece of the partition containing a_j". But also $q$ and $\pi(q)$ are compatible and here we run into trouble, because $\pi[q]$ q$forces that$a_j$is in a_{i_0}$ and $f^{-1}[m]$ a_{i_1}$are in the same piece of the partition, and$q$\pi(q)$ forces that it 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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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.

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