Does every non-empty set admit a group structure (in ZF)? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T18:22:44Z http://mathoverflow.net/feeds/question/12973 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/12973/does-every-non-empty-set-admit-a-group-structure-in-zf Does every non-empty set admit a group structure (in ZF)? Konrad Swanepoel 2010-01-25T21:49:02Z 2011-11-02T21:30:59Z <p>It is easy to see that in ZFC, any non-empty set $S$ admits a group structure: for finite $S$ identify $S$ with a cyclic group, and for infinite $S$, the set of finite subsets of $S$ with the binary operation of <em>symmetric difference</em> forms a group, and in ZFC there is a bijection between $S$ and the set of finite subsets of $S$, so the group structure can be taken to $S$. However, the existence of this bijection needs the axiom of choice.</p> <p>So my question is</p> <blockquote> <p>Can it be shown in ZF that for any non-empty set $S$ there exists a binary operation $\ast$ on $S$ making $(S,\ast)$ into a group?</p> </blockquote> http://mathoverflow.net/questions/12973/does-every-non-empty-set-admit-a-group-structure-in-zf/12976#12976 Answer by Zev Chonoles for Does every non-empty set admit a group structure (in ZF)? Zev Chonoles 2010-01-25T22:01:54Z 2010-01-25T22:01:54Z <p>I'm not sure what to make of the part about the subsets of $S$ under symmetric difference - yes, they form a group, but I don't believe that imparts a group structure on $S$ itself - but the answer to your main question is yes. Simply declare one element to be the identity and make the rest equal to powers of some other element - in other words, the set $S$ can made into the cyclic group of order $|S|$.</p> http://mathoverflow.net/questions/12973/does-every-non-empty-set-admit-a-group-structure-in-zf/12984#12984 Answer by Justin Palumbo for Does every non-empty set admit a group structure (in ZF)? Justin Palumbo 2010-01-25T22:54:34Z 2011-11-02T21:30:59Z <p>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.</p> <p>See e.g at <a href="http://groups.google.com/group/sci.math/msg/06eba700dfacb6ed" rel="nofollow">http://groups.google.com/group/sci.math/msg/06eba700dfacb6ed</a></p> <hr> <p>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:</p> <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/12973/does-every-non-empty-set-admit-a-group-structure-in-zf/12988#12988 Answer by Ashutosh for Does every non-empty set admit a group structure (in ZF)? Ashutosh 2010-01-26T00:07:02Z 2010-02-01T04:47:10Z <p>In ZF, the following are equivalent:</p> <p>(a) For every nonempty set there is a binary operation making it a group</p> <p>(b) Axiom of choice</p> <p>Non trivial direction [(a) -> (b)]:</p> <p>The trick is Hartogs construction which gives for every set $X$ an ordinal $\aleph(X)$ such that there is no injection from $\aleph(X)$ into $X$. Assume for simplicity that $X$ has no ordinals. Let $o$ be a group operation on $X \cup \aleph(X)$. Now for any $x \in X$ there must be an $\alpha \in \aleph(X)$ such that $x o \alpha \in \aleph(X)$ since otherwise we get an injection of $\aleph(X)$ into $X$. Using $o$, therefore, one may inject $X$ into $(\aleph(X))^{2}$ by sending $x \in X$ to the &lt;-least pair $(\alpha, \beta)$ in $(\aleph(X))^{2}$ such that $x o \alpha = \beta$. Here, &lt; is the lexic well ordering on the product $(\aleph(X))^{2}$. This induces a well ordering on $X$.</p>