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 *symmetric difference* 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.


So my question is



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> 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?
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Does every non-empty set admit a group structure (in ZF)?
In ZF, the following are equivalent:


(a) For every nonempty set there is a binary operation making it a group


(b) Axiom of choice


Non trivial direction [(a) $\to$ (b)]:


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 $\circ$ 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 $x \circ \alpha \in \aleph(X)$ since otherwise we get an injection of $\aleph(X)$ into $X$. Using $\circ$, therefore, one may inject $X$ into $(\aleph(X))^{2}$ by sending $x \in X$ to the $<$-least pair $(\alpha, \beta)$ in $(\aleph(X))^{2}$ such that $x \circ \alpha = \beta$. Here, $<$ is the lexicographic well-ordering on the product $(\aleph(X))^{2}$. This induces a well-ordering on $X$.


