4 group structures only exist on nonempty sets (-:

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) -> (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 $o$ 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 <-least pair $(\alpha, \beta)$ in $(\aleph(X))^{2}$ such that $x o \alpha = \beta$. Here, < is the lexic well ordering on the product $(\aleph(X))^{2}$. This induces a well ordering on $X$.

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In ZF, the following are equivalent:

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

(b) Axiom of choice

Non trivial direction [(a) -> (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 $$o be a group operation on X \cup \aleph(X). Now for any x \in X there must be an \alpha \in \aleph(X) such that x o \alpha \in \aleph(X) since otherwise we get an injection of \aleph(X) into X. Using$$, o$, therefore, one may inject$X$into$(\aleph(X))^{2}$by sending$x \in X$to the <-least pair$(\alpha, \beta)$in$(\aleph(X))^{2}$such that$x o \alpha = \beta$. Here, < is the lexic well ordering on the product$(\aleph(X))^{2}$. This induces a well ordering on$X\$.

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