A group structure implies the axiom of choice

Statement

In this section it is assumed that every set $X$ can be endowed with a group structure $(X, \bullet)$.

Let $X$ be a set. Let $\aleph(X)$ be the Hartogs number of $X$. This is the least cardinal number such that there is no injection from $\aleph(X)$ into $X$. It exists without the assumption of the axiom of choice. Assume here for technical simplicity of proof that $X$ has no ordinal. Let $\bullet$ denote multiplication in the group $(X \cup \aleph(X), \bullet)$.

For any $x \in X$ there is an $\alpha \in \aleph(X)$ such that $x \bullet \alpha \in \aleph(X)$. Suppose not. Then there is an $y \in X$ such that $y \bullet \alpha \in X$ for all $\alpha \in \aleph(X)$. But by elementary group theory, the $y \bullet \alpha$ are all different as α ranges over $\aleph(X)$ (i). Thus such a $y$ gives an injection from $\aleph(X)$ into $X$. This is impossible since $\aleph(X)$ is a cardinal such that no injection into $X$ exists.

Now define a map $j$ of $X$ into $\aleph(X) \times \aleph(X)$ endowed with the lexicographical wellordering by sending $x \in X$ to the least $(\alpha, \beta) \in \aleph(X) \times \aleph(X)$ such that $x \bullet \alpha = \beta$. By the above reasoning the map $j$ exists and is unique since least elements of subsets of wellordered sets are unique. It is, by elementary group theory, injective.

Finally, define a wellordering on $X$ by $x < y$ if $j(x) < j(y)$. It follows that every set $X$ can be wellordered and thus that the axiom of choice is true.

For the crucial property expressed in (i) above to hold, and hence the whole proof, it is sufficient for $X$ to be a cancellative magma, e.g. a quasigroup. The cancellation property is enough to ensure that the $y \bullet \alpha$ are all different.