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  "body": "\n## Statement\n\nIn this section it is assumed that every set $X$ can be endowed with a group structure $(X, \\bullet)$.\n\nLet $X$ be a set. Let $\\aleph(X)$ be the Hartogs number of $X$. This is the least [[cardinal number|0EBACQFAZW]] 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|0RDJK7EQYY]]. Let $\\bullet$ denote multiplication in the group $(X \\cup \\aleph(X), \\bullet)$.\n\nFor 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.\n\nNow 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.\n\nFinally, define a wellordering on $X$ by $x \u003c y$ if $j(x) \u003c j(y)$. It follows that every set $X$ can be wellordered and thus that the axiom of choice is true.\n\nFor 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.\n",
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