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  "body": "\n## Statement\n\nAny nonempty finite set has a group structure as a [[cyclic group|0B85Q41RP6]] generated by any element. Under the assumption of the axiom of choice, every infinite set $X$ is equipotent with a unique cardinal number $\\lvert X\\rvert$ which equals an aleph. Using the axiom of choice, one can show that for any family $S$ of sets $\\lvert \\bigcup S\\rvert \\le \\lvert S\\rvert \\times \\sup \\{ |s| : s \\in S\\}$ (**A**). Moreover, by Tarski's theorem on choice, another equivalent of the axiom of choice, $\\lvert X\\rvert^{n} = \\lvert X\\rvert$ for all finite $n$ (**B**).\n\nLet $X$ be an infinite set and let $F$ denote the set of all finite subsets of $X$. There is a natural multiplication $\\bullet$ on $F$. For $f, g \\in F$, let $f \\bullet g = f \\Delta g$, where $\\Delta$ denotes the [[symmetric difference|0ME21QN125]]. This turns $(F, \\bullet)$ into a group with the empty set, $\\text{Ø}$, being the identity and every element being its own inverse; $f \\Delta f = \\text{Ø}$. The associative property, i.e. $(f \\Delta g) \\Delta h = f \\Delta (g \\Delta h)$ is verified using basic properties of union and set difference. Thus $F$ is a group with multiplication $\\Delta$.\n\nAny set that can be put into [[bijection|0GG9TVW90M]] with a group becomes a group via the bijection. It will be shown that $\\lvert X\\rvert = \\lvert F\\rvert$, and hence a one-to-one correspondence between $X$ and the group $(F, \\bullet)$ exists. For $n = 0,1,2, ...$, let $F_{n}$ be the subset of $F$ consisting of all subsets of cardinality exactly $n$. Then $F$ is the disjoint union of the $F_{n}$. The number of subsets of $X$ of cardinality $n$ is at most $\\lvert X\\rvert^{n}$ because every subset with $n$ elements is an element of the $n$-fold [[cartesian product|09CEDJYBKP]] $X^{n}$ of $X$. So $\\lvert F_{n}\\rvert \\le \\lvert X\\rvert^{n} = \\lvert X\\rvert$ for all $n$ (**C**) by (**B**).\n\nPutting these results together it is seen that $\\lvert F\\rvert = \\lvert \\bigcup_{n \\in \\omega}F_{n}\\rvert \\le \\aleph_{0} \\cdot \\lvert X\\rvert = \\lvert X\\rvert$ by (**A**) and (**C**). Also, $\\lvert F\\rvert \\ge \\lvert X\\rvert$, since $F$ contains all singletons. Thus, $\\lvert X\\rvert \\le \\lvert F\\rvert$ and $\\lvert F\\rvert \\le \\lvert X\\rvert$, so, by the [[Schröder–Bernstein theorem|0A9V6RKQG6]], $\\lvert F\\rvert = \\lvert X\\rvert$. This means precisely that there is a bijection $j$ between $X$ and $F$. Finally, for $x, y \\in X$ define $x \\bullet y = j^{-1}(j(x) \\Delta j(y))$. This turns $(X, \\bullet)$ into a group. Hence every set admits a group structure.\n",
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