---
tag: PV59XGVR27
kind: statement
role: proposition
name: The axiom of choice implies a group structure
sources:
  - wikipedia: en:37039918@r1336022484
    locator: Group structure and the axiom of choice, section The axiom of choice implies a group structure
    note: adopted
    mention: wikipedia://en/37039918#9XV0
    sections: [Statement]
inventory: [DQ129P338A]
---

## Statement

Any 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**).

Let $X$ 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 $\Delta$.

Any 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**).

Putting 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.