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In mathematics a group is a set together with with a binary operation on the set called multiplication that obeys the group axioms. The axiom of choice is an axiom of ZFC set theory which in one form states that every set can be wellordered.

In ZF set theory, i.e. ZFC without the axiom of choice, the following are equivalent:


 * For every nonempty set $X$ there exists a binary operation · turning (X,·) into a group.
 * The axiom of choice is true.

A Group Structure implies the Axiom of Choice
In this section it is assumed that every set X can be endowed with a group stucture (X,·).

Let X be a set. Let ℵ(X) be Hartogs number for X. This is the least cardinal number such that there is no injection from ℵ(X) into X. It exists without the assumption of the axiom of choice. Assume for simplicity that X has no ordinals. Let · denote multiplication in the group (X∪ℵ(X),·).

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

Now define a map j of X into $ℵ(X)×ℵ(X)$ endowed with the lexicographical wellordering by sending x∈X to the least (α,β) ∈ ℵ(X)×ℵ(X) such that x·α=β. By the above reasonong the map j exists and is 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.

The Axiom of Choice implies a Group Structure
Under the assumption of the axiom of choice, every set X is equipotent with a unique cardinal number |X| which equals an aleph. One can show that for any family S of sets |⋃S|≤|S| · sup{|s| : s∈S}. This is done in the same fashion that one shows that a countable union of countable sets is countable. Moreover, |X|n=|X| for all finite n.

Let X be a nonempty set and let F denote the set of all finite subsets of X. There is a natural group operation · on F. For f,g∈F, let f·g = fΔg where Δ denotes the symmetric difference. This turns F into a group with the empty set, Ø, being the identity and every element being it's own inverse; fΔf= Ø. The associative property, i.e. (fΔg)Δh = fΔ(gΔh} is verified using basic properties of union and set difference.

Thus F is a group with multiplication Δ. Any set that can be put into bijection with a group becomes a group via the bijection. It will be shown that |X|=|F|, and hence a one-to-one correspondence between X and the group F exists.

For $n = 0,1,2, ...$, Let Fn be the subset of F consisting of all subsets of cardinality exactly n. Then F is the disjoint union of the Fn.

For any n, the number of subsets of X of cardinality n is at most |X|n=|X| because every subset with n elements is an element of the n-fold cartesian product Xn of X. So |Fn|≤|X|n=|X| for all n.

Putting these results together it is seen that |F|≤ℵ0 · |X| = |X|

The cardinality |F| of F is at least |X|, since F contains all singletons.

Thus, |X|≤|F| and |F|≤|X|, so, by the Schroeder-Bernstein theorem, |F|=|X|. This means precisely that there is a bijection j between X and F. Finally, for x,y∈X define x·y = j-1(j(x)Δj(y)) This turns X into a group. Every group admitts a group structure.