In mathematics a group is a set together 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 statements are equivalent:
In this section it is assumed that every set can be endowed with a group structure .
Let be a set. Let be the Hartogs number of . This is the least cardinal number such that there is no injection from into . It exists without the assumption of the axiom of choice. Assume here for technical simplicity of proof that has no ordinal. Let denote multiplication in the group .
For any there is an such that . Suppose not. Then there is an such that for all . But by elementary group theory, the are all different as α ranges over (i). Thus such a gives an injection from into . This is impossible since is a cardinal such that no injection into exists.
Now define a map of into endowed with the lexicographical wellordering by sending to the least such that . By the above reasoning the map 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 by if . It follows that every set 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 to be a cancellative magma, e.g. a quasigroup. The cancellation property is enough to ensure that the are all different.
Any nonempty finite set has a group structure as a cyclic group generated by any element. Under the assumption of the axiom of choice, every infinite set is equipotent with a unique cardinal number which equals an aleph. Using the axiom of choice, one can show that for any family of sets (A). Moreover, by Tarski's theorem on choice, another equivalent of the axiom of choice, for all finite (B).
Let be an infinite set and let denote the set of all finite subsets of . There is a natural multiplication on . For, let, where denotes the symmetric difference. This turns into a group with the empty set,, being the identity and every element being its own inverse; . The associative property, i.e. is verified using basic properties of union and set difference. Thus is a group with multiplication .
Any set that can be put into bijection with a group becomes a group via the bijection. It will be shown that, and hence a one-to-one correspondence between and the group exists. For, let be the subset of consisting of all subsets of cardinality exactly . Then is the disjoint union of the . The number of subsets of of cardinality is at most because every subset with elements is an element of the -fold cartesian product of . So for all (C) by (B).
Putting these results together it is seen that by (A) and (C). Also,, since contains all singletons. Thus, and, so, by the Schröder–Bernstein theorem, . This means precisely that there is a bijection between and . Finally, for define . This turns into a group. Hence every set admits a group structure.
There are models of ZF in which the axiom of choice fails. In such a model, there are sets that cannot be well-ordered (call these "non-wellorderable" sets). Let be any such set. Now consider the set . If were to have a group structure, then, by the construction in first section, can be well-ordered. This contradiction shows that there is no group structure on the set .
If a set is such that it cannot be endowed with a group structure, then it is necessarily non-wellorderable. Otherwise the construction in the second section does yield a group structure. However these properties are not equivalent. Namely, it is possible for sets which cannot be well-ordered to have a group structure.
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