Английская Википедия:Group structure and the axiom of choice

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Файл:Ernst Zermelo.jpeg

Ernst Zermelo in 1904 proved the wellordering theorem using what was to become known as the axiom of choice.

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:

• For every nonempty set Шаблон:Math there exists a binary operation Шаблон:Math such that Шаблон:Math is 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 Шаблон:Math can be endowed with a group structure Шаблон:Math.

Let Шаблон:Math be a set. Let Шаблон:Math be the Hartogs number of Шаблон:Math. This is the least cardinal number such that there is no injection from Шаблон:Math into Шаблон:Math. It exists without the assumption of the axiom of choice. Assume here for technical simplicity of proof that Шаблон:Math has no ordinal. Let Шаблон:Math denote multiplication in the group Шаблон:Math.

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

Now define a map Шаблон:Math of Шаблон:Math into Шаблон:Math endowed with the lexicographical wellordering by sending Шаблон:Math to the least Шаблон:Math such that Шаблон:Math. By the above reasoning the map Шаблон:Math 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 Шаблон:Math by Шаблон:Math if Шаблон:Math. It follows that every set Шаблон:Math 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 Шаблон:Math to be a cancellative magma, e.g. a quasigroup. The cancellation property is enough to ensure that the Шаблон:Math are all different.

The axiom of choice implies a group structure

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 Шаблон:Math is equipotent with a unique cardinal number Шаблон:Abs which equals an aleph. Using the axiom of choice, one can show that for any family Шаблон:Math of sets Шаблон:Math (A). Moreover, by Tarski's theorem on choice, another equivalent of the axiom of choice, Шаблон:Math for all finite Шаблон:Math (B).

Let Шаблон:Math be an infinite set and let Шаблон:Math denote the set of all finite subsets of Шаблон:Math. There is a natural multiplication Шаблон:Math on Шаблон:Math. For Шаблон:Math, let Шаблон:Math, where Шаблон:Math denotes the symmetric difference. This turns Шаблон:Math into a group with the empty set, Шаблон:Math, being the identity and every element being its own inverse; Шаблон:Math. The associative property, i.e. Шаблон:Math is verified using basic properties of union and set difference. Thus Шаблон:Math is a group with multiplication Шаблон:Math.

Any set that can be put into bijection with a group becomes a group via the bijection. It will be shown that Шаблон:Math, and hence a one-to-one correspondence between Шаблон:Math and the group Шаблон:Math exists. For Шаблон:Math, let Шаблон:Math be the subset of Шаблон:Math consisting of all subsets of cardinality exactly Шаблон:Math. Then Шаблон:Math is the disjoint union of the Шаблон:Math. The number of subsets of Шаблон:Math of cardinality Шаблон:Math is at most Шаблон:Math because every subset with Шаблон:Math elements is an element of the Шаблон:Math-fold cartesian product Шаблон:Math of Шаблон:Math. So Шаблон:Math for all Шаблон:Math (C) by (B).

Putting these results together it is seen that Шаблон:Math by (A) and (C). Also, Шаблон:Math, since Шаблон:Math contains all singletons. Thus, Шаблон:Math and Шаблон:Math, so, by the Schröder–Bernstein theorem, Шаблон:Math. This means precisely that there is a bijection Шаблон:Math between Шаблон:Math and Шаблон:Math. Finally, for Шаблон:Math define Шаблон:Math. This turns Шаблон:Math into a group. Hence every set admits a group structure.

A ZF set with no 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 Шаблон:Math be any such set. Now consider the set Шаблон:Math. If Шаблон:Math were to have a group structure, then, by the construction in first section, Шаблон:Math can be well-ordered. This contradiction shows that there is no group structure on the set Шаблон:Math.

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.

For example, if <math>X</math> is any set, then <math>\mathcal P(X)</math> has a group structure, with symmetric difference as the group operation. Of course, if <math>X</math> cannot be well-ordered, then neither can <math>\mathcal P(X)</math>. One interesting example of sets which cannot carry a group structure is from sets <math>X</math> with the following two properties:

1. <math>X</math> is an infinite Dedekind-finite set. In other words, <math>X</math> has no countably infinite subset.
2. If <math>X</math> is partitioned into finite sets, then all but finitely many of them are singletons.

To see that the combination of these two cannot admit a group structure, note that given any permutation of such set must have only finite orbits, and almost all of them are necessarily singletons which implies that most elements are not moved by the permutation. Now consider the permutations given by <math>x\mapsto a\cdot x</math>, for <math>a</math> which is not the neutral element, there are infinitely many <math>x</math> such that <math>a\cdot x=x</math>, so at least one of them is not the neutral element either. Multiplying by <math>x^{-1}</math> gives that <math>a</math> is in fact the identity element which is a contradiction.

The existence of such a set <math>X</math> is consistent, for example given in Cohen's first model. Surprisingly, however, being an infinite Dedekind-finite set is not enough to rule out a group structure, as it is consistent that there are infinite Dedekind-finite sets with Dedekind-finite power sets.

References

• Шаблон:Cite journal
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