Past exam of the logic and set theory course of the University of Oxford / 2009 / B / 6

Zorn's lemma states that a nonempty partially ordered set in which every chain has an upper bound has a maximal element. Order the partial group operations by inclusion of their graphs. The collection is a set, since its members are subsets of $X\times X\times X$. It is nonempty: choose $x_0\in X$ and put the one-element group operation on $\{x_0\}$.

To construct a chain union of partial group operations, for the given nonempty chain, let $m={\bigcup }_{n\in \mathcal{C}}n$ and let $A$ be the union of the corresponding group carriers. Comparability guarantees agreement wherever two operations are both defined, so $m$ is a function. Its domain is $A\times A$: any two elements of $A$ lie together in one carrier, by taking the larger of the two chain members containing them. Their product then lies in that carrier and hence in $A$.

Any three elements similarly lie together in one chain member, where associativity holds; therefore associativity holds for $m$. The identities of nested nonempty groups agree: the smaller identity remains an idempotent under the larger group operation, whose only idempotent is its identity. Hence the whole chain has a common identity. Each element's inverse in a carrier remains its inverse in every larger carrier, so it is an inverse in the union as well. Thus $(A,m)$ satisfies all group axioms andIt is an upper bound for the chain. The empty chain also has an upper bound because $\mathcal{G}$ is nonempty, completing the hypothesis needed for Zorn's lemma.

Apply Zorn's lemma using the chain upper bounds just proved, obtaining a maximal partial group operation $m$ on a nonempty carrier $A\subseteq X$. Suppose $|X\setminus A|\ge |A|$. Then choose a subset $B\subseteq X\setminus A$ and a bijection $b:A\to B$. The set $A\cup B$ can be identified with $A\times C_2$ by sending $(a,0)$ to $a$ and $(a,1)$ to $b(a)$.

Use group extension by a duplicate carrier to transport the direct-product group operation along this bijection. Its restriction to $A\times A$ is exactly the original $m$, because $(a,0)(a^{\prime },0)=(m(a,a^{\prime }),0)$. It therefore gives a strictly larger member of $\mathcal{G}$, contradicting maximality. Thus $|A|\nleqq |X\setminus A|$. Cardinal comparability, allowed as a consequence of Zorn's lemma, leaves the strict opposite inequality:The argument uses choice through Zorn's lemma and cardinal comparability; the extension itself is an explicit transported direct-product operation once the bijection with $B$ is chosen.

If $X$ is finite and nonempty, let $n=|X|\ge 1$, choose a bijection with the cyclic group $\mathbb{Z}/n\mathbb{Z}$, and transport its addition to $X$. This includes the one-element case.

If $X$ is infinite, use the maximal partial group $(A,m)$ from the preceding part. Its carrier must be infinite: otherwise $|X\setminus A|<|A|$ would make the complement finite too, forcing $X$ finite. Since the complement injects into $A$,where the last equality is the permitted infinite-cardinal doubling identity. Cantor-Schröder-Bernstein supplies a bijection $j:X\to A$. DefineAssociativity follows by applying $j$ and using associativity of $m$; the identity is $j^{-1}(e_A)$ and the inverse of $x$ is $j^{-1}(j(x{)}^{-1})$. Hence $(X,\star )$ is a group.

This final transport need not extend the earlier partial operation on its original subset. That stronger claim is neither needed nor generally possible: for example a group of order three cannot occur as a subgroup of a group of order four. The proof establishes exactly the existence of some group structure on every nonempty set, using the allowed choice consequences in the infinite case.