Does every set have a group structure?

Solution 1:

The trivial answer is "no": the empty set does not admit a group structure.

The statement

If $X$ is a nonempty set, then there is a binary operation $\cdot$ such that $(X,\cdot)$ is a group.

is equivalent to the Axiom of Choice.

It is not needed for finite or countable sets: if $X$ is finite, with $n$ elements, then let $f\colon X\to\{0,1,\ldots,n-1\}$ be a bijection, and use transport of structure to give $X$ the structure of a cyclic group of order $n$. If $X$ is denumerably infinite, biject with $\mathbb{Z}$ and use transport of structure.

For uncountable sets, we can use the Axiom of Choice: let $|X|=\kappa$. Then the direct sum of $\kappa$ copies of $\mathbb{Z}$ has cardinality $\kappa$, so there is a bijection $$f\colon X\to \bigoplus_{i\in\kappa}\mathbb{Z}.$$ Use transport of structure again to make $X$ into a group.

That the converse holds (the statement implies the Axiom of Choice), is proven in this Math Overflow post.

Solution 2:

Assuming the axiom of choice, then the universe is "well-behaved" and we are finding ourselves in two different situations given a non-empty set $X$:

• If $X$ is finite, then there is a bijection between $X$ and $\{0,\ldots,n-1\}$ which then implies there is a group structure similar to $X$ and $\mathbb Z/n\mathbb Z$ with $+^{\!\!\mod n}$.
• The set $X$ is infinite, then by we can take $G=\bigoplus_{i\in X}\mathbb Z$. We can consider $G$ as finite functions from $X\times\mathbb Z$. This means that: $$|G|\le|\{f\subseteq X\times\mathbb Z\mid f\text{ is a finite set}\}|=|X\times\mathbb Z|=|X|\cdot\aleph_0=|X|\le|G|$$ Where the first $=$ sign follows from the axiom of choice: every infinite set is equinumerous with the collection of all its finite subsets; and the last $\le$ follows from the injective map $x\mapsto\{\langle x,1\rangle\}$, thus by Cantor-Bernstein we have that $X$ and $G$ have the same cardinality therefore we can use a bijection between them to define the group structure on $X$.

On the other hand, it appears that if every set has a group structure then the axiom of choice holds. The proof appears in this MathOverflow thread, and requires a mild familiarity with constructs which relate to the axiom of choice.

The nutshell of the proof is this:

• Given an infinite set $X$ we define $H(X)$ to be the least ordinal $\alpha$ that there is no injection $g:\alpha\to X$ (this is known as the Hartog number of $X$)
• If $X$ can be injected into $H(X)$ then $X$ can be well ordered, since being injected into an ordinal means that $X$ inherits a well order.
• Using the assumption that every set can be given a group structure we give a group structure to $X\cup H(X)$, and from this we deduce that there exists an injection from $X$ into $H(X)$.
• Therefore if every set can be given a group structure, every set can be well ordered and therefore the axiom of choice holds.

Lastly, a somewhat natural example of a set which cannot be given a group structure in a model contradicting the axiom of choice:

We say that $A$ is Dedekind-finite if every proper subset $B$ of $A$ has cardinality strictly less than the cardinality of $A$. Every finite set, then, is a Dedekind-finite set. Equivalently $A$ is Dedekind-finite if and only if it does not have a countably infinite subset.

When not assuming the axiom of choice it is consistent that infinite Dedekind-finite sets exist (infinite means not in bijection with $\{0,\ldots,n\}$ for any $n\in\mathbb N$).

The first Cohen model which exhibited the independence of the axiom of choice from ZF was one in which he added a Dedekind-finite set of real numbers. This set cannot be given a group structure.

Why? If $X$ has a group structure if there is an element of infinite order we can define an injection from $\mathbb N$ into $X$ using $n\mapsto x^n$; if all the elements have finite order then we can partition the set into infinitely many parts of finite size defined by the order of each element.

The set $A$ in the first Cohen model is Dedekind finite, therefore every group structure would have that all elements have finite order; however its construction gives us that every partition into finite subsets is almost entirely singletons. This means that if a group structures was endowed almost every element would have to be of order $1$. This is of course impossible, therefore this set is an example of how a set might not have a group structure definable on without the axiom of choice.

(Interestingly enough, not all infinite Dedekind-finite sets are counterexamples. It is perfectly consistent to have a group structure on an infinite Dedekind finite set in some cases!)