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Let $\kappa$ be a fixed cardinality. Consider the isomorphism classes of groups that are generated by a set of cardinality at most $\kappa$, and pick a representative for each isomorphism class. Do these representatives form a set? (That is to say, not a proper class).

I think we may equivalently show that the (representatives of isomorphism classes of) groups of rank equal to $\kappa$ form a set. Then we could take the union of these sets over cardinalities $\le \kappa$.

Honestly, I don't really know where to start here – it's a fair bit out of my comfort zone. Can we perhaps bound the cardinality of a group depending on its rank?

I'll also be happy with a reference or a rough outline, since I didn't manage to google my way to an answer.

This question came about from seeing that this construction is used to construct solution sets for the forgetful functor $\mathrm{Group} \to \mathrm{Set}$ (e.g. in Category Theory in Context, Riehl), which in turn let's Quoting Riehl:

Let $\Phi'$ be the set of representatives for isomorphism classes of groups that can be generated by a set of elements of cardinality at most $|S|$. Let $\Phi$ be the set of functions $S \to UG$, with $G \in \Phi'$, whose image generates the group $G$. This is [...] a solution set for $U : \mathrm{Group} \to \mathrm{Set}$.

In this context, it is imporatant that the solution sets are indeed small sets, in order for the adjoint functor theorem to work.

Marius S.L.
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Milten
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Yes, they form a set (I'm assuming the Axiom of Choice here, for cardinality arguments; but you need it to select one representative from each equivalence class anyway, so...). Indeed, you can bound the size of a group in terms of the size of a generating set for the group.

(You can also find the argument in George Bergman's exposition of how to construct the free group on three elements as a subgroup of a big enough direct product in his Invitation to General Algebra and Universal Constructions, Section 3.3, pp 33. Available from his website, page 30 of the PDF there.

I've also given the argument below in slightly less generality, e.g. here )

Lemma. Let $G$ be a group, and let $X\subseteq G$ be a generating set (that is, $\langle X\rangle = G$). Then $|G|\leq |X|\aleph_0$. In particular, if $X$ is countable then $G$ is countable, and if $X$ is uncountable then $|G|=|X|$.

Proof. The elements of $G$ are finite products of elements of $X$ and their inverses. The set of all possible finite products has at most $$2|X| + (2|X|)^2 + \cdots +(2|X|)^n+\cdots$$ elements, hence at most $|X|\aleph_0$ elements.

If $|X|$ is infinite, then $|X|\aleph_0=|X|$, so $|X|\leq |G|\leq |X|$, giving the equality. $\Box$

Theorem. Let $S$ be a set of cardinality $\kappa$. Then every group of cardinality at most $\kappa$ is isomorphic to a group whose underlying set is a subset of $S$.

Proof. Let $G$ a group of cardinality at most $\kappa$. Then there exists an one-to-one (set) function $f\colon G\to S$. Let $H$ be the group with underlying set $f(G)$ and with operation $f(x)*f(y) = f(xy)$, where $xy$ is the product in $G$. (In other words, use transport of structure. Then $f$ is an isomorphism between $G$ and the group $H$, whose underlying set is a subset of $S$. $\Box$

In particular, if $\kappa$ is a cardinal, then every group $G$ of cardinality at most $\kappa$ is isomorphic to a group whose underlying set is a subset of $\kappa$ itself.

In addition, the collection of all possible group structures on a set $S$ is itself a set, since you can think of them as a subset of the collection of all tuples $(S,f,i,x)$, where $f\colon S\times S\to S$ is a binary operation, $i\colon S\to S$ is a unary operation, and $x\in S$ is a nullary operation (by identifying the multiplication of $S$ with $f$, the inversion map with $i$, and the identity element with $x$).

So, let $\kappa$ be a cardinal. Let $S$ be a set of cardinality $\kappa\aleph_0$. Then every group generated by a set of cardinality at most $\kappa$ is isomorphic to an element from the set $$\{(X,f,i,x)\mid X\subseteq S, f\colon X^2\to X, i\colon X\to X, x\in X\}.$$

Hence there is a collection of representatives of the isomorphism class which is a set. By the Axiom of Replacement, any collection of representatives of isomorphism classes of groups generated by a set of cardinality at most $\kappa$ forms a set.

Arturo Magidin
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    The axiom of choice is really not necessary here, you just get a clunkier cardinal bound without it (you can still find some set $S$ such that every group generated by $\kappa$ elements embeds in $S$; $S$ will be something like $P(P(\kappa^\mathbb{N}))$). And the application in question doesn't rely on choosing exactly one representative of each isomorphism class; you just need a set which contains at least one representative of each isomorphism class. – Eric Wofsey Apr 19 '23 at 18:16
  • @EricWofsey: You are right, of course, but I didn't really want to worry too much about the AoC issues here. – Arturo Magidin Apr 19 '23 at 18:19
  • Thank you! That was rather simple I see That lemma was the key that I should have noticed. – Milten Apr 19 '23 at 18:32
  • @Milten Analogues of this lemma are the key to applications of the general adjoint functor theorem to construct free gadget functors for nearly any kind of algebraic gadget, so you should find many more uses for it in the future. – Kevin Carlson Apr 20 '23 at 16:04