Forming a group over an arbitrary collection $C$. Does this imply $C$ is a set (under ZFC)?

Question

I was puzzling over a question this morning that arose from a category theoretical problem and don't think I have the set-theoretic background to answer it myself. The question arose from trying to demonstrate that some weird category was abelian - I constructed a rule for the group structure, but wasn't sure that the Hom collections were sets in the first place.

Given an arbitrary collection $C$ of objects which is not assumed to be a set, if we can construct a binary operation $\cdot$ which satisfies all group axioms, does this imply that $C$ is a set (under ZFC axioms)?

A pedantic answer could be "yes" since then we have a group, and groups are defined over sets, but this does not sit well with me. On the other hand, to me, it seems like a group operation is "local" in that it shouldn't have to consider the size of the universal collection.

Or alternatively, is there anything in ZFC that restricts us from defining groups over arbitrary collections rather than restricting to sets? What goes wrong?

Answer 1

Take the class of all sets, and define the binary operation as symmetric difference. Then all group axioms are satisfied.

There is no reason a group has to be a set, but there are too many obstacles, for example the binary operation cannot be defined as a function, but a proposition $\forall x \forall y \exists !z, p(x,y,z)$. As so, we cannot even formulate this trivial fact in ZFC: for any group $G$, $G$ has an identity element -- because we cannot quantify over propositions, hence cannot quantify over groups. So it's difficult to formulate any nontrivial theorem about groups if the underlying class is not a set.

Also, if we pick a model of ZFC, then the above construction indeed gives a genuine group in that model (as models are always sets), though the symmetric difference in the model isn't necessarily the real one. So being too large is more a theoretical annoyance than essentially bad.

Answer 2

No. You can even put an abelian group structure on the universe of sets (usually called $V$), the binary operation is symmetric difference, the identity element is $\emptyset$, and every element is its own inverse.

Now, what does this mean from a foundational point of view? Well, you can write a sentence $\sigma$ in the language of set theory which says $$ \forall A,B,C [A\Delta (B\Delta C)=(A\Delta B)\Delta C]\wedge \forall A,B[A\Delta B=B\Delta A]\wedge \forall A[A\Delta \emptyset=A] $$ Then the statement "$(V,\Delta)$ is a group" is really an shortened way of saying that $\sf ZFC$ (or whatever your ambient set theory is) proves the sentence $\sigma$.

In the meta theory, we have a relative interpretation of the theory of (abelian) groups into $\sf ZFC$: to the binary operation symbol (from the theory of groups), we associate a formula $\varphi(x,y,z)$ in the language of set theory such that $$ {\sf ZFC}\vdash \forall x,y\exists !z \varphi(x,y,z) $$ (where $\varphi(x,y,z)$ "says" $z=x\Delta y$) and $\sf ZFC$ proves every group axiom, with each instance of the binary operation adequately replaced by $\varphi$ so that you actually have a formula in the language of set theory.