If every definable class admits a group structure, must global choice hold?

Question

It is a remarkable fact, due to Hajnal and Kertész and explained very well in this MathOverflow answer by user Ashutosh, that the axiom of choice is equivalent to the assertion that every nonempty set carries a group structure.

My question is about the class analogue of this statement.

Questions.

1. In models of ZFC, is global choice equivalent to the assertion that every definable nonempty class carries a definable group structure?
2. In Gödel-Bernays set theory GB+AC, dropping the definability requirement on the classes, is global choice equivalent to the assertion that every nonempty class carries a group structure?

By global choice in the ZFC context I intend to refer to the existence of a definable global choice function, allowing parameters, and this is equivalent to the existence of a (parametrically) definable well ordering of $V$ in order type Ord. In the GB context, the concept of class is provided explicitly by the Henkin semantics.

In the set case, the Hajnal-Kertész proof (same as Ashutosh's) makes a key use of Hartog's theorem, asserting that every set $x$ admits an ordinal that does not inject into it. But the class analogue of this theorem is simply not true for classes, since for example every class injects into $V$, being a subclass of $V$. So the proof simply doesn't generalize directly from sets to classes.

This makes me expect that the answers will be negative, and the challenge is to find a model of ZFC where every definable class admits a group structure, but nevertheless global choice fails. I have no idea how to make that happen.

Alternatively, perhaps there is another proof of the set-based result, which does not use Hartog's theorem, and which generalizes naturally to classes, in which case we could expect positive answers.

Other natural questions in the vicinity would be:

• Does $V$ admit a definable group structure in ZFC? (Yes, using symmetric difference, answered in comments by Asaf.)
• What about the class of all sets of ordinals? Or $[\text{Ord}]^\omega$? (Yes, answered in the comments)
• In general, which classes in ZFC admit a group structure?

Of course if global choice holds, then it is easy to define suitable group structures on any nonempty class, the same as one would do it for sets of any nonzero size.

Answer 1

Yes. By the argument in the answer you linked, for any class $A$, either $\text{Ord}$ injects into $A$, or $A$ is well-orderable. If $A$ is well-orderable and a proper class, then $\text{Ord}$ injects into $A$ (send $\alpha$ to the $\alpha$th element), so in either case $\text{Ord}$ injects into $A$. You proved in this answer that this implies global choice.

Answer 2

ZFC proves that $V$ admits a definable group structure (and actually ZF does too). First note that the Schröder–Bernstein theorem still holds in the context of definable classes (i.e., for any definable classes $A$ and $B$, if there is a definable injection from $A$ into $B$ and from $B$ into $A$, then there is a definable bijection between $A$ and $B$). This means that it's sufficient to exhibit a definable class group $G$ with a definable injection $f: V \to G$.

We can now just take $G$ to be the 'free group generated by $V$.' In other words, the elements of $G$ are finite sequences of pairs $(x,\pm)$ satisfying that there are no consecutive pairs of the form $(x,+)(x,-)$ or $(x,-)(x,+)$ (where $x$ is an arbitrary set and $\pm$ is one of two fixed tokens indicating whether $(x,\pm)$ is meant to be interpreted as $x$ or $x^{-1}$). Multiplication on $G$ is defined in the obvious way. The map taking $x$ to the singleton sequence $(x,+)$ is obviously an injection of $V$ into $G$ and the identity map is an injection of $G$ into $V$, so we can define a group law on $V$ via the bijection given to us by the Schröder–Bernstein theorem for definable classes.