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I have learned that the Axiom of Choice is equivalent to the statement that every set can be endowed with a group structure. Now, in searching for answers to the question asked in the title, I have found that the canonical explanation for why the collection of all groups is a proper class is that, for any set (or really for any cardinality) there is a free group on the elements of that set. I'm wondering:

1)Is the fact that we have so many free groups related to/dependent on AC?

2)Are the groups that we lose in switching from AC to $\neg$ AC enough that the collection of groups becomes a set?

Perry Bleiberg
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  • In addition to what Noah wrote about the second question, you might want to read http://math.stackexchange.com/questions/317028/ and the answer I've posted there, which sort of asked a similar thing. – Asaf Karagila Feb 19 '17 at 01:49

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No, Choice is not needed to show that the class of groups is not a set. There are several ways to see this.

  • The class of one-element sets is not a set - and each one-element set can be given a group structure (the trivial group), so there are proper-class-many trivial groups.

  • If you want a proper class of nonisomorphic groups, this can still be done: as long as $X$ is an infinite well-orderable set, we can construct (without Choice!) a free group with domain $X$. The claim now follows from Burali-Forti.

  • And even this is unnecessary: For $X$ a set, we may - again, without choice - form the free group generated by $X$. The domain of this group will be the set of appropriate equivalence classes of finite strings from $X$, which exists without Choice. And now we just need to find a proper class of sets which yield nonisomorphic free groups in this way, and this is not hard to do. In particular, this is the method you mention in your question - and so the answer to your sub-question is No, Choice is not needed to construct the free group on a set. Note that this group is different from a free group with domain equal to a given set: showing that every infinite set is the domain of a free group does indeed require Choice.

These general principles apply to practically every kind of structure: except in very artificial cases, the class of structures of a certain type is a proper class, and unless there is a bound on the size of the cardinalities of these structures (e.g. finite groups) then there is a proper class of non-isomorphic structures of this type.

Also, your conception of the role of Choice in the size of the universe is incorrect. You can think of a model with Choice as being small, actually, since it doesn't have any non-well-orderable sets; and this makes as much sense as thinking of a model without choice as being small, since it doesn't have enough choice functions. And remember that choice holds in the constructible universe, so any model of ZF contains a submodel of ZFC; so in a precise sense, you can gain choice by losing sets.

Noah Schweber
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  • Thanks for this great answer. I am having trouble finding the definition of a domain of a group, which I am not familiar with. – Perry Bleiberg Feb 18 '17 at 19:20
  • @user52969 The domain is just the underlying set. E.g. the domain of the group of integers under addition is just the set of integers. Note that isomorphic groups can have different domains. – Noah Schweber Feb 18 '17 at 20:05
  • @AkivaWeinberger Yes, and Asaf gave this example in his answer below. – Noah Schweber Feb 21 '17 at 16:33
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Consider, in $ZF$, Godel's proper class $L$ of constructible sets. $(L,\epsilon)$ satisfies $ZFC.$

As for $On,$ the ordinals, we have $x\in On\iff (x\in On)^{(L,\epsilon)}.$

For $x\in On$ let $G_x$ be such that $(G_x$ is a group-structure on $x)^{(L,\epsilon)}.$

Observe that $(G_x$ is a group-structure on $x)^{(L,\epsilon)}\implies (G_x$ is a group structure on $x)$.

So regardless of $AC$ or $\neg AC$, there is a group-structure on every ordinal.

DanielWainfleet
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Every singleton can be endowed with a group structure. Even if you want them to be non isomorphic, every power set can be given a [commutative] group structure, using the symmetric difference as the addition operation.

No choice is needed to show there is a proper class of singletons or a proper class of power sets.

Asaf Karagila
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  • +1 for the powerset example, which I think is more natural than the wellorderable-sets example (though I think it would help the OP if you explained how powersets can be given a group structure . . .). – Noah Schweber Feb 18 '17 at 16:26