Question

The collection of all groups is a proper class, since every set gives rise to a group. But what about the collection of all isomorphism classes of groups? By which argument do I see, that it is a set or a proper class?

Answer 1

Since free groups are isomorphic if and only if they have a basis of the same cardinality (probably assuming some choice axiom here), the collection of all isomorphism classes of groups has at least the size of the collection of all isomorphism classes of sets, hence is not a set.

Answer 2

I think the hard part of the question is showing that the isomorphism classes of sets do not form a set, i.e., the cardinals form a proper class. I'm not a set theorist, but I think the Burali-Forti paradox is applicable here.

Once we have a proper class of cardinals, there are many constructions of groups corresponding to sets of the appropriate size. Examples:

1. For each set, its power set has a natural structure as a vector space over a field with two elements, and these can be distinguished by their cardinality.
2. Similarly, you can make vector spaces of chosen dimension over your favorite field, and if the dimension is larger than the cardinality of the field, the underlying sets can be distinguished.
3. Automorphism groups of sets.
4. Automorphism groups of suitable vector spaces.
5. Free groups, as in Andrew Stacey's answer.
6. Free abelian groups.
7. Amalgams using your favorite groups (assuming suitably nontrivial choices).

Answer 3

If you don't like Andrew Stacey's choice of free groups, consider the group of (finite or infinite, as you like) permutations on a set G. As in Andrew's example, a group isomorphism is a set isomorphism, so again there are at least X-many group isomorphism classes, where X "counts" the number of set-isomorphism classes, e.g. cardinals. I prefer using Choice to simplify arguments, but I suspect the above example can be done in ZF without choice, especially if you restrict to torsion permutation groups.

Gerhard "Ask Me About System Design" Paseman, 2010.01.25

Answer 4

All the other answers are more than satisfactory. I have some lecture notes on basic set theory which also answer these questions, so I might as well post links to them:

To see that the cardinals do not form a set, see Fact 20 on page 10 of

http://math.uga.edu/~pete/settheorypart1.pdf

This first handout is super-naive set theory (countable choice is assumed without comment) which is meant to be accessible to an undergraduate math major. In particular I don't say "cardinal" there but rather spell things out in a more explicit way.

This seems a little simpler than the Burali-Forti paradox (which on the other hand is manifestly choice-free), for which see Section 1.4 of

http://math.uga.edu/~pete/settheorypart3.pdf

To see that there is an X of any given infinite cardinality (where X is: a field, a noncommutative group, etc.) see

http://math.uga.edu/~pete/settheorypart4.pdf

My argument for the case of fields -- which implies that of groups by taking the additive group of the field -- uses (only) that for any infinite cardinal $\kappa$, $\kappa \times \aleph_0 = \kappa$. I'm pretty sure that this is a lot weaker than AC.

If you know the Skolem-Lowenheim theorem in model theory, then it is silly to do all of these cases individually: a consistent theory in a language of cardinality $\kappa$ admits models of any infinite cardinality which is greater than or equal to $\kappa$. This is equivalent to AC (see Bell and Slomson), but the special case of countable languages is presumably not.