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?

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. 


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 BuraliForti 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:



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 Xmany group isomorphism classes, where X "counts" the number of setisomorphism 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 


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 supernaive 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 BuraliForti paradox (which on the other hand is manifestly choicefree), 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 SkolemLowenheim 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. 

