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?
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asked Jan 25, 2010 at 8:49
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10$\begingroup$ Technically, the question is not properly posed, since the isomorphism class of any single group (even the trivial group) is itself a proper class, and not a set. Thus, you have a "collection" of proper classes already, and this is neither a set nor a class. Rather, what you want to ask about is: Is there a proper class of pairwise non-isomorphic groups? And this is the question that the answerers answered. $\endgroup$– Joel David HamkinsCommented Jan 25, 2010 at 13:06
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$\begingroup$ I appreciate this clarification, thank you. At least I had something sensible in mind, even if I spelled it out incorrectly. Sorry for that. $\endgroup$– Hans-Peter StrickerCommented Jan 25, 2010 at 14:19
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4 Answers
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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.
answered Jan 25, 2010 at 8:53
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1$\begingroup$ You don't need AC if you just use the cardinals: you have Z^kappa for different cardinals kappa. $\endgroup$ Commented Jan 25, 2010 at 13:10
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$\begingroup$ Thanks! I'm always wary of such issues so phrase my answers to "play safe" so it's nice to know when such is warranted or not. $\endgroup$ Commented Jan 25, 2010 at 13:46
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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:
- 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.
- 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.
- Automorphism groups of sets.
- Automorphism groups of suitable vector spaces.
- Free groups, as in Andrew Stacey's answer.
- Free abelian groups.
- Amalgams using your favorite groups (assuming suitably nontrivial choices).
answered Jan 25, 2010 at 9:20
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$\begingroup$ I agree with you, I immediately stumbled over this after having read Andrew's answer. $\endgroup$ Commented Jan 25, 2010 at 9:24
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$\begingroup$ I should clarify: I'm pretty sure set theorists would say that the claim "cardinals form a proper class" is a standard result, e.g., in ZFC. I happen to think that a careful proof of this is trickier than any of the listed group constructions, but that may say more about my background than anything else. $\endgroup$– S. Carnahan ♦Commented Jan 25, 2010 at 11:34
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$\begingroup$ If the cardinals were a set, then their supremum would be a largest cardinal. But there is no largest cardinal, since (in ZFC) Cantor proved kappa < 2^kappa. (In ZF, one can similarly show kappa+ exists, and kappa < kappa+, so AC is not needed.) $\endgroup$ Commented Jan 25, 2010 at 13:17
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$\begingroup$ @JDH: yes, that's pretty much the argument of the first link I provided below (not including the parenthetical part of your remark). $\endgroup$ Commented Jan 25, 2010 at 13:28
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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
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$\begingroup$ An my thought was: direct sum of copies of Q. I.e., the dimension of the vector space. For every cardinal there is a vector space with that dimension, and spaces with different dimension (over Q) are not isomorphic as groups. $\endgroup$ Commented Jan 25, 2010 at 13:10
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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://alpha.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://alpha.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://alpha.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.
answered Jan 25, 2010 at 13:14
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$\begingroup$ About your AC remark and cardinal arithmetic: If ZF, you can prove that kappa^2=kappa for any infinite (well-ordered) cardinality, so yes, kappa\times\aleph_0 = kappa for such kappa. If you mean "cardinal" in the not-AC sense, however, then if kappa is infinite but Dedekind finite, it cannot have \kappa\times\aleph_0 = \kappa, since this implies a countably infinite subset. But to get the proper class of models, as you suggest, one needs to use only the well-ordered cardinals. $\endgroup$ Commented Jan 25, 2010 at 13:23