In group theory, the single most important piece of information about a group is its cardinality, which is of course either finite, countably infinite, or uncountably infinite. Usually, however, uncountably infinite simply means a cardinality of $\aleph_{1}$, the same as $\mathbb{R}$. My question is: is there anywhere that groups with cardinality strictly greater than $\aleph_{1}$ arise naturally? Of course, it is easy enough to construct groups with arbitrarily large cardinality, but I cannot recall ever seeing them used.

31$\begingroup$ The cardinality of $\mathbb{R}$ is not necessarily $\aleph_1$. $\endgroup$ – François G. Dorais♦ Aug 13 '10 at 0:45

23$\begingroup$ What gave you the impression that the single most important piece of information about a group is its cardinality? It misses nearly all of the richness of group theory. For instance, many interesting classes of finite simple groups can be studied using a lot of the same tools as are used in the study of simple complex Lie algebras and semisimple linear algebraic groups. $\endgroup$ – BCnrd Aug 13 '10 at 0:54

15$\begingroup$ Turning this remark around, can one give a "natural" example of a group which has cardinality $\aleph_1$, independent of CH? $\endgroup$ – Pete L. Clark Aug 13 '10 at 0:55

8$\begingroup$ Pete, very good question! Here is one answer: it is consistent with ZFC that there are no Borel sets, and even no analytic sets, of size $\aleph_1$. In such a model of set theory, the only groups (built out of real numbers) of size $\aleph_1$ would have high descriptive settheoretic complexity. This could be taken as a negative answer to your question. But a positive answer could still arise by builiding a group directly out of the countable ordinals, rather than by using reals. $\endgroup$ – Joel David Hamkins Aug 13 '10 at 1:05

3$\begingroup$ You can push it up to $\Sigma^1_2$ using the Mansfield Solovay theorem, and if you assume PD, then there are no projective sets of size $\aleph_1$. In those models, there are arguably no natural examples of sets of reals of size exactly $\aleph_1$. $\endgroup$ – Joel David Hamkins Aug 13 '10 at 3:48
In line with Joel's answer, my favorite "outrageously large group" is the group $G = \operatorname{Aut}(\mathbb{C})$ of field automorphisms of the complex numbers. It has cardinality $2^{2^{\aleph_0}}$, which is pretty scary. But that's just the beginning of how large it is. For instance, from the study of realclosed fields, one can deduce that the number of conjugacy classes of order $2$ elements of $G$ is also $2^{2^{\aleph_0}}$. It is also an extension of the absolute Galois group of $\mathbb{Q}$ (a profinite group which is conjectured to have among its quotients every finite group, up to isomorphism) by the huge simple group $\operatorname{Aut}(\mathbb{C}/\overline{\mathbb{Q}})$.

5$\begingroup$ A nice reference for some of the basic facts on Aut(C) is the 1957 Lester R. Ford prizewinning article: Paul B. Yale, Automorphisms of the complex numbers, Math. Mag. 39 (1966), 135141. jstor.org/stable/2689301 $\endgroup$ – Skip Aug 13 '10 at 0:59

1$\begingroup$ Why is $\operatorname{Aut}(\mathbb C/\overline{\mathbb Q})$ simple? $\endgroup$ – Emil Jeřábek Mar 1 '11 at 17:26

3$\begingroup$ @Emil: That is a special case of a theorem of Daniel Lascar: see the end of math.uga.edu/~pete/galois.pdf for a precise reference. $\endgroup$ – Pete L. Clark Mar 1 '11 at 17:30
I would expect that automorphism groups of natural structures would count as natural groups in your sense. But automorphism groups of uncountable structures often have size larger than the continuum. In general, the size of the automorphism group of a structure of size $\kappa$ is bounded above by $2^\kappa$, which is strictly larger than $\kappa$, and this upper bound is often reached, when the structure is insufficient to restrict the general nature of automorphisms. For example, the number of bijections of an infinite set of size $\kappa$ with itself is $2^\kappa$.
I am sure that you will be able to construct many other natural structures of uncountable size $\kappa$, whose automorphism groups have size $2^\kappa$, and these would seem to the sort of examples you seek.
P.S. Let me also note that your remark that the reals have size $\aleph_1$ is only correct when the Continuum Hypothesis holds. In general, the size of the reals, also known as the continuum, is $2^{\aleph_0}$, which is also denoted $\beth_1$, whereas $\aleph_1$ is simply the first uncountable cardinal.
Does a group showing up in a College Algebra (precalculus) course count as arising naturally? I'm pretty sure we teach students to add two functions (from the reals to the reals) pointwise to get a new function, even there. Of course, on the one hand really we only ask them to deal with the countable subset of functions with a finite description, and on the other hand Abelian groups are not as interesting, but technically that defines a group with cardinality greater than the continuum. (We also define inverse functions and composition, but at first glance it seems that strictly monotone functions must have only continuum cardinality.)

$\begingroup$ But if you are truly in precalculus, you likely only find piecewise continuous functions, and even if you allow countably many pieces, you still have only continuum many such functions, the same size as $\mathbb{R}$, since on each piece they are determined by their values on the rationals. $\endgroup$ – Joel David Hamkins Aug 13 '10 at 1:38

1$\begingroup$ @Joel: Hence my caveat about only asking them to grok the countable subgroup of finitely described functions. $\endgroup$ – Tracy Hall Aug 13 '10 at 1:53

$\begingroup$ Yes, that is a nice way to say it. (+1). $\endgroup$ – Joel David Hamkins Aug 13 '10 at 1:55

1$\begingroup$ Every Banach space is an Abelian group, and many of the usual Banach spaces we study have larger cardinality than the reals. $\endgroup$ – Carl Mummert Aug 13 '10 at 12:55

$\begingroup$ @CM: Really? (Meaning: "I know very little about Banach spaces", not "I doubt that very much.") What is such an example? $\endgroup$ – Pete L. Clark Aug 13 '10 at 23:24
Any product group like $\{0, 1\}^I$ for index sets $I$, using mod 2 addition coordinatewise. It's just (isomorphic to) the power set of $I$ using symmetric difference as the addition. It's of course also a ring (pointwise multiplication / intersection ). These Boolean groups often come up in general topology.