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Can you give me some examples of groups of cardinality greater than continuum? All the (infinite) examples that I've been taught are countable or of cardinality continuum.

Are there any ''natural examples''? (Examples that you come across when you are researching some problem or investigating some theory)

ante.ceperic
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    Cartesian products with many many factors can give you examples. – OR. Jul 31 '13 at 08:08
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    Another quick answer would be the free group generated by the reals or something bigger. – Scaramouche Jul 31 '13 at 08:09
  • Any direct sum or direct product of more than $2^\omega$ non-trivial groups. – Brian M. Scott Jul 31 '13 at 08:10
  • You said natural. Spaces of not-too-nice real functions, say on the reals. – OR. Jul 31 '13 at 08:11
  • @Scaramouche: the cardinality of the free group on $\mathbb{R}$ is "only" $# \mathbb{R} = \mathfrak{c}$: see my answer. (But yes: take something bigger...) – Pete L. Clark Jul 31 '13 at 08:32
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    It is a curious fact that all countably-generated groups are countable. Thus, an uncountable group must be uncountably generated! Which is, perhaps, why people shy away from studying arbitrary uncountable groups. (It is also perhaps because countable groups are rich and wild enough - we needn't increase cardinality to get our fix of excitement!) – user1729 Jul 31 '13 at 09:06
  • @user1729: I like your first sentence very much. About the last part: people do study arbitrary uncountable groups. I am just now finishing a paper with the goal to determine exactly which groups arise in a certain way (if you care: as the group of rational points on some elliptic curve over some field). What seems "shy-worthy" is studying uncountable groups using generators and relations. (Also, my result is just as hard to prove for arbitrary countable groups, so I also agree with you there.) – Pete L. Clark Jul 31 '13 at 09:15
  • @Pete Sorry, you are right. My comment was poorly written (I didn't want to leave it as the first sentence, which came across as a bit of a non-sequitur, so I added the last bit). I was not trying to imply that noone studied groups of cardinality greater than the continuum/uncountable groups. I suppose my point was that the study of arbitrary uncountable groups will gain you little which cannot be gained from studying arbitrary countable groups (but people still study specific uncountable groups). However, I am a biased generators and relations person so I am not sure if that is true. – user1729 Jul 31 '13 at 09:29
  • @user1729: Model theory can provide some formal justification for the richness of at-most-countable groups vis-a-vis all groups. In contrast, there are areas of group theory where nontrivial set theory intervenes, and there is no nontrivial set theory for countable sets. (That sounds too categorical, but I hope the meaning is clear: none of that hyper-Mahlo-weakly ineffable cardinal stuff...) The latter is certainly not something that has ever come up in my own work (if it did, I would probably turn and run), but it's good to know what's out there... – Pete L. Clark Jul 31 '13 at 09:36

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My favorite group of cardinality greater than the continuum is the group of field automorphisms of the complex numbers.

This is a very large and interesting group: not only does it have cardinality $2^{\mathfrak{c}} = 2^{2^{\aleph_0}}$ (to the AC patrol: yes, I'm assuming the Axiom of Choice here), it has this many conjugacy classes of involutions, which follows from (the affirmative answer I received to) this MO question. (This is also climbing my list of most frequently made errors by veteran mathematicians. I have watched many smart people claim that it follows from the Artin-Schreier Theorem that every index $2$ subfield of $\mathbb{C}$ is isomorphic to $\mathbb{R}$. The cardinal number by which this statement is off is pretty staggering.)

It's sort of a less fun answer, but: there are certainly groups of every infinite cardinality. If you want to be slick about it, this follows from the Lowenheim-Skolem Theorem in model theory (to the AC patrol...), since the theory of groups has a countable language and admits infinite models. One example is that the free abelian group (and also the free group) on an infinite set $S$ has cardinality equal to that of $S$. Moreover, the group $\operatorname{Sym} S$ of all bijections on an infinite set $S$ has cardinality $2^{\# S} > \# S$.

You can go on to construct your own favorite examples. E.g. there is a field $F$ of every infinite cardinality $\kappa$ (e.g. a rational function field over $\mathbb{Q}$ in $\kappa$ indeterminates) and for all $n \geq 2$, the group $\operatorname{PSL}_n(F) = \operatorname{SL}_n(F)/\text{center}$ of $n \times n$ matrices in $F$ with determinant $1$ modulo scalar matrices with determinant $1$ is a simple group of cardinality $\# F$. (This is a fun example because we also know all possible orders finite simple groups...but that's just a little bit harder!) All of these are fairly natural examples, I think.

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Pete L. Clark
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You can construct a group (ring) structure on any non empty set $E$. This is trivial if E is finite, so assume that $E$ is infinite. The free module $Z_2[E]$ over $Z/2Z$ with basis $E$, is equipotent to $P_f(E)$ the set of all finite subsets of $E$. We have only to show that $E$ and $P_f(E)$ have the same cardinality ( so we can trnsport any structure of one of them to the other).

Now you have only to see The cardinality of the set of all finite subsets of an infinite set

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Yassine Guerboussa
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Consider $\mathbb{R}$ as a $\mathbb{Q}$-vectorspace and look at $\rm{Hom}_{\mathbb{Q}}(\mathbb{R},\mathbb{Q})$, the set of $\mathbb{Q}$-linear maps from $\mathbb{R}$ to $\mathbb{Q}$.

Since the dimension of $\mathbb{R}$ is $|\mathbb{R}|$, the above vectorspace has strictly larger dimension. This then means that the cardinality of the above space is strictly larger than $|\mathbb{R}|$ and hence so is the order of the underlying abelian group.

Tobias Kildetoft
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