Groups of cardinality greater than the continuum 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) 
 A: 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 
A: 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.
A: 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.
