Can a non abelian group structure be given to $\Bbb{R}$? Can a non abelian group structure be given to the set of real numbers. I was told that such structure can be given to $\Bbb{R}$ , as we can get a bijection from $\Bbb{R}$ to $GL_2({\Bbb{R}})$ and since the operation on $GL_2({\Bbb{R}})$ is non abelian we can thus define a non abelian structure on $\Bbb{R}$.
Are there any other ways of giving $\Bbb{R}$ a non abelian group structure instead of trying to get a bijection from $\Bbb{R}$ to another infinite (uncountable) group where the operation is non abelian. Is this the only way sets like $\Bbb{R}$(which seem to have an inate property of being abelian) be given a non abelian group structure
 A: As explained in the other answers, if you only consider $\mathbb{R}$ as a set than there are a lot of group structures which make it a non-abelian group, since it is just a set of cardinality $2^{\aleph_0}$ with no added structure. If you want a constructive definition for the group operation, then it shouldn't be hard to construct a bijection of $\mathbb{R}$ with $\mathbb{R}^4$, and on $\mathbb{R}^4$ you can take the group structure of the quaternions I think you can just construct a bijection of $\mathbb{R}$ with $$\left\{\left(
 \begin{matrix}
  a & b \\
  0 & 1
 \end{matrix}
\right):a,b\in\mathbb{R}\right\}.$$
However, if you want to consider the topology of $\mathbb{R}$, then there is no group structure making it into a non-abelian topological group. Every topological group homeomorphic to $\mathbb{R}$ is a one-dimensional Lie group and hence is abelian (and in fact is isomorphic to $\mathbb{R}$ itself).
A: In a way, yes. 
As you notice, the natural group structure on $\Bbb R$ is that of addition (induced from the addition of rationals and ultimately the addition of integers). Thus there is no canonical or even easily describable non-abelian group structure on $\Bbb R$, which means that any somewhat easily describable non-abelian group structure comes from a group structure that has a simple description on a different underlying set - and we transfer this to $\Bbb R$ via a bijection. This is all a bit hand-wavy, of course, because things like "easily describable" are not well-defined. After all, one could just take some effort to hide the bijection to describe the group operation as if it were defined o $\Bbb R$ directly.
Thus if we want to define a non-abelian group structure on $\Bbb R$ "directly" (whatever that means) we'd certainly end up with something that defies "simple" descriptions, for example by adapting a proof that shows that groups of all cardinalities exist.
A: Considered purely as a set, the only property that $\Bbb{R}$ has is its cardinality i.e. it is an uncountable set. So your question is whether a non-abelian group structure can be defined on an uncountable set. The example of $GL_2(\Bbb{R})$ shows that the answer is yes. However, describing an explicit bijection between $\Bbb{R}$ and $GL_2(\Bbb{R})$ that maps multiplication on $GL_2(\Bbb{R})$ to an explicit function on $\Bbb{R}$ is, I think, very difficult. The reason that $\Bbb{R}$ appears to have an "innate" preference for abelian group structures is that the typical examples of group structures defined on $\Bbb{R}$ preserve it's topology i.e. they consider $\Bbb{R}$ (or a subset of $\Bbb{R}$) as a topological space (not just as a set) and create a topological group with a group function that is continuous.
