How can we make $\mathbb{R}^n$ into a multiplicative group? Are there any sorts of multiplicative group structures we can put on set $\mathbb{R}^n$? By multiplicative, I mean that it should be compatible with the natural scalar multiplication on $\mathbb{R}^n$; that is, for any $x,y \in \mathbb{R}^n$ and $c \in \mathbb{R}$, we should have
$$ c(x \star y) = cx \star y$$
For $n > 2$, can such structures be abelian? The only example I'm familiar with is $n = 4$, in which we have the quaternions; this is clearly non-abelian.
 A: For the question, as asked, one may simply define:
$ (a_1,\dots,a_n) * (b_1,\dots,b_n) = (a_1+b_1,\dots,a_n+b_n)$.
This is a multiplicative structure on $\mathbb{R}^n$, turning it into an abelian group.
A: It is not entirely clear what you mean by a multiplicative group structure. Notice that the product of quaternions only turns the set $\mathbb{R}^4\setminus\{(0,0,0,0)\}$ into a group - not all of $\mathbb{R}^4$.
In an attempt to coerce a more precise question out of you, I lead off by proffering the following multiplicative structure on $\mathbb{R}^3$. Identify the vector $(a,b,c)$ with the upper triangular matrix
$$
\left(
\begin{array}{ccc}
1&a&b\\
0&1&c\\
0&0&1
\end{array}\right).
$$
The usual matrix product turns this copy of $\mathbb{R}^3$ into a multiplicative group, doesn't it?
Using larger matrices, we can cover other real vector spaces.
If you want the group operation to be distributive over the usual vector addition, you should say so! Sorry about being a bit cranky.
A: I suppose you are searching for a finite-dimensional, associative division algebra over the reals of dimension $n>2$? According to the Frobenius theorem none of those is abelian and the only such division algebra that exists, is given by the (non-abelian) Quaternions.
