# 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.

-
I'm guessing you want some group structure compatible with the vector space structure. In which case, 0 having an inverse should cause problems. If you want $\mathbb{R}^n$ to be a field, $\mathbb{R}^2 \simeq \mathbb{C}$ as a vector space, and there cannot be any other. – ronno Oct 26 '12 at 7:23
As I recall the "best" you can do in this regard gets worse and worse as you go along. The quaternions in $\mathbb{R}^4$ lose commutativity, the octonions in $\mathbb{R}^8$ lose associativity. – AsinglePANCAKE Oct 26 '12 at 7:30
The multiplicative structures one can put on $\mathbb{R}^n$ compatible with the $\mathbb{R}$ vector space structure are completely classified by the Artin–Wedderburn theorem. en.wikipedia.org/wiki/Artin%E2%80%93Wedderburn_theorem – jspecter Oct 26 '12 at 8:17
I have updated my question to be a little bit more specific about what I mean about "multiplicative" group structure. – Christopher A. Wong Oct 26 '12 at 18:12

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.

-

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.

-
Sorry about the ambiguity; I updated my question to mention that I meant that the group structure should be compatible with scalar multiplication. – Christopher A. Wong Oct 27 '12 at 1:15
To answer the modified question, the elementwise 'multiplication' achieves $c(x\star y) = cx \star y$, in contrast to $cx \star cy$ as here for the elementwise addition. But of course 'elementwise' is somehow trivial, since it simply represents $n$ separated $1$-dimensional operations. – cubic lettuce Oct 27 '12 at 8:22

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.

-