Let ($\mathbb{R}^n$,*) be a division algebra over $\mathbb{R}$, $n>1$, $\Rightarrow n$ is even I want to prove: Let ($\mathbb{R}^n$,*) be a division algebra over $\mathbb{R}$, $n>1$, $\Rightarrow n$ is even.
I'm stuck. My thoughts are: I want to use the hairy ball theorem and I want to construct  a vector field such that the vector field vanishes nowhere: $v:S^{n-1}\to \mathbb{R}^n$, so $v$ has to be continuous and $\langle v(x),x\rangle=0$ for every $x\in S^{n-1}$ and $v(x)\in T_xS^{n-1}$, $v(x)\not= 0$ for every $x\in S^{n-1}$.If n is even, $n-1$ is odd and the hairy ball theorem tells me that i can construct v such that v vanishes nowhere.
Someone gave me the hint that i should use that $n_a: \mathbb{R}^n \setminus \{0\} \to \mathbb{R}^n\setminus \{0\},\; x\mapsto ax$ is a homeomorphism, n>1 and $a\in \mathbb{R}^n\setminus \{0\}$,  because  $\mathbb{R}^n$ is a division algebra. My read is to use this for the construction of $v$. Furthermore i know: $ \mathbb{R}^n\setminus \{0\} $ is homotopy equivalent to $S^{n-1}$ but i'm not sure if I need this. My question is: How to construct this $v$ with the properties as stated above? I'm stuck.  My only read is that v has to be something like a "multiplication-map" as my homeomorphism, but I dont know what the map $v$ has to do exactly. And if the map $v$ is constructed, what to do next? Then I'm done? I don't need to apply homology I think, is it correct? Help would be greatly appreciated. Regards 
Edit: 
1)I know the definition: a "division algebra" A is a finite-dimensional algebra (wich could be non-associative) such that A is an integral domain.
2)This is a homework of a algebraic topology lecture. We talked about homology in lecture and now we talk about applications of homology. We only had the definition what a vector field is and the hairy ball theorem. I dont know Lie groups for example. 
Ok my problem now is to find the inverse function of $M_a:S^{n-1}\to S^{n-1} x\mapsto \frac{a\star x}{\|a\star x\|_2}$, $a\in S^{n-1}$ as you can see in the comments below. I want to express its inverse function with the inverse function of $L_a:S^{n-1}\to S^{n-1}, x\mapsto a\star x$.  My aim is to prove that the map $M_a$ is a diffeomorphism.
 A: Original Answer
Here is a hint. For the two obvious cases of division algebras over $\mathbb{R}$, namely the complex numbers and the quaternions, we know that the set of unit vectors (more generally, the set of non-zero vectors) is a Lie group. Moreover, for any Lie group, we know that they are parallelizable---that is, the tangent bundle to the Lie group is trivial.
Can you use this to get the result that you're looking for?
Expanded given more context
(For reference, a Lie group is just a group that is also a smooth manifold. That said, I will try and extract that from the outline of a proof)
The key idea is that on your collection of non-zero vectors in your division algebra $A$ you have a translation map: that is, for each $a \in A$, you have a map $T_a : A \to A$ which is just multiplication by $a$. Moreover, you can check that this will induce an isomorphism on tangent spaces (this is important!).
So consider the tangent space to the origin $1 \in A$ (though any fixed point will do). This is a non-zero vector space, so choose a non-zero vector in this vector space.
What happens if you then look at the translates of this non-zero tangent vector across the whole collection of non-zero vectors?
