TITLE: Why there is no $5$-dimensional analogous of Quaternions?
QUESTION [5 upvotes]: Why there is no $5$-dimensional analogous of Quaternions? Why the following definition is not well-defined?
$$i^2=j^2=k^2=\ell^2=ijk=jk\ell=k\ell i=-1,\quad ijkl=1.$$

REPLY [10 votes]: If you are looking for a real vector space $V$ with basis $\{1,i,j,k,\ell\}$ and an associative product such that $V$ becomes an $\Bbb{R}$-algebra, then we run into the following difficulty. 
The rule that $i^2=-1$ means that $V$ also has a structure as a vector space over the field $\Bbb{C}=\Bbb{R}(i)$.
But a vector space over $\Bbb{C}$ necessarily has an even dimension as a vector space over $\Bbb{R}$.

The above is probably not the shortest route to a contradiction (see the comment by verret). But it also rules out many modifications to the suggested relations defining the product.