Given the Quaternion group $$Q_8=\{\pm1,\pm i,\pm j,\pm k\}$$ with the product: $1$ is the unit, $-1$ commutes with all elements, $(-1)^2=1$ and $i^2=j^2=k^2=ijk=-1.$ The group ring $\mathbb{R}[Q_8]$ has dimension $8$ (as an algebra over $\mathbb{R}$). The elements of $Q_8$ all have multiplicative inverses. Why is this not a division algebra over $\mathbb{R}$?
The easy solution is: Such an division algebra has dimension $1$, $2$ or $4$ (theorem of Frobenius).
Can this be shown without using Frobeniu's theorem about division algebras ober the real numbers?