$\mathbb{R}[Q_8]$ is not a division algebra 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?
 A: There are extremely simple ways to see this.
You could observe that group rings are almost never simple rings.
Another good thing for your toolbox for finite groups $G$ is that if $G$ has a nontrivial normal subgroup, $\mathbb R[G]$ isn't even a domain. The quaternion group has, to put it mildly, lots of normal subgroups.
To see this, note that the sum of elements in a finite nontrivial normal subgroup creates a nontrivial central idempotent that splits the ring in two pieces. If the normal subgroup $H$ had order $k$, then $\frac1k\sum_{h\in H}h$ is a central idempotent. Consequently $eR\oplus (1-e)R$ splits the ring into the product of two rings. Or more simply, $e(1-e)=0$ proves there are zero divisors.
This only relies on $|H|$ being a unit in the coefficient ring. There are still some related tricks to this in rings of positive characteristic, but it's more than sufficient already for this problem. If the order of the subgroup is a divisor of the characteristic of the ring, then the sum of the elements in the subgroup is nilpotent, and it is not a domain (or division ring) in that case either.
