It is known that, for any group $G$, there exists a group $H$ containing $G$ such that the group ring $F[H]$ for some field $F$ is primitive, see Formanek, Edward; Snider, Robert L., Primitive group rings, Proc. Am. Math. Soc. 36 (1972), 357-360 (1973). ZBL0223.16007.
The group and faithful, simple module is constructed as follow:
$G_1 = G$, $M_1 = F[G_1]$
$G_2 = \textrm{Aut}_F(M_1)$, $M_2 = F[G_2] \oplus M_1$
and so on with $H = \bigcup G_i$ and $M = \bigcup M_i$.
We can consider $M$ both a left- and a right-module. Is it simple and faithful on both sides?
If so, then consider $\textrm{End}_{F[H]} M$ (where $M$ is a left-module) as it is left-simple we know that this is a division ring by Schur's lemma. However, there is also a ring homomorphsim from $F[G]$ into $\textrm{End}_{F[H]} M$ by the action of right scalar multiplication. And as $M$ is right-faithful, it has no non-zero annihilators and thus the homomorphsim is injective.
It seems, therefore, we have embedded $F[G]$ into a division ring?