  We prove that the group algebra $KG$ of a group $G$ over a field $K$ is
primitive, provided that $G$ has a free subgroup with the same cardinality as
$G$, and that $G$ satisfies the following condition $(\ast)$: for each subset
$M$ of $G$ consisting of a finite number of elements not equal to $1$, and for
any positive integer $m$, there exist distinct $a$, $b$, and $c$ in $G$ so that
if $(x_{1}^{-1}g_1x_{1}) \cdots (x_{m}^{-1}g_mx_{m})=1$, where $g_i$ is in $M$
and $x_i$ is equal to $a$, $b$, or $c$ for all $i$ between $1$ and $m$, then
$x_{i}=x_{i+1}$ for some $i$. This generalizes results of \cite{Bal},
\cite{For}, \cite{Ni07}, and \cite{Ni11}, and proves that, for every countably
infinite group $G$ satisfying $(\ast)$, $KG$ is primitive for any field $K$. We
use this result to determine the primitivity of group algebras of one relator
groups with torsion.
