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.

24 pages, Already published in J. Algebra, Minor typos have been corrected in this version