Let $\kappa$ be an inaccessible cardinal, and let $G$ be a group with $|G| \geq \kappa$. For any cardinal $\lambda \le \kappa$ (regular, say, but not necessary), say $G$ is $\lambda$-simple if for all normal subgroups $N \lt G$ we have $|N| \lt \lambda$. Clearly a group is simple iff it is $2$-simple. Can we force (or even construct without forcing) a $\lambda$-simple group $G$?

Let $\kappa$ be an inaccessible cardinal, and let $G$ be a group with $|G| \geq \kappa$. For any cardinal $\lambda \le \kappa$ (regular, say, but not necessary), say $G$ is $\lambda$-simple if for all normal subgroups $N \lt G$ we have $|N| \lt \lambda$. Clearly a group is simple iff it is $2$-simple. Can we force (or even construct without forcing) a $\lambda$-simple group $G$?

EDIT: Yes, take a simple group of the required cardinality, this exists by Goldstern's comment. Note that $\lambda$-simple implies $\alpha$-simple for all $\alpha \gt \lambda$.

A more interesting question (pointed out by Benjamin Steinberg) is whether we can bound the size of the normal subgroups from below as well. Given a second cardinal $\lambda' \lt \lambda$, can we find a group of cardinality $\ge \kappa$ such that all normal $N \lt G$ have $\lambda' \lt |N| \lt \lambda$?

Let $\kappa$ be an inaccessible cardinal, and let $G$ be a group with $|G| \geq \kappa$. For any cardinal $\lambda \le \kappa$ (regular, say, but not necessary), say $G$ is $\lambda$-simple if for all normal subgroups $N \lt G$ we have $|N| \le \lambda$. Clearly a group is simple iff it is $2$-simple. Can we force (or even construct without forcing) a $\lambda$-simple group $G$?

Let $\kappa$ be an inaccessible cardinal, and let $G$ be a group with $|G| \geq \kappa$. For any cardinal $\lambda \le \kappa$ (regular, say, but not necessary), say $G$ is $\lambda$-simple if for all normal subgroups $N \lt G$ we have $|N| \lt \lambda$. Clearly a group is simple iff it is $2$-simple. Can we force (or even construct without forcing) a $\lambda$-simple group $G$?