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 $G$? 




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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 $\lambda' \lt |N| \lt \lambda$?



Universe-sized groups with only set-sized normal subgroups, their cardinality in a certain range
A few answers with references: 


The answer to the original question is "yes": There are simple groups of any infinite cardinality. For example, the set of all permutations of $\kappa$ with finite support has normal subgroup $A\_\kappa$ of index 2, the even permutations; $A\_\kappa$ has cardinality $\kappa$ and is simple. 


More examples of simple groups of any infinite cardinality can be found in Lang's book "Algebra", chapter XIII, sections 8 and 9. 


For the special case of $\kappa = 2^{\aleph\_0}$, $\lambda=\aleph\_1$,
[wikipedia](http://en.wikipedia.org/wiki/Symmetric_group#Normal_subgroups) mentions Ch 11.3 in Scott's 1987 book on Group Theory and Ch. 8.1 in Dixon-Mortimer's 1996 book on permutation groups as references for the following theorem: 


* The symmetric group on a countable set has only 2 nontrivial normal subgroups, both of them countable: even permutations, and permutations with finite support.


