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

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    $\begingroup$ Do you mean $\le \lambda$ or $<\lambda$? Anyway, it seems to me that every simple group is $\lambda$-simple, or have I misunderstood the question? There are simple groups of any cardinality $\kappa$, e.g. the set of "even" permutations on a set of size $\kappa$. (The smallest normal subgroup containing all 3-cycles.) $\endgroup$
    – Goldstern
    Nov 20, 2011 at 12:47
  • $\begingroup$ Maybe he means $\lambda$-simple and not $\lambda'$-simple for $\lambda'<\lambda$. Some motivation would make this clearer. $\endgroup$
    – Benjamin Steinberg
    Nov 20, 2011 at 14:19
  • $\begingroup$ Benjamin Steinberg's version of this question sounds interesting. $\endgroup$
    – S. Carnahan ♦
    Nov 20, 2011 at 16:31
  • $\begingroup$ Whoops, I did mean |N| < lambda. You've answered my question Goldstern, thanks. $\endgroup$
    – David Roberts ♦
    Nov 20, 2011 at 20:53
  • $\begingroup$ @Benjamin, I'm thinking of a pair of groups $L \lt G$ of cardinality $\geq \kappa$ such that $L$ contains no normal $N \lt G$ of cardinality $\kappa$. But if I can just take $G$ simple, then I am done. $\endgroup$
    – David Roberts ♦
    Nov 21, 2011 at 0:53

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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 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.
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