TITLE: Three questions on large simple groups and model theory
QUESTION [21 upvotes]: Yesterday, in the short course on model theory I am currently teaching, I gave the following nice application of downward Lowenheim-Skolem which I found in W. Hodges A Shorter Model Theory:
Thm: Let $G$ be an infinite simple group, and let $\kappa$ be an infinite cardinal with $\kappa \leq |G|$.  Then there exists a simple subgroup $H \subset G$ with $|H| = \kappa$. 
(The proof, which is short but rather clever, is reproduced on p. 10 of http://www.math.uga.edu/~pete/modeltheory2010Chapter2.pdf.)
This example led both the students and I (and, course mechanics aside, I am certainly still a student of model theory) to ask some questions:
$1$. The theorem is certainly striking, but to guarantee content we need to see an uncountable simple group without, say, an obvious countable simple subgroup.  I don't know that many uncountable simple groups.  The most familiar examples are linear algebraic groups like $\operatorname{PSL}_n(F)$ for $F$ an uncountable field like $\mathbb{R}$ or $\mathbb{C}$.  But this doesn't help, an infinite field has infinite subfields of all infinite cardinalities -- as one does not need Lowenheim-Skolem to see!  (I also mentioned the case of a simple Lie group with trivial center, although how different this is from the previous example I'm not sure.)  The one good example I know is supplied by the Schreier-Ulam-Baer theorem: let $X$ be an infinite set.  Then the quotient of $\operatorname{Sym}(X)$ by the normal subgroup of all permutations moving less than $|X|$ elements is a simple group of cardinality $2^{|X|}$.  (Hmm -- at least it is when $X$ is countably infinite.  I'm getting a little nervous about the cardinality of the normal subgroup in the general case.  Maybe I want an inaccessible cardinal or somesuch, but I'm getting a little out of my depth.)  So:

Are there there other nice examples of uncountable simple groups?

$2$. At the beginning of the proof of the theorem, I remarked that straightforward application of Lowenheim-Skolem to produce a subgroup $H$ of cardinality $\kappa$ which is elementarily embedded in $G$ is not enough, because it is not clear whether the class of simple groups, or its negation, is elementary.  Afterwards I wrote this on a sideboard as a question:

Is the class of simple groups (or the class of nonsimple groups) an elementary class?

Someone asked me what techniques one could apply to try to answer a problem like this.  Good question!
$3$. The way I stated Hodges' result above is the way it is in my lecture notes.  But when I wrote it on the board, for no particular reason I decided to write $\kappa < |G|$ instead of $\kappa \leq |G|$.  I got asked about this, and was ready with my defense: $G$ itself is a simple subgroup of $G$ of cardinality $|G|$.  But then we mutually remarked that in the case of $\kappa = |G|$ we could ask for a proper simple subgroup $H$ of $G$ of cardinality $|G|$.  My response was: well, let's see whether the proof gives us this stronger result.  It doesn't.  Thus:

Let $G$ be an infinite simple group.  Must there exist a proper simple subgroup $H$ of $G$ with $|H| = |G|$?

Wait, I just remembered about the existence of Tarski monsters.  So the answer is no.  But what if we require $G$ to be uncountable?

REPLY [9 votes]: Another rather nice example: the group of automorphisms of $[0,1]$ endowed with the Lebesgue measure (i.e, bi-measurable maps preserving the measure, identified when they coincide outside a set of Lebesgue measure zero). In a related vein, the full group of any countable, ergodic, probability measure preserving Borel equivalence relation on $[0,1]$ is simple.

REPLY [3 votes]: More uncountable simple groups:  The automorphism group of the random graph is of size $2^{\aleph_0}$ and simple (Truss), the same holds for the automorphism group of the unique countable atomless Boolean algebra, i.e., the group of auto-homeomorphisms of the Cantor space $\{0,1\}^{\mathbb N}$.