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

propersimple 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?