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://alpha.math.uga.edu/~pete/modeltheory2010Chapter2.pdf.](http://alpha.math.uga.edu/%7Epete/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?
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$2$. 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?
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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](http://en.wikipedia.org/wiki/Tarski_monster_group). So the answer is no. But what if we require $G$ to be uncountable?



Three questions on large simple groups and model theory
The class of simple groups isn't elementary. To see this, first note that if it were, then an ultraproduct of simple groups would be simple. But an ultraproduct of the finite alternating groups is clearly not simple. (An $n$-cycle cannot be expressed as a product of less than $n/3$ conjugates of $(1 2 3)$ and so an ultraproduct of $n$-cycyles doesn't lie in the normal closure of the ultraproduct of $(1 2 3)$. ) 


It turns out that an ultraproduct $\prod\_{\mathcal{U}} Alt(n)$ has a unique maximal proper normal subgroup and the corresponding quotient $G$ is an uncountable simple group. This group $G$ has the property that a countable group $H$ is sofic if and only if $H$ embeds into $G$. For this reason, $G$ is said to be a universal sofic group.


As for your third question, Shelah has constructed a group $G$ of cardinality $\omega\_{1}$ which has no uncountable proper subgroups. Clearly $Z(G)$ is countable. Consider $H = G/Z(G)$. Then $H$ also has no uncountable proper subgroups.Furthermore, every nontrivial conjugacy class of $H$ is uncountable and it follows that $H$ is simple.


