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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?

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6 Answers 6

up vote 17 down vote accepted

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.

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Thanks, Simon. Your ultraproduct construction will be a satisfying answer for my class (once I discuss ultraproducts). What you say seems to leave open the question of whether a simple group can be elementarily equivalent to a nonsimple group. Do you know the answer to that? –  Pete L. Clark Jul 22 '10 at 7:12
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Consider the simple group $S$ of finite even permutations of $\mathbb{N}$. Arguing as above, a (nontrivial) ultrapower of $S$ won't be simple but will be elementarily equivalent to $S$. –  Simon Thomas Jul 22 '10 at 7:15
    
Gotcha: work with the "constant sequence" $A_{\infty}$ rather than the sequence $A_n$. Thanks again. –  Pete L. Clark Jul 22 '10 at 7:21
    
You mention that the ultraproduct has a unique maximal proper normal subgroup. Is there a reference for that? –  Andreas Thom Aug 5 '10 at 16:49
    
A suitable reference is: G. Elek and E. Szabo, Hyperlinearity, essentially free actions and $L^{2}$-invariants. The sofic property, Math. Ann. 332 (2005), 421--441. –  Simon Thomas Aug 10 '10 at 19:32

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.

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Another example of an ultraproduct of simple groups which is not simple (somewhat sillier than Simon's): take some (any) nonprincipal ultraproduct of the finite cyclic groups $Z_p$ as $p$ ranges over all primes. This ultraproduct must be infinite, and abelian (since commutativity is an elementary property), but any infinite abelian group has a proper subgroup, hence it can't be simple.

(I say "sillier" than the alternating groups example since it seems like nobody cares about abelian simple groups, so Simon's answer deals with the natural follow-up question.)

OK, as for, "what are some general techniques for tackling a problem like this one," I'd first look at Theorem 4.1.12 of Chang and Keisler's Model Theory textbook: a class of structures is elementary if and only if it is closed under both ultraproducts and elementary equivalence. This seems relevant because ultraproducts, at least, seems like a relatively natural concept for algebraists, and depending on the particular class, you might hope to also find an algebraic definition of "elementarily equivalent."

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Thanks, John. I actually had the ultraproduct of the finite fields of order $p$ in mind (like any good "field arithmetician"). Why I gave up on this construction as showing that the simple groups do not form an elementary class is beyond me -- I guess I was focusing more on the harder question of whether the class is preserved by elementary equivalence. (And, as a jouryneyman model theorist, I still occasionally get confused by the fact that "class preserved by elementary equivalence" $\neq$ "elementary class". Couldn't the terminology be a little better here?!?) –  Pete L. Clark Jul 27 '10 at 7:09

I just remembered another family of simple groups of arbitrarily large cardinality.

Theorem (D. Lascar, 1995) Let $L/K$ be a field extension with both $L$ and $K$ algebraically closed and $\operatorname{trdeg}(L/K) > \aleph_0$. Then $\operatorname{Aut}(L/K)$ is a simple group.

Since it is also true that for every algebraically closed field $K$, $|\operatorname{Aut}(K)| = 2^{|K|}$ (e.g. Theorem 80 of http://math.uga.edu/~pete/FieldTheory.pdf), it follows that for every uncountable algebraically closed field $K$ of characteristic $0$, $\operatorname{Aut}(K/\overline{\mathbb{Q}})$ is a simple group of cardinality $2^{|K|}$.

I find this result to be interestingly reminiscent of the Schreier-Ulam-Baer theorem, and I have at times idly wondered whether there is some common explanation or a more general result that specializes to these two theorems. (Idle indeed because I have barely looked at the proof of either theorem.)

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This is an instance of a common phenomena. If $G$ is an automorphism group of a "rich" structure and $N$ is the normal subgroup of "bounded" automorphisms, then $G/N$ is simple. Examples include an infinite set with no extra structure, the linear order $\mathbb{Q}$, the countable random graph (which has no "bounded" automorphisms), etc –  Simon Thomas Jul 23 '10 at 7:35

In the Lascar theorem (posted by Pete) there is an assumption that the transcendence degree of $L$ over $K$ is uncountable (MR1689847 (2000g:20009) ). For a countable transcendence degree something weaker is known. I asked some years ago Lascar about finite degree case, and he said that it is open and seems to be very hard (i.e. if $tr deg_K(L)<\omega$ then $Aut(L/K)$ is simple).

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Thank you, Jakub. I have corrected the statement in my answer accordingly. –  Pete L. Clark Aug 4 '10 at 15:54

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 $\{0,1\}^{\mathbb N}$.

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