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I am teaching a first-semester course in abstract algebra, and we are discussing group isomorphisms. In order to prove that two group are not isomorphic, I encourage the students to look for a group-theoretic property satisfied by one group but not by the other. I did not give a precise meaning to the phrase "group-theoretic property", but some examples of the sort of properties I have in mind are $$ \forall g,h\in G:\exists n,m\in\mathbb{Z}:(n,m)\neq (0,0)\wedge g^n=h^m,\\ \forall H\leq G:\exists g,h\in G:H=\langle g,h\rangle,\\ \forall g,h\in G:\exists i\in G: \langle g,h\rangle = \langle i\rangle $$ One of my students asked if, give two non-isomorphic groups, there is always a group-theoretic property satisfied by one group but not the other. In a sense, "being isomorphic to that group over there" is a group-theoretic property. But this is not really what I have in mind.

To pin down the class of properties I have in mind, let's say we allow expressions involving

  • quantification over $G$, subgroups of $G$, and $\mathbb{Z}$,
  • group multiplication, inversion, and subgroups generated by a finite list of elements
  • the symbol $1_G$ (the group identity element),
  • addition, subtraction, multiplication, exponentiation (provided the exponent is non-negative), and inequalities of integers ,
  • the integer symbols $0$ and $1$,
  • raising a group element to an integer power, and
  • equality, elementhood, and logical connectives.

I do not know much about model theory or logic, but my understanding is that this is not the first-order theory of groups. In particular, this MSE question indicates that there exist a torsion and a non-torsion group which are elementarily equivalent (meaning they cannot be distinguished by a first-order statement in the language of groups), but these groups can be distinguished by a property of the above form. I have also heard that free groups of different rank are elementarily equivalent, but these can also be distinguished by a property of the above form.

My questions are:

(1) Is there a name for the theory I am considering? Or something closely (or distantly) related?

(2) Are there examples of non-isomorphic groups that cannot be distinguished by a property of the above form? Are there examples where the groups involved could be understood by an average first-semester algebra student?

user577215664
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Julian Rosen
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3 Answers3

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First, let's start with the silly answer. Your language only has countably many different expressions, so can only divvy groups up into continuum-many classes - so there are definitely non-isomorphic groups it can't distinguish! In general this will happen as long as your language has only set-many expressions: you need a proper class sized logic like $\mathcal{L}_{\infty,\infty}$ to distinguish between all pairs of non-isomorphic structures.

That said, you're right that you're looking at something much stronger than first-order logic. Specifically, you're describing a sublogic of second-order logic, the key difference being that second-order logic lets you quantify over arbitrary subsets of the domain, and indeed functions and relations of arbitrary arity over the domain, and not just subgroups. Second-order logic doesn't have an explicit ability to refer to (say) integers built in, but it can do so via tricks of quantifying over finite configurations.

While the exact strength of the system you describe isn't clear to me, second-order logic is known to be extremely powerful. In particular, I believe there are no known natural examples of non-isomorphic second-order-elementarily-equivalent structures at all, although as per the first paragraph of this answer such structures certainly have to exist! So second-order-equivalence is a pretty strong equivalence relation, and in practice will suffice to distinguish all the groups your students run into.

Noah Schweber
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  • Out of curiosity, how do you state having infinite cyclic subgroups without going into infinitary logic? – tomasz Oct 09 '20 at 20:14
  • Where can I read more about proper class sized logics? – user76284 Oct 09 '20 at 20:31
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    @tomasz For example, we can say that an element $g$ of a group has infinite order (this is basically the same) by saying "The smallest subgroup containing $g$ has a non-surjective self-injection." This is of course second-order, but not infinitary. In general, for most things we run into normally second-order logic is actually vastly stronger than infinitary logic, with combinatorial tricks like the above being used to get around apparent reliance on infinitary Boolean combinations. – Noah Schweber Oct 09 '20 at 20:40
  • @user76284 The huge collection Model-theoretic logics is freely available online and has some relevant material. The most important proper-class-sized logic is definitely $\mathcal{L}{\infty,\omega}$; it's certainly much better-behaved than the "full infinitary logic" $\mathcal{L}{\infty,\infty}$. For example, $\mathcal{L}{\infty,\infty}$-equivalence is just isomorphism, whereas $\mathcal{L}{\infty,\omega}$-equivalence is potential isomorphism (= existence of a "back-and-forth system"; equivalently, isomorphism in some forcing extension). – Noah Schweber Oct 09 '20 at 20:42
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    Regarding the insufficiency of a countable number of sentences: I've found myself wondering whether "ordinal definable" classes would suffice to overcome that, e.g. you use a sentence with some number of free variables which are intended to represent ordinal parameters plus one for the group, and then consider the classes which are formed by assigning all possible tuples of ordinals to those free variables, say over a ZFC foundation. That would certainly be enough to distinguish groups of different cardinalities, without feeling cheap like "isomorphism to this given group". – Daniel Schepler Oct 09 '20 at 21:38
  • @DanielSchepler So basically you're talking about isomorphism-closed ordinal-definable subclasses of the class of groups, right? It's consistent that every set is ordinal definable, so it's consistent that this is in fact equivalent to the cheap solution; however, I also think that it's consistent that this doesn't work, that there are nonisomorphic groups which aren't ordinal definably separated. (Also, note that this is an "external" approach: we're talking about evaluating a formula in $V$, rather than within the group itself.) – Noah Schweber Oct 09 '20 at 21:41
  • Right. (Actually, I first thought of this question in the context of topological spaces, where there are already standard topological properties of this sort, e.g. whether the space has a basis of cardinality $< \kappa$. Though I guess also for groups, it's easy to come up with ordinal classifiers, e.g. the minimum cardinality of a generating set.) – Daniel Schepler Oct 09 '20 at 22:01
  • I was trying to write down the argument in your first paragraph formally, and I got worried about something. This is far from my area, so more likely than not I am just confused (which tends to happen when I think about logic), but here goes. As I understand it, one might make your argument in the following way: Let $X$ be the countable set of (Gödel numbers of) group-theoretic properties of the shape I consider. Cook up some set $Y$ of pairwise non-isomorphic groups, with cardinality strictly larger than the continuum. Then define $f:Y\to\mathcal{P}(X)$ to be the function... – Julian Rosen Oct 10 '20 at 00:54
  • (cont.) taking a group $G\in Y$ to the set of formulas that group satisfies. Then $f$ cannot be injective since $|Y| > |\mathcal{P}(X)|$, so there must exist $G_1\neq G_2\in Y$ with $f(G_1)=f(G_2)$. This means that $G_1$ and $G_2$ are non-isomorphic groups satisfying precisely the same set of properties. My worry is that: why does the function $f$ exist? Tarski's undefinability theorem seems to say that there is no formula to capture "the property with Gödel number $n$ is true the group $G$", and my definition of $f$ implicitly invokes such a formula. – Julian Rosen Oct 10 '20 at 00:54
  • @JulianRosen Tarski says that a truth predicate for a structure (of an appropriate type) cannot be defined within the structure itself. But it can certainly be defined in a larger "ambient universe." This is what is happening here: letting $V$ be the set-theoretic universe as usual, $Th(V)$ is undefinable in $V$ but in $V$ we can definably send each set-sized structure to its theory. The appropriate counting argument takes place in $V$, not in any particular group. (Note that if this sort of idea didn't work, model theory wouldn't make sense!) – Noah Schweber Oct 10 '20 at 01:03
  • Now when we try to talk about $V$'s theory inside $V$, things do indeed get weird. For example, surprising though it may seem we can have a model of $\mathsf{ZFC}$ in which every set is definable, despite that seeming to result in a paradox - since the relevant counting argument can't actually be performed inside that model. (See here for details.) But everything relevant to this question is just about working with set-sized structures, which $V$ can do. – Noah Schweber Oct 10 '20 at 01:06
  • @NoahSchweber ordinal defiability is definable in first order set theory, strangely enough. – Christopher King Oct 10 '20 at 16:51
  • @PyRulez That's true (I don't think I said it wasn't?). – Noah Schweber Oct 10 '20 at 17:26
  • @NoahSchweber you mentioned this being an external approach. I thought that was what you meant. – Christopher King Oct 10 '20 at 17:42
  • @PyRulez External to the group - that is, taking place in $V$ as opposed to within the group itself (or something "near" the group, like its powerset): "we're talking about evaluating a formula in , rather than within the group itself." – Noah Schweber Oct 10 '20 at 17:43
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Here are some simple examples where you at least need to make some decisions about what you believe about set theory to determine whether two groups are isomorphic. Assuming the axiom of choice every vector space has a basis, so $\mathbb{R}$ is isomorphic (as a group) to some direct sum of copies of $\mathbb{Q}$ (in fact necessarily to a direct sum of $|\mathbb{R}|$ copies of $\mathbb{Q}$). The existence of such a basis for $\mathbb{R}$ over $\mathbb{Q}$ allows you to construct Vitali sets, which are non-measurable, and there are models of $ZF \neg C$ in which every subset of $\mathbb{R}$ is measurable, so $\mathbb{R}$ fails to have a basis in such models.

Another example along the same lines is $\left( \prod_{\mathbb{N}} \mathbb{Q} / \bigoplus_{\mathbb{N}} \mathbb{Q} \right)^{\ast}$, taking the dual as a $\mathbb{Q}$-vector space. Assuming the axiom of choice this is a direct sum of $|\mathbb{R}|$ copies of $\mathbb{Q}$ again, but without at least enough choice to construct something like non-principal ultrafilters on $\mathbb{N}$ it's not clear how to write down a single nonzero element of this group!

Qiaochu Yuan
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    It's an interesting question to ask what are the hardest groups to distinguish up to isomorphism that you would run into in a first-year class in abstract algebra. Here's an example I ran into in recently in another math.SE question: is it true that if $K$ is any field then $PSL_n(K) \cong PSL_m(K)$ iff $n = m$? In the question it sufficed to distinguish any $n \neq m$ where $n, m \ge 3$ and I could distinguish $PSL_3$ and $PSL_9$. Details: https://math.stackexchange.com/questions/3843808/does-the-association-v-mapsto-glv-define-a-functor/3845230#3845230 – Qiaochu Yuan Oct 09 '20 at 02:47
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Another example that's basic and probably counts as cheating: Consider $G = \mathbb{Z}^{\kappa}$ for cardinal numbers $\kappa$ starting with $\aleph_0$. I don't know enough model theory to prove it, but I can't imagine there is a group-theoretic property that distinguishes between these.

Jakub Konieczny
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    In first-order logic there certainly isn't, but the OP isn't limiting themselves to first-order logic (see the role of the true integers or quantification over subgroups). Second-order logic certainly can distinguish between for example $\mathbb{Z}^{\aleph_0}$ and $\mathbb{Z}^{\aleph_1}$, although I'm not sure if the OP's fragment of SOL can. – Noah Schweber Oct 09 '20 at 17:33
  • My intuition is that it's impossible to make distinction using OPs methods, but like I said - I have no proof. If there exists a group property that distinguishes between these, it has to be very unnatural one. – Jakub Konieczny Oct 09 '20 at 17:41
  • Yes, I think I have the same intuition. Towards a proof we have the following fact: two groups $G,H$ are equivalent w/r/t the OP's logic iff their associated structures $[G],[H]$ are elementarily equivalent in the usual sense, where for a group $A$ the structure $[A]$ is the two-sorted structure consisting of $(i)$ $A$ itself and $(ii)$ the lattice of subgroups of $A$, with elementhood and the "shift" operation $(x,Y)\mapsto {xy: y\in Y}$ relating them. – Noah Schweber Oct 09 '20 at 17:47
  • It's known that $\text{Hom}(\mathbb{Z}^I, \mathbb{Z}) \cong \bigoplus_I \mathbb{Z}$ when $I$ is countable (a very strange result for sure, see: https://en.wikipedia.org/wiki/Baer%E2%80%93Specker_group); does anyone know if it has a chance of holding in general? If so we can do it. I'm not quite sure if this is expressible in the OP's language but in any case "the dimension of $\text{Hom}(G, \mathbb{Z}) \otimes \mathbb{Q}$" qualifies as a group-theoretic property by any stretch of the imagination, I think. – Qiaochu Yuan Oct 10 '20 at 21:02