(Soft Question) What kinds of properties are transferred by isomorphisms?

Question

In many, many different areas of math, we define abstract objects and structure-preserving maps between them, and then come across a suitable notion of "isomorphism." The idea is always that isomorphic objects share every property that a mathematician in that field would care about: Homeomorphic topological spaces have all their topological properties in common, group isomorphisms have all their group-theoretic properties in common, and so on.

But at the end of the day, we still have to verify by hand that any given property is preserved by an isomorphism in the category we're working in. For example, say we're working in group theory. It is intuitively obvious that under an isomorphism $\phi : G \to H$, corresponding elements have the same order, compose in the same way, $\phi$ takes subgroups to subgroups, normal subgroups to normal subgroups, centers to centers, and any group constructed out of $G$ (direct/semidirect products, quotients, etc) should yield an isomorphic result when $G$ is replaced by $H$. (These are just a few of many examples.) It feels like it should be the case that all of these properties correspond between $G$ and $H$, but nonetheless we have to verify each of them one at a time. The proofs are never hard, and they suggest that there must be a more general way to think about them--some sort of big theorem that says that all "group-theoretic properties" can be transferred from one group to another by an isomorphism.

How could we make this idea rigorous? How would we even define a "group-theoretic property", or analogously a "topological property" or a "linear-algebraic property" or a "manifold property"? I would think the definition would stem from the idea that such properties are those that are phrased using only the structure of a group, (or a topological space, or a vector space, or a manifold); but this still seems imprecise. Assuming we could make this notion precise, could we then proceed to prove a general theorem that all such properties/objects are preserved by isomorphisms in the category we're working in, and then we don't have to tediously prove, for example, that group isomorphisms carry centers to centers, or homeomorphic spaces have the same number of conencted components, etc., because these would all fall out as special cases?

Or is this a futile task? Maybe it so happens that there are weird examples of properties that seem like they should be preserved by isomorphisms but aren't, even though they are phrased using only the structure of the category. Math is full of pathologies, and at this point I can't seem to trust 100% that isomorphisms are these magic structure-preserving identifications that they are always made out to be.

Answer 1

Mathematical logic (specifically model theory) provides a partial answer. Let $M$ and $N$ be structures for a first-order language $L$. $M$ and $N$ are elementarily equivalent if every closed formula satisfied by one is satisfied by the other. $M$ and $N$ are isomorphic if there is a 1-1 map between $M$ and $N$ that preserves all the relations and functions mentioned in the signature of $L$. Theorem: if $M$ and $N$ are isomorphic, then they're elementarily equivalent. See, say Marker Model Theory: An introduction, §1.1, or Hodges A Shorter Model Theory, §1.2.

I think this serves as a reasonable candidate for "a general theorem that all such properties/objects are preserved by isomorphisms in the category we're working in".

I say a partial answer, because choosing the language in each case remains an issue. Let me elaborate for your example of groups. We want to show that being a subgroup, or a normal subgroup, or the center, is preserved by isomorphisms, all in one shot. For $L$, we include the following in its signature: the constant symbol 1, the function symbols $\cdot,{}^{-1}$, and a unary relation symbol $S$ for the subset under discussion. (There are other signatures that would also serve.) Here are the closed formulas that express "$S$ is a subgroup", etc. I'm going to be a bit sloppy for increased readability, using juxtaposition for the operation and omitting parentheses. Also, when I write "$S$ is a subgroup" in the second two bullets, just imagine the first bullet being repeated in full.

• $S(1)\wedge\forall x\forall y[S(x)\wedge S(y)\rightarrow S(x^{-1})\wedge S(xy)]$
• $S$ is a subgroup and $\forall x\forall y[S(x)\rightarrow S(y^{-1}xy)]$
• $S$ is a subgroup and $\forall x[\forall y(yx=xy)\rightarrow S(x)]\wedge \forall x[S(x)\rightarrow\forall y(yx=xy)]$

So if $M$ and $N$ are isomorphic, then $M$ satisfies one of these formulas if and only if $N$ does—that's what elementary equivalence says. And if $M$ and $N$ are isomorphic groups, then the subsets defined by the relation symbol $S$ correspond, and therefore one is a subgroup (or normal, or the center, or anything expressible by a closed formula in this language) if and only if the other is.

If you're familiar with first-order logic, you'll be aware of various hurdles to overcome. For example, to define "commutator subgroup" with a closed formula, you'd need to expand the language to allow for sequences of arbitrary finite length, since the commutator subgroup is generated by the commutators. That means incorporating $\mathbb{N}$ into the structure in some manner. I don't mean that $\mathbb{N}$ would be a subset of the group, rather that the structure would be (implicitly) an ordered tuple $(G,\mathbb{N},\ldots)$. For "derived series" you'd need to expand the language some more. But all these obstacles can be mastered with standard techniques.

A fuller answer would discuss connecting the category theory with the model theory. I plead limitations of both space and my expertise.

Answer 2

I would argue that a "group theoretic property" or a "topological property" etc. is precisely defined to be a property that is invariant under group isomorphism, topological isomorphism (also called "homeomorphism" : as pointed out in the comments, there is only one notion of isomorphisms, it just so happens that for algebraic structures, there are equivalent formulations using bijections, i.e. isomorphisms in $\mathbf{Set}$), etc.

In this sense, the answer is tautological : group theoretic properties are preserved under isomorphism...because they are.

Of course that's not a satisfying answer, because this doesn't reduce the amount of proofs we have to do (we still have to prove that such property is preserved under isomorphism to prove that it's a group theoretic property).

But the advantage of taking this point of view is that it comes with a natural way of checking that something is, in fact, a group theoretic property (I'm using the example of groups here because it's easier to just use one example), so it shifts the focus on something else, and that something else is easier to make sense of.

Indeed, to check that something is preserved under isomorphism, that is, is a group theoretic property, it suffices to check that it can be defined internally to the category of groups.

For instance, "an element of order $\mid n$ of $G$" can be defined as a morphism $\mathbb Z/n \to G$; and an element of order $n$ as such a morphism that cannot be factored as $\mathbb Z/n\to \mathbb Z/d \to G$ for any $d<n$ (or you could say "an element of order $\mid n$ which is a monomorphism". Or you could say that an element is a morphism $\mathbb Z\to G$ and that it has order $n$ if and only if it can be factored as $\mathbb Z\to\mathbb Z/n\to G$ and no lower $d$, for instance.

Or else, an abelian group can be defined to be an object that admits a "group object" structure in the category of groups (this point of view is actually helpful in other regards), so it's invariant under isomorphism as well.

There are various ways of seeing that such and such definition can be defined categorically, but in the end it always allows you to see that it is invariant under isomorphism.

The reason is that properties that are defined internal to a category are invariant under isomorphism. To get a feel of why this is true, you may want to check out my other answer here, which attempts to explain that.

Let me add that, regardless of the philosophical question of whether something is a group theoretic property, or what that even means, the idea of expressing various concrete notions categorically can be extremely interesting.

Just to give an example : over a ring $R$, there's a notion of "finitely presented module". Now this is defined purely in terms of arrows and so on, so it's easy to see that it's invariant under isomorphism. But in fact, more is true : you can define it internal to the category of $R$-modules without using any specific $R$-module : finitely presented $R$-modules are precisely the compact objects of that category. Now the notion of a compact object is purely categorical (so it doesn't even refer to $R$-modules), and so it's transported along equivalences of categories. This can be helpful in setting up the bases of Morita theory.

This is one categorical level higher ("invariant under equivalence of categories"), so it's not entirely relevant to your question, but it shows that the more you can define things categorically, the more invariant they become; so it's a good argument in favour of the point of view I've tried to portray here.

But, as I pointed out (if I recall correctly) in my other answer, coming up with a precise (and useful !!) statement about this sort of thing, that applies in all contexts (the accepted answer's statement is certainly simple and precise; unfortunately it only applies in cases where you're dealing with categories of first order structures on a certain language - of course you can expand to higher orders etc. but it will nonetheless be limited) would actually be very difficult; and in the end, knowing what kind of things are invariant under isomorphism and what kind of things aren't is mostly a matter of experience.

You just know that being of order $n$ is preserved under an isomorphism; whereas $\pi\in G$ isn't. These things become obvious with experience - and sometimes, it is a problem because sometimes our intuition fails. For instance, sometimes you forget that things are invariant under isomorphism in another category, but might not be in the category you're actually considering.

I've never seen a blatantly wrong example though, of something that you would be convinced is preserved under isomorphism, even if you thought about it for a long long time; but that actually isn't. I think it's one of the most robust non-precise notions there is.

Answer 3

This is just a point of view on group isomorphisms. Maybe it touches some periphery of your question.

---

In the definition of isomorphic groups$^{(1)}$ there seems to be much of hindsight: why a bijection with the operation-preserving property should make two groups "equally structured" (i.e. isomorphic)? Or, equivalently: why such a bijection should be rightly named isomorphism?

Let's try to settle the definition of isomorphic groups on a slightly different basis. The structure of a group is the outcome of the group operation coming into play; thus, it sounds reasonable to define structure of the group $G$ the image of $G$ in $\operatorname{Sym}(G)$ via left multiplication. With these premises, a problem arises if we want to draw any conclusion about the "isomorphicity" of two groups $G$ and $\tilde G$, since in general $\operatorname{Sym}(G)\cap \operatorname{Sym}(\tilde G)=\emptyset$. But there's a way to come up to one same arena where the comparison between the two structures can actually take place in terms of an equality ("$=$"): to "transport" the structure of $G$ into $\operatorname{Sym}(\tilde G)$ by means of the bijection$^{(2)}$ $\varphi^{(\psi)}\colon \operatorname{Sym}(G)\to \operatorname{Sym}(\tilde G)$, $\sigma\to\psi\sigma\psi^{-1}$, induced by a bijection $\psi\colon G\to \tilde G$. In line with this standpoint, let's consider the following diagram:

where $\theta$ and $\tilde\theta$ are Cayley's injections$^{(3)}$. So, we set forth the following:

Definition.

The groups $G$ and $\tilde G$ are said to be isomorphic if there is a bijection $\psi\colon G\to \tilde G$ such that:

$$\varphi^{(\psi)}\theta\psi^{-1}=\tilde\theta \tag 1$$

i.e. such that the diagram commutes.

This definition means that two groups are isomorphic if there is a bijection between them, such that it allows to transport the structure of one onto precisely the structure of the other. As a characterization of such "nice" bijections, we have the following:

Lemma.

Two groups $G$ and $\tilde G$ are isomorphic if and only if there is a bijection $\psi\colon G\to \tilde G$ such that:

$$\psi(gh)=\psi(g)\psi(h),\space\forall g,h\in G\tag 2$$

Proof. In fact:

\begin{alignat}{1} &(1) \iff \\ &((\varphi^{(\psi)}\theta\psi^{-1})(\tilde g))(\tilde h)=(\tilde\theta(\tilde g))(\tilde h), \space\forall \tilde g,\tilde h\in \tilde G \iff \\ &(\varphi^{(\psi)}(\theta(\psi^{-1}(\tilde g))(\tilde h)=\tilde g\tilde h, \space\forall \tilde g,\tilde h\in \tilde G \iff \\ &(\psi(\theta(\psi^{-1}(\tilde g)))\psi^{-1})(\tilde h)=\tilde g\tilde h, \space\forall \tilde g,\tilde h\in \tilde G \iff \\ &\psi(\theta(\psi^{-1}(\tilde g))(\psi^{-1}(\tilde h)))=\tilde g\tilde h, \space\forall \tilde g,\tilde h\in \tilde G \iff \\ &\psi(\psi^{-1}(\tilde g)\psi^{-1}(\tilde h))=\tilde g\tilde h, \space\forall \tilde g,\tilde h\in \tilde G \iff \\ &\psi(\psi^{-1}(\tilde g)\psi^{-1}(\tilde h))=\psi(\psi^{-1}(\tilde g))\psi(\psi^{-1}(\tilde h)), \space\forall \tilde g,\tilde h\in \tilde G \iff \\ &(2) \\ \end{alignat}

$\Box$

Therefore, property $(2)$ characterizes the bijections which make two groups isomorphic (according to the given definition), and thence they are rightly named isomorphisms.

---

$^{(1)}$Two groups are said to be isomorphic if there is an isomorphism from one group to the other.

$^{(2)}$We still don't know the word "isomorphism".

$^{(3)}$We still don't know the word "embedding".