In what sense of “structure” do group homomorphisms “preserve structure”?

It is commonly said that group homomorphisms “preserve the structure of the group”, e.g., from Wikipedia:

The purpose of defining a group homomorphism as it is, is to create functions that preserve the algebraic structure.

Now, my default notion of structure is the one that holds that isomorphisms are structural identities—two isomorphic mathematical objects have the same structure. But that notion of structure requires preserving not just the operations on the elements of the domain but also requires the two sets of objects be equipollent.

But homomorphisms aren’t, in general, reversible—or else they’d be isomorphisms. Since homomorphisms don’t preserve size, in what sense does it preserve “structure”? Is there any more to the notion of structure here than the simple definition of algebraic structure as a set endowed with certain operations?

Toy example illustrating the problem: suppose you have a group homomorphism between groups $G$ and $H$ (but no isomorphism). Suppose also that you have an isomorphism between $G$ and some further group $J$. Since there is a homomorphism between $G$ and $H$, the structure in $G$ is preserved in $H$. Since there is an isomorphism between $G$ and $J$ they have the same structure. But presumably if $H$ preserves the structure of $G$ then they have the same structure. That can’t be true, though, since they aren’t isomorphic. It seems then that either “structure” means two different things when discussing isomorphisms and homomorphisms, or “preserves” doesn’t mean that the homomorphic sets have the same structure (and then I have no idea what “preserves” actually means in this context).

I’m clearly confused about something. I’m just not sure what.

TL;DR What is the difference between the notions of “structure” when you say that isomorphic structures have the same structure and when you say that a group homomorphism preserves the group structure?

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It preserves the structure in the sense that $f(e_G)=e_H$ and $f(a\cdot_Gb)=f(a)\cdot_Hf(b)$, assuming that $f\colon G\to H$ is a homomorphism of course.

This is a weak preservation, but it preserves some structure nonetheless. Consider, for example, any bijection between $\Bbb Z$ and $\Bbb Q$. Does it preserve any additive structure? It does not.

Saying that two groups “have the same structure” means that they are isomorphic; but just having a map which preserve, or rather does not destroy, the group structure is often sufficiently interesting.

The “structure” preserved is how the elements relate to each other under the operations of the algebraic structure. For example, for commutative rings, if $\ a = b^2 – c^2 = (b-c)(b+c)\ $ is a difference of squares then applying a ring hom $\,h\, :\, a\mapsto \bar a\,$ shows that it remains a difference of squares in the image ring, viz. $\ \bar a = \bar b^2-\bar c^2 = (\bar b – \bar c)(\bar b +\bar c),\,$ so this particular ring-theoretic structure is preserved. Similarly preserved are polynomial expressions, i.e. expressions composed of basic ring operations (addition, multiplication) and constants $\,0,1.\,$ For example, a root of a polynomial remains a root of the image of the polynomial. So we can prove that a polynomial has no roots by showing it has no roots in a simpler homomorphic image, e.g. a ring with size so small that we can easily test all elements to see if they are roots, e.g. parity arguments: $\,f(x) = x(x\!+\!1)+2n\!+\!1\,$ has no integer roots since $\,x(x\!+\!1)\,$ is even, so adding $2n\!+\!1$ yields an odd hence $\ne 0;\,$ equivalently $\,f(0)\equiv 1\equiv f(1),\,$ i.e. $\,f\,$ has no roots mod $\,2,\,$ hence no integer roots. This generalizes as follows

Parity Root Test $\ $ A polynomial $\,f(x)\,$ with integer coefficients
has no integer roots when its $\rm\,\color{#0a0}{constant\,\ coefficient}\,$ and $\,\rm\color{#c00}{coefficient\,\ sum}\,$ are both odd.

Proof $\ $ If so then $\ \color{#0a0}{f(0)} \equiv 1\equiv \color{#c00}{f(1)}\,\pmod 2,\ $ i.e. $\:f\:$ has no roots in $\,\Bbb Z/2 = $ integers mod $\,2,\,$ therefore $\,f\,$ has no integer roots. $\ $ QED

In the same way, we can often reduce problems to “smaller”, simpler problems in modular images. Because homs preserve the ambient algebraic structure, as above, we can often deduce information about the original problems from information gleaned in the simpler modular images. Such problem solving by modular reduction is an algebraic way of “dividing and conquering”.

Let’s let $f:G\rightarrow H$ be a group homomorphism. We might say that $f$ preserves the structures of equalities (without quantifiers) – that is, we can say that, if, in $G$
then we can embed this system into $H$ by setting
More generally, we could embed any set of equalities into $H$ similarly – for instance, if we have that two specific elements commute, then we can find two (not necessarily distinct) elements in $H$ which commute – in this sense, we find the structure of $G$ in $H$ by using $f$, even if the structure shows up in some trivial way (i.e. the identity element satisfies every system of equation writable with group operations, hence the trivial homomorphism is a real thing). That is, we consider “structure” to mean “sets of equalities involving group operations”

Notably, the structure that a homomorphism does not preserve but an isomorphism does includes inequality and anything to do with quantifiers. For instance, if in $G$ we have
$$\forall x,y [xy=yx]$$
then an isomorphism lets us translate this same statement over to $H$ – quantifier and all, meaning $H$ is abelian. It also lets us embed statements involving inequality, like how the system
$$x\neq e$$
might be solvable in $\mathbb Z/2\mathbb Z$, but not in quotient groups (i.e. images of homomorphisms), like the identity group – so the additional structure of inverses is required here. Now, we can’t use the above to define an isomorphism vs. a homomorphism, but it does serve as an example of how homomorphisms preserve some very specific structure, limited to using only the group operations and equality, where isomorphisms give the stronger conditions necessary to make more general manipulations equivalent in both groups.

As you tagged the question (model-theory), here is an answer from a model-theoretical viewpoint.

The sentence you highlight is wrong. Morphisms do not preserve structure they preserve truth.

Once two structure A and B are given (groups, rings, ordered rings, etc.) they come with their own notion of truth. Then

[i] homomorphisms preserve the truth of atomic formulas,

[ii] embeddings are homomorphisms that preserve also the truth of negative atomic formulas (it follows that they are injective)

[iii] isomorphisms are surjective embeddings (it follows that they preserve the truth of every fist-order formula).

The Wikipedia’s editor you quote was probably speaking informally, so we shouldn’t make a fuss about it.

Another model-theoretic answer, in some way a rephrasing of @Primo Petri’s answer.

If you are considering first-order structures (meaning a base set with some finitary operations and relations defined on it), the properties preserved by a homomorphism are exactly those expressed by existential positive formulas. These are the ones that you can construct by using atomic formulas, $\wedge$, $\vee$, and $\exists x$. Moreover, these are exactly the first-order properties preserved by homomorphisms.

Likewise, the first-order properties preserved by surjective homomorphisms are exactly the positive formulas (same as before but you can use $\forall x$ as well). Both of these results can be traced back to the work of Łos, Tarski, and Lyndon.

For algebraic structures, I like to think of homomorphisms as functions that send true equations to true equations. For example, if we have the group $G = C_p = \langle x \mid x^p = 1 \rangle$, the cyclic group of order $p$, then we know that every element $g \in G$, in particular, the generator, satisfies $g^p = 1$. So if we have a homomorphism $\phi: G \to H$, then by properties of group homomorphisms, $\phi(g)^p = \phi(g^p) = \phi(1_G) = 1_H$, so we must map $g \in G$ to an element $\phi(g) = h$ that satisfies the same equations that $g$ does. That is, we must have $h^p = 1$, which we just proved must be the case.

So what does it look like when a homomorphism doesn’t preserve structure? Well, we have obvious examples, like non-trivial maps $C_p \to \mathbb Z$, but what about something that preserves some structure, but not others? The sorts of equations that you can make with commutative rings are polynomials, because you have addition and multiplication, so ring homomorphisms must preserve polynomial equations. Take for example the map $\phi : \mathbb Z \to \mathbb Z, \, 1 \mapsto -2$. It is not hard to show that this uniquely defines a group homomorphism, but since it does not send $1$ to $1$, it is not a ring homomorphism. Therefore, we should expect it to preserve linear equations, but not polynomial equations. Indeed, we have that if

$$ \sum x_i = 0,$$


$$ \sum \phi(x_i) = \sum -2x_i = -2 \sum x_i = 0. $$

But if we have an arbitrary polynomial,

$$ \sum_i \prod_j^{k_i} x_{ij},$$

($k_i$ is the number of factors in the $i$th term) then

$$ \sum_i \prod_j^{k_i} \phi(x_{ij}) = \sum_i \prod_j^{k_i} -2x_{ij} = \sum_i -2^{k_i} \prod_j^{k_i} x_{ij}, $$

but unless all the $k_i$ are equal, we cannot factor out the $2^{k_i}$ term, and if the $k_i$ all happen to be equal, then we can just multiply one of the terms by $1$, increasing the $k_i$ for that term, but not changing the equation. Therefore, $\phi$ preserves the abelian group structure of $\mathbb Z$, but not the ring structure.