# 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?

• See if the answers to this question help. Feb 21, 2015 at 1:33
• @BrianM.Scott Your answer there was exactly what I needed. I was missing that the kernel represents how much structure has been lost. I'll leave it up to the mods that be whether to close as a duplicate or leave this here so as to not bury Asaf's answer (not sure the etiquette on this SE). Feb 21, 2015 at 1:42
• The way I like to think of it is as if the groups have the same multiplication table (if you properly arrange the rows and columns). Then the isomorphism is just a relabeling of the elements. So really isomorphic groups really are the same, up to labeling. Feb 21, 2015 at 1:44
• @Dennis: Glad it helped. There’s no harm in leaving it: anyone who comes across it will get Asaf’s answer and a link to the earlier question and its answers. Feb 21, 2015 at 1:47
• @Dennis I added a simple example to my answer which might help lend some intuition. Feb 21, 2015 at 2:41

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\, :\, r\mapsto \bar r\,$$ shows that it remains a difference of squares in the image ring, viz. $$\ \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 of its elements to see if they are roots, e.g. common 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 $$\!\bmod 2\!:\ f(0)\equiv 1\equiv f(1),\,$$ i.e. $$\,f\,$$ has no roots mod $$\,2,\,$$ hence it has no integer roots. This idea 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".

• A correction, not to be pedantic, but to make sure I'm understanding properly. Should "The 'structure' being preserved is how the elements relate to each other under the operations of the algebraic structure." be instead "The 'structure' being preserved is how some of the elements relate to each other under the operations of the algebraic structure."; I'm just asking because if the structure you start with has a larger cardinality than the structure you end up with, won't it only preserve part of this relational structure? Or am I missing something? Feb 21, 2015 at 1:53
• @Dennis Such preservation remains true no matter how much the target collapses (possibly even to a one element structure). If we consider more general relational structures (vs. algebraic = equational structures), then we can have inequalities as axioms, and inequations like $\,1\ne 0\,$ needn't be preserved under homs; e.g. the one-element zero ring $\{0\}$ is an image of every ring, but not of any domain, due to the axiom $\,1\ne 0.\ \$ Feb 21, 2015 at 2:17

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 $G$ $$xy=z$$ then we can embed this system into $H$ by setting $$f(x)f(y)=f(z).$$ 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 $$x^2=e$$ $$x\neq e$$ 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.

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.

• +1 for the nice distinction between groups with the same structure, and maps that preserve structure. Feb 21, 2015 at 1:45
• Rephrasing "preserves" as "does not (totally) destroy" makes the whole thing make a lot more sense to me. My ordinary notion of preserves would seem to imply sameness, whereas clearly this notion does not (since then, like I worried, homomorphisms would just be isomorphisms). This notion of preservation is a bit funny actually. In this sense, since the barbarian raiders didn't totally destroy Rome, they "preserved" Rome! Kinda makes me want to start a "Historical Preservation Society" that reduces historic buildings to near-rubble. Feb 21, 2015 at 1:59
• Zero structure is some structure. Feb 21, 2015 at 4:10
• @Denis: It is unfortunate when we let the natural meaning of words to interfere with their mathematical meaning. Feb 21, 2015 at 9:57

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.

• I'm not acquainted with model theory. But I take it that preserving truth does not imply preserving falsity as well, so that a false statement might validly become true. Anyway, I don't think that in general morphisms can be taken to preserve some form of truth, though maybe in algebra they do. Morphisms of topological spaces (a.k.a. continuous maps) have the property that the associated "inverse image" map preserves openness of subsets, which is hard to construe as preserving truth. Yet in some loose sense continuous maps do preserve (in a one-sided way) topological structure. Feb 21, 2015 at 11:06
• @Marc-van-Leewen Indeed, the notion of structure in model-theory is similar to (a generalization of) the one used in algebra. (It can adapted to continuous structures, but it less straightforward.) Feb 21, 2015 at 12:23

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

$$\sum x_i = 0,$$

then

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

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.

Functors are the most general case of structure preserving function. A functor is a function that sends objects to objects and arrows to arrows in such a way that the structure is preserved which means 3 things: 1)an identity arrow is sent into an identity arrow 2)given an arrow between two objects, the image of the arrow is another arrow between the images of the objects 3)operation of arrow is preserved (composition is the general case of an associative operation)

A group is a one object category, so that we really only have to pay attention to 1) and 3) which means that the function preserves unit and operation between elements. The last condition means that the function ¨commutes¨ with the operation in the sense that you get the same result from operating two elements and mapping the result, or if you map the original elements and then operate them in the image structure.

One thing that seems to confuse also is the fact that a structure preserving function does not have to bijective. For example, if the morphism $\phi:G\rightarrow H$ is not surjective this means we are mapping the group $G$ into a subgroup of $H$ (that is why the structure is preserved even if we dont cover all tof the image). When the morphism is not injective this means we dont preserve distinctness (there will be at least one pair of different elements in G mapped into the same element of H).