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Could somebody give a concrete example of a group structure being preserved in a isomorphism, et cetera? I always hear this 'preserve structure' thing. Ok, could somebody give me a rigorous definition of what it means to preserve structure? I'm tired of hearing things like 'they behave the same way under some operation bla bla bla'

Zev Chonoles
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    The property of being abelian ($xy = yx$ for all $x$ and $y$ in the group), the property of having an element of order 9, the property of having a subgroup of order 10, the property of having a normal subgroup of index 10, and the property of having a nontrivial commutator subgroup are all preserved by an isomorphism. You should sit down and try to check these yourself, e.g., if $f \colon G \rightarrow H$ is an isomorphism of groups then $G$ if abelian iff $H$ is abelian, $G$ has an element of order 9 iff $H$ does, and so on. – KCd Jul 06 '15 at 00:50
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    You're not going to get a fully rigorous definition of "preserve structure" because it's a mnemonic device or a sort of euphemism; it is supposed to motivate the various types of [insert prefix]-morphisms you encounter. To put another way, it isn't a fully rigorous statement. – pjs36 Jul 06 '15 at 01:07
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    If isomorphisms preserve structure, and you can't name any structural features that isomorphisms preserve, it's probably because you don't know any examples of structural features. It strikes me as a super weird question - like asking for a concrete example of a bachelor that isn't married. Anyway, what "structure" does a group have you ask? Well, the way in which a group differs from a barren set devoid of any structure is the binary operation, so it's only natural that "preserve structure" should be "preserve the binary operation"! – anon Jul 06 '15 at 05:42
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    @pjs36: As a minor and arguable nitpick, I'd say that it's a perfectly rigorous statement, but not a formal one. – anomaly Jul 06 '15 at 14:11
  • Is is safe to say that the operator of a group is what imposes the structure on the group (pretty much by definition)? – David R Tribble Jul 06 '15 at 22:03

8 Answers8

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First let's consider a one-to-one correspondence that does not preserve structure. Let each real number $x$ correspond to $x+1$. Say $\varphi(x)=x+1$. Let the group operation be addition. Notice that $6+7=13$. But $$\varphi(6)+\varphi(7) = 7 + 8 = 15 \ne 14 = \varphi(13).$$

Now let's consider one that does preserve structure:

Consider the group of permutations of $A,B,C$.

This group contains

  • the identity permutation,
  • three transpositions: $A\leftrightarrow B$, $A\leftrightarrow C$, $B\leftrightarrow C$,
  • two rotations: $$\begin{array}{ccccc} A & & \leftarrow & & C \\ & \searrow & & \nearrow \\ & & B \end{array}\quad\text{and}\quad \begin{array}{ccccc} A & & \rightarrow & & C \\ & \nwarrow & & \swarrow \\ & & B \end{array}.$$

Now consider a group of six rational functions: \begin{align} x & \mapsto x \\[6pt] x & \mapsto 1/x \\[6pt] x & \mapsto 1-x \\[6pt] x & \mapsto x/(x-1) \\[6pt] x & \mapsto (x-1)/x \\[6pt] x & \mapsto 1/(1-x) \end{align} The group operation is composition of functions. You can check that this set of functions is closed under composition. Notice that each of the second, third, and fourth ones is its own inverse. Notice the fifth and sixth ones are squares of each other and inverses of each other and the cube of each is the identity.

Now let $\iota$ be the identity element. Let $\alpha, \beta,\gamma$ be the three self-inverse elements. Let $\delta,\varepsilon$ be the two that are squares of each other. We have a multiplication table: $$ \begin{array}{c|cccccc} & \iota & \alpha & \beta & \gamma & \delta & \varepsilon \\ \hline \iota & \iota & \alpha & \beta & \gamma & \delta & \varepsilon \\ \alpha & \alpha & \iota & \varepsilon & \delta & \gamma & \beta \\ \beta & \beta & \delta & \iota & \varepsilon & \alpha & \gamma \\ \gamma & \gamma & \varepsilon & \delta & \iota & \beta & \alpha \\ \delta & \delta & \beta & \gamma & \alpha & \varepsilon & \iota \\ \varepsilon & \varepsilon & \gamma & \alpha & \beta & \iota & \delta \end{array} $$

Q: Is this the table for the first group described above or the second?

A: It's either. Or both. One can label the members of the first group with the six Greek letters in such a way that $\iota$ is the identity, $\alpha,\beta,\gamma$ are the three self-inverse elements, and $\delta,\varepsilon$ are each other's squares, and if done right, this will be the multiplication table, and one can do the same with the second group. The one-to-one correspondence is this: the element labeled with a certain Greek letter in one group corresponds to the one labeled with the same Greek letter in the other group. Preserving the structure means getting the same multiplication table either way.

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I'd say the formal definition is the following. (I don't assume that this means anything to you, but I thought you might want to see it anyways.) Fix some language $L$.

A function $$ \pi \colon \mathcal M \rightarrow \mathcal N $$ between two $L$-structures $\mathcal M, \mathcal N$ preserves their $L$-structure iff it is an ismorphism, i.e.

  • $f$ is bijective
  • for all constant symbols $c \in L$ $$\pi(c^{\mathcal M}) = c^{\mathcal N}$$
  • for all $n$-ary function symbols $f \in L$ and $x_1, \ldots, x_n \in M$ $$\pi ( f^{\mathcal M} (x_1, \ldots, x_n) ) = f^{\mathcal N} (\pi(x_1), \ldots, \pi(x_n))$$
  • for all $n$-ary relation symbols $R \in L$ and $x_1, \ldots, x_n \in M$ $$R^{\mathcal M} (x_1, \ldots, x_n) \text{ iff } R^{\mathcal N}(\pi(x_1), \ldots, \pi(x_n))$$

Let my try to explain why we would even want to define such a thing: Let $(G, \cdot_G)$ be your favorite group and fix any bijection $$ \pi \colon G \rightarrow X $$ (for a suitable set $X$. For example $(G, \cdot_G) = (\mathbb Z,+)$, $X = \mathbb N$.)

We may define a group $(X, \cdot_X)$ by setting $$ x \cdot_X y = z \text{ iff } \pi^{-1}(x) \cdot_G \pi^{-1}(y) = \cdot_G \pi^{-1}(z) $$

Then $$ \pi \colon (G, \cdot_G) \rightarrow \colon(X, \cdot_X) $$ preserves their $\{\cdot \}$-structure, where $\cdot$ is a $2$-ary function symbol and $\cdot^G = \cdot_G$, $\cdot^X = \cdot_X$.

If we now consider the example $(G, \cdot_G) = (\mathbb Z, +)$ and $X = \mathbb N$, then $(\mathbb Z, +)$ and $(X, \cdot_X)$ certainly "look different" (as sets). But as groups, they "are the same". We just "changed the names of the elements", but we don't actually care what these elements are. We care about the group structure of $G$ and that hasn't been effected.

If you insist, that you actually care for the elements of a given group, ask yourself the following question: What exactly is a real number? (There many different ways to construct the reals, but no one cares (when thinking of it as a group) which definition you have in mind, because the structure is always the same. Some mathematician don't even care, if you don't know how to construct the reals at all.)

Stefan Mesken
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Assume that we have two groups $(G,*),(H,*^\prime)$, and assume $\phi$ is a mapping from $G \to H$

That is $\phi:G \to H$

Now $\phi$ is said to preserve the structure if and only if $$\phi(a*b) = \phi(a) *^\prime \phi(b)$$

Notice that $*$ is the operation for group $G$ and $*^\prime$ is the operation of group $H$

Also if $\phi$ preserves the structure, It is also said to be a group homomorphism

In plain English, This is what it means to preserve the structure.

It doesn't matter whether I take the product $ab$ and apply the function $\phi$ to it or I take $a$ apply $\phi$ then take $b$ apply $\phi$ and then take their product. The result is the same

Notice here I used product, I mean Group operation.

alkabary
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    lol you just defined homomorphism as being 'something that preserve structure'. Yes, the definition of homomorphism is rigorous, but I want to know which 'structures' are preserved and also what does that even mean – Guerlando OCs Jul 06 '15 at 00:54
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    There are some rigorous definitions of 'structure-preservation' using universal algebra, but nobody actually thinks of structure preserving functions in that way. What is meant exactly by "structure?" I don't even know. Just think about it intuitively. If you can appreciate the significance of the sentence "It doesn't matter whether I take the product ab and apply the function ϕ to it or I take a apply ϕ then take b apply ϕ and then take their product. The result is the same" then you should see what the fuss is about "structure preservation." – D_S Jul 06 '15 at 01:30
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A group $(G,\ast)$ is a set $G$ "with a group structure". What is the group structure? It is the function $\ast:G\times G\to G$ that defines the operation of the group, $$\ast(g_1,g_2)=g_1\ast g_2$$ If $(G,\ast)$ is a group, and $(H,\star)$ is a group, then we say a function $\varphi$ from the set $G$ to the set $H$ "preserves group structure" when it "takes the group structure of $G$ to the group structure of $H$". What does that mean? It means the following two group structures on the set $\varphi(G)$ (that is, the image of $G$ under the function $\varphi$) agree:

  • the operation $\diamond:\varphi(G)\times\varphi(G)\to\varphi(G)$ defined by $$\varphi(g_1)\diamond\varphi(g_2)\overset{\text{def}}{=}\varphi(g_1\ast g_2)$$ This is intuitively the operation on $\varphi(G)$ that the function $\varphi$ "sent" the operation $\ast$ to. We would say that $\diamond$ is the operation on $\varphi(G)$ "induced by" the operation $\ast$ from $G$.

  • the operation $\bullet:\varphi(G)\times\varphi(G)\to\varphi(G)$ defined by $$\varphi(g_1)\bullet\varphi(g_2)\overset{\text{def}}{=}\varphi(g_1)\star\varphi(g_2)$$ This is intuitively the operation on $\varphi(G)$ that it "already has" from the operation $\star$ on $H$. (Technically I am glossing over something here. If you want I can expand.)

When $\varphi$ "sends the operation on $G$ to the operation on $H$", we say that $\varphi$ has preserved group structure. This happens precisely when $\varphi:G\to H$ is a group homomorphism.

Now, this all guarantees that we're "preserving" (in the intuitive sense) group operations as $\varphi$ sends elements from $G$ to $H$. However, if we make a stricter interpretation of the word "preserve" to mean "preserve in a reversible way", then we might also say that we want there to exist a function $\psi:H\to G$, with $\psi$ "preserving group structure", with the property that $\psi$ "undoes" $\varphi$ and $\varphi$ "undoes" $\psi$. Then we've arrived at the definition of a group isomorphism.

Zev Chonoles
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    Do you actually say that a non-injective, non-surjective homomorphism preserves the group structure? This seems odd,as it's structure (in the intuitive sense) is violated, especially if I consider injections of the trivial group into some $G$ or the surjection from $G$ into the trivial group. – Stefan Mesken Jul 06 '15 at 01:40
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    Yes, if we have agreed to interpret "preserve group structure" as preserve operation. No, if we have agreed to interpret "preserve group structure" as preserve operation in a reversible way or alternatively preserve operation and underlying set. – Zev Chonoles Jul 06 '15 at 01:42
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    The fact that "preserve" could be interpreted with two different levels of "strictness" was something that explicitly tried to address in my answer. – Zev Chonoles Jul 06 '15 at 01:44
  • Sure, I don't claim that this is "wrong". It just doesn't agree with the terminology I'm used to. (In a case of a homomorphism that isn't necessarily an isomorphism I'd say "it respects the operation".) – Stefan Mesken Jul 06 '15 at 01:50
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    @Stefan I find it slightly peculiar to say that a non-isomorphic homomorphism "respects the operation" but does not "respect the structure", as the structure of a group, more or less by definition, is determined by the operation. I don't profess an opinion about the original question, except that when it's important to be precise about the notion of preservation, it's better simply to say a little more than "preserves structure", exactly because of this ambiguity. – Travis Willse Jul 06 '15 at 02:56
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The idea of "structure-preserving function" comes from model theory and universal algebra.

We formalize the idea of a "structure" by considering a set $S$ with some operations $f_1, \ldots, f_n$, where each $f_i$ can have some number of arguments ranging over elements of $s$. We usually require these functions to satisfy some axioms, depending on which structure we're considering. For example, a group can be defined as a set $G$ with two operations $e$ (identity), $*$ (multiplication). $e$ takes no arguments, while $*$ takes two. (A $0$ argument function is a constant; here $e$ means the identity element of the group). It is required to satisfy the familiar axioms of groups:

  • For all $g \in G$, $g * e = e * g = e$
  • For all $g_1, g_2, g_3 \in G$, $g_1 * (g_2 * g_3) = (g_1 * g_2) * g_3$
  • For all $g \in G$ there exists some $h \in G$ such that $g * h = h * g = e$

We may define a ring similarly (we have two constants, $0, 1$ representing additive and multiplicative identities as well as the two operations of addition and multiplication), and so on with fields, etc. A structure-preserving map is required to preserve all of the operations $f_i$.

This means that if we want $\phi$ to be a structure preserving homomorphism on the structure $(S, f_1, \ldots, f_n)$ to a structure $(T, g_1, \ldots, g_n)$ ($T$ is required to be the same type of structure with the same functions, axioms, etc.), then for each $i = 1, \ldots, n$, if $f_i$ takes $r_i$ arguments, then for all $s_1, \ldots, s_{r_i} \in S$, $\phi(f_i(s_1, \ldots, s_{r_i})) = g_i(\phi(s_1), \ldots, \phi(s_{r_i}))$.

In the case of groups, this means that if $\phi: G \rightarrow H$ is a group homomorphism, then we require:

  • for all $g, g' \in G$, $\phi(g *_G g') = \phi(g) *_H \phi(g')$ (where $*_G$, $*_H$ are the multiplication operations in $G,H$ respectively).
  • $\phi(e_G) = e_H$

The second requirement is usually left out in textbooks, since it is an elementary consequence of the group axioms that a map between groups that satisfies the first requirement will also satisfy the second (see if you can show this).

Note: We can actually define the idea of a structure preserving homomorphism a little more generally. We can allow our structures to also have relations: the easiest example is the structure of a partially ordered set. This is a set $S$ with a single binary relation $\leq$ which satisfies the usual axioms of a partially ordered set. A map of posets is required to satisfy the following requirement: if $a \leq b$, then $\phi(a) \leq \phi(b)$, and similarly, given a structure with a relation $R$, then a structure preserving map satisfies: if $aRb$ then $\phi(a) R \phi(b)$.

An isomorphism is a bijective structure preserving map whose inverse is a structure preserving map. It is a theorem of model theory that given an isomorphism of any structure $\phi: A \rightarrow B$, and any sentence $S$ which only mentions things that can be defined in terms of the structure (this can be formalized: but basically think of any statement about groups - it is a group-theoretic statement that there is an element of order 267, but it is not a group theoretic statement that the group is a subset of the natural numbers, as the latter is not about groups), then $S$ is true of $A$ if and only if $S$ is true of $B$.

Dorebell
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To complement Dorebell's excellent answer, I'll describe the same universal algebra ideas in the language of category theory rather than set theory. Well, I won't describe them, because other people have described them very well already, so I'll link existing expositions.

If you're not already familiar with the idea of a category, I recommend reading the opening pages of Awodey's Category Theory or Freyd and Scedrov's Categories, Allegories to become acquainted.

In any category that has finite products (and therefore, automatically, a terminal object), you can define a group object as an object $G$ equipped with a certain system of arrows between $G$ and products of $G$. Certain diagrams involving these arrows are required to commute. The arrows are the processes of multiplication, inversion, and producing the identity element. The commutative diagrams are the requirements that multiplication is associative, multiplying with the inverse gives the identity, and multiplying with the identity does nothing.

A very nice feature of this desciption is that it describes many different flavors of group in one breath. For example, a group object in the category of sets is an ordinary group, a group object in the category of smooth manifolds is a smooth Lie group, a group object in the category of supermanifolds is a supergroup, a group object in the category of algebraic varieties is an algebraic group, and a group algebra in the opposite category of commutative rings is a Hopf algebra (which doesn't even sound like it's a group). There are lots more examples where most of those game from.

Describing the "certain system of arrows" that defines a group object is pretty tedious, so people save time by calling this system of arrows a group structure. Then they can say, "Oh, group objects are simple! A group object is just an object equipped with a group structure."

Now we can finally say what it means for an arrow to "preserve structure." An arrow $f \colon X \to Y$ between two objects naturally induces arrows between their products: for example, we get an arrow $f \times f \colon X \times X \to Y \times Y$. An arrow $f \colon G \to H$ between two group objects preserves the group structure if $f$ and $f \times f$ and so on commute with the arrows giving the group structures of $G$ and $H$.

I've been talking about group objects as an example, but you can do the same thing with other kinds of algebraic structures. For example, you can study about ring objects and natural numbers objects. Natural numbers objects are super cool: just like group objects in different categories give different flavors of group theory, natural numbers objects in different categories give weird alternate-universe versions of number theory.

Vectornaut
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Perhaps a negative example (CONCRETE) would be in order to illustrate this.

Let $G$ be the group of positive rational numbers under multiplication. Let $f\colon G\to G$ be the function $$f\bigg(\frac mn\bigg)= \frac nm$$ Check that this is a homomorphism actually an isomorphism.

If $\frac mn < \frac rs$ you can also check that $f(\frac mn) \not< f(\frac rs)$. SO this homomorphism does not preserve the hierarchy of numbers by their size. But to define the group operation we had no use of this hierarchy: and so not preserved. However all other things that you brush aside with contempt with bla bla will be preserved.

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The best way I heard isomorphic described was "the same thing, just renamed". I feel like this is the way of looking at things. Copied from another post:

A good example to add. "There is only one group of order 2 up to isomorphism". The group is $G=\{e,a\}$ with $ea=ae=a$ and $a^2=e$. Now you could argue "I have another group, $H=\{e,b\}$ with $eb=be=b$ and $b^2=e$". Your group IS different than mine, but your group is just the renamed version of mine. "Renaming" is isomorphism.

The structure, what things actually are doing is the same. Whether you use $a$s, or $b$s doesn't matter. What does matter is how these things actually behave. Nontrivial isomorphisms are interesting. Things that may a priori different things might actually just be renamed.