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Def. of groups being isomorphic - some motivation.

  1. Jun 23, 2013 #1

    Stephen Tashi

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    For another thread https://www.physicsforums.com/showthread.php?p=4420542 , I want to give simple motivations for the definition of two groups being isomorphic and related topics. My explanations have become so lengthy that I decided to post them here in 3 parts. ( They are probably too elementary for most readers of the other thread and that thread is already long.)

    I've read an explanation similar to mine somewhere. It made the reader work harder to find the isomorphisms. I picked notation that makes one of them blatantly obvious.

    The 3 messages will follow. Maybe they will be helpful to someone -and maybe they need some fixing.
     
    Last edited: Jun 23, 2013
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  3. Jun 23, 2013 #2

    Stephen Tashi

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    1) Becoming Comfortable With Definition Of Being Isomorphic

    A mathematical definition is not a statement that is proven. It is statement explaining terminology that people have chosen to use. Whether a definition becomes widely accepted is question of sociology not a question of mathematics. This little essay is not a "proof" for the definition of being isomorphic. It's only an effort to explain why many mathematicians like it.

    For the more complicated mathematical structures, caution is needed in using simple phrases such as "is", "is identical to", "is the same as" or "is equal to". If we split hairs, caution is needed even with simple structures. For example "1 + 1/2" and "3/2" are equal if we are talking about arithmetic, but in a computer programming class discussing matching strings of symbols, they aren't. When studying things with considerable structure like a vector space, students have trouble with questions like "Is the zero scalar equal to the zero vector?" because their previous courses are habitually careless about using the term "equal".

    Technically, there is no mathematical definition of the phrase "is equal" or the symbol [itex] = [/itex] without reference to a particular equivalence relation. In other words you should say that two things "are equal in the aspect of ..." or "equal with respect to...". But human beings (including mathematicians) toss the word "equal" around freely and the meaning is usually clear from the context.

    If [itex]G[/itex] and [itex] H [/itex] are groups, the natural interpretation of "[itex] G = H [/itex] " is that the two groups are equal in every aspect that constitutes a group.

    An abstract group is a set of things and a binary operation on that set that satisfies some properties (those that I remember by the chant "closed, associative,identity, inverse"). I carelessly use the notation "[itex] G[/itex]" for both the group and the set of elements in the group. Careful writers denote the group as "[itex]G[/itex]" and the set of things in the group as "[itex]\{G\}[/itex]. I will reform momentarily just to say that the natural interpretation of "[itex] G=H [/itex]" would be that [itex]\{G\} = \{H\} [/itex] (as sets) and that the two groups have the same binary operation defined on the set.

    The above definition of "[itex] = [/itex]" for groups turns out to be a useless concept. After all, what makes "[itex]=[/itex]" a useful concept for other mathematical objects (such as numbers) is that we study equations when two different expressions are equal to each other ( such as [itex] x^2 + 3 = 4x - 7 )[/itex]. There are expressions in group theory that denote taking groups and making other groups from them (such as direct sums, quotient groups), but these methods require creating new groups that have a different set of elements than the groups they are applied to. It is rare for two different processes on groups to produce groups with the same set of elements. Thus when working with groups there are hardly any situations where the above definition of "[itex]=[/itex]" is a useful.

    The most useful equivalence relation between groups is the relation "is isomorphic to". I know of no book where the author is bold enough to use the symbol "[itex]= [/itex] to mean "is isomorphic to". Many use "[itex]\cong[/itex]". (Authors fearlessly use "[itex]=[/itex]" for the principal equivalence relation between numbers, vectors, matrices, sets, functions, all of which are different equivalence relations. Perhaps the reason they chicken out when it comes to groups is that "[itex]G=H[/itex]" might be mistaken to mean only that [itex]\{G\}=\{H\}[/itex], and this is not even required for being isomorphic.)

    The easiest way to motivate the definition of being isomorphic is to consider finite groups. Much information about a finite group [itex]G[/itex] can be displayed by showing its multiplication table. [On the forum, I am writing the table as a matrix since the Latex for tables doesn't work in posts.] The table below is for a "Klein four-group", also called a "viergruppe".

    A Multiplication Table For The Group [itex]G[/itex]

    [tex] \begin{pmatrix} \ & \ & a & b & c & i \\ - & - & - & - & - & - \\ a & \ & i& c & b & a \\ b &\ & c & i & a & b \\ c & \ & b & a & i & c \\ i & \ & a & b & c & i \end{pmatrix} [/tex]


    The above table tell us: [itex] (a)(a) = i, \ (a)(b) = c,\ (a)(c) = b, \ (a)(i) = a [/itex] etc. ( I prefer that multiplication tables be written so that you read the leftmost factor from the column on the left and the rightmost factor from the row at the top. The 4-group is commutative so the order of factors doesn't matter in this particular table.) You can see that [itex] "i" [/itex] is the identity element. One unusual feature of this group is that elements are their own inverses. For example [itex] (a)(a) = i [/itex].

    The multiplication table doesn't tell us everything about the group. It doesn't tell how the set of elements in the group was specified so we don't know what "[itex] i [/itex] ","[itex] a[/itex] " etc. represent. They could be numbers or matrices or functions, or just symbols. It also doesn't tell how the table was created. For example, was the binary operation on the group implemented by matrix multiplication, or modular arithmetic, composing functions, or just writing symbols into the table without any systematic procedure in mind?

    In another respect the multiplication suggests too much information about a group. In the definition of a group, there is no requirement that the set of elements be ordered in any way. But when you make multiplication table you have to pick an order. The table below lists the elements of [itex]G[/itex] in a different order and has the same information about multiplication as the table above..

    Another Format Of A Multiplication Table For [itex]G[/itex]

    [tex] \begin{pmatrix} \ & \ & i & a & b & c \\ - & - & - & - & - & - \\ i & \ & i & a & b & c \\ a &\ & a & i & c & b \\ b & \ & b & c & i & a \\ c & \ & c & b & a & i \end{pmatrix} [/tex]


    Consider a multiplication table for a different group [itex]H[/itex]

    A Multiplication Table For The Group [itex]H[/itex]

    [tex] \begin{pmatrix} \ & \ & I & A & B & C \\ - & - & - & - & - & - \\ I & \ & I & A & B & C \\ A &\ & A & I & C & B \\ B & \ & B & C & I & A \\ C & \ & C & B& A & I \end{pmatrix} [/tex]

    A natural reaction is to say "[itex] H [/itex] and [itex] G [/itex] have the same multiplications tables, so [itex] H [/itex] and [itex] G [/itex] are equal groups."

    But, as pointed out above, we must be careful about asserting equivalences like "equal" and "is the same" without specifying what aspects of two things are the same. For example, the two tables don't use the same set of elements. The purpose of the definition of isomorphism is to capture the sense in which the groups [itex] G [/itex] and [itex] H [/itex] "are essentially the same".

    One thought is that "[itex]G[/itex] is isomorphic to [itex]H[/itex]" ought to mean that three is a way to match up the names of one group with the names of the other group so the multiplication tables match up entry-by-entry. However, if we knew [itex]G[/itex] only from first multiplication table that was given for it, then it's not clear that matching up that table to the table for [itex]H[/itex] entry-by-entry can work. We need to focus our attention on matching the information in the tables, not the format of the tables.

    A vague way to describe the sameness between [itex]G[/itex] and [itex]H[/itex] is to say that for each element (such as [itex] a [/itex] )in [itex]G[/itex] there is an element (such as [itex] A [/itex]) in [itex]H[/itex] that "corresponds" to it. The word "corresponds" has a satisfying sound in ordinary speech, but it doesn't have a deep or specific meaning in mathematics. Usually when we speak of things "corresponding", it means that we have established a 1-to-1 mapping between two sets that accomplish some purpose. So we will begin to define "[itex]G[/itex] is isomorphic to [itex]H[/itex]" by writing:

    Notice that the definition says " there exists a 1-to-1 function [itex] \phi [/itex]". It doesn't say " for each 1-to-1 function [itex] \phi [/itex]". Intuitively, there are wrong ways to match up the elements of the two groups. For example [itex] i [/itex] is the identity of [itex]G[/itex] and [itex] I [/itex] is the identity of [itex]H[/itex]. Whatever we are going to do with [itex] \phi [/itex] might not work for a function that maps [itex] i [/itex] to a non-identity element ( such as [itex] A [/itex]).

    A comforting way to say what we want the function [itex]\phi [/itex] to accomplish is to say that "corresponding products are mapped to the corresponding answer". But how can we say this precisely?

    In the example at hand, I've used symbols that make it obvious that there exists a function [itex] \phi [/itex] that satisfies our intuitive notion of "the correct" correspondence. It is the function [itex] \phi [/itex] such [itex]\phi(i) = I [/itex], [itex]\phi( a) = A [/itex], [itex] \phi(b) = B [/itex], etc.

    A result like [itex] (a)(b) = c [/itex] (which is computed in group [itex]G[/itex]) should be mapped to a result that "corresponds" with what happens in group [itex] H [/itex].

    In the example at hand, we could express this by saying [itex] \phi(a) \phi(b) = \phi(c) [/itex] because we want [itex] (a)(b) = c [/itex] to "correspond" to [itex]\phi(a)\phi(b) = \phi(c) [/itex] which evaluates to [itex](A)(B) = C [/itex]. However, this uses a fact that is particular to the example, namely that we know [itex] (a)(b) = c [/itex]. A more generally applicable statement is to say that [itex] \phi(a) \phi(b) = \phi( (a)(b) ) [/itex], without claiming to know a specific symbol for the result of the product [itex] (a)(b) [/itex].

    You can state the previous reasoning as consistency requirement. It says that if [itex] \phi [/itex] correctly matches up the elements of the groups then we can see what element "corresponds" to the product [itex](a)(c) [/itex] in two ways. One way is evaluate the product and get a single symbol as the answer and then see what that symbol corresponds to. The other way is to see what symbols the individual facators of the product correspond to and multiply those two symbols together.

    Adding the requirement that [itex] \phi((x)(y)) = \phi(x)\phi(y) [/itex] is how we we define [itex] \phi [/itex] to be a "correct" way establish a correspondence between the elements of the two groups.

    An amusing but less direct way to motivate the definition is this scenario: Suppose we want to find the result of the product [itex] (a)(b) [/itex] in group [itex] G [/itex], but someone spilled chocolate syrup on the multiplication table for [itex] G [/itex] and we can't read the answer. We can see the multiplication table for group [itex] H [/itex] and we know a function [itex] \phi [/itex] that gives a correct correspondence between the elements of the two groups. Being clever, we use [itex] \phi [/itex] to find what elements in [itex] H [/itex] correspond to [itex] a[/itex] and [itex] b [/itex]. We look up the result of the product of those elements in the multiplication table for [itex] H [/itex]. Then we see what element in [itex] G [/itex] corresponds to that result by using the inverse function of [itex] \phi [/itex].

    In symbolic form, what we have done is utilize [itex] (a)(b) = \phi ^{-1} ( \phi(a) \phi(b) ) [/itex]

    Applying the function [itex] \phi[/itex] to both sides of that equation gives [itex] \phi(ab) = \phi(a) \phi(b)[/itex].

    The definition of being isomorphic doesn't contain any explicit guarantees that it is an equivalence relation. As the saying goes, "it can be shown" that it is. The fact than being isomorphic is an equivalence relation is a theorem, not a definition.
     
  4. Jun 23, 2013 #3

    Stephen Tashi

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    Now that you understand the definition for two groups being isomorphic, let me shake your faith in it a little.

    Notice the definition says "there exists a 1-to-1 function [itex] \phi ... [/itex]. It doesn't say "there exists a unique 1-to-1 function [itex] \phi [/itex]". It leaves open the possibility that there might be several different functions that work.

    In the example, symbols have been chosen to make it easy to see that that the particular function [itex] \phi [/itex] that maps [itex]i [/itex] to [itex] I [/itex], [itex] a [/itex] to [itex] A [/itex] etc. satisfies [itex] \phi( (x)(y) ) = \phi(x) \phi(y) [/itex].

    It turns out that there are other functions that also have that property.

    For example, consider the function [itex] \zeta [/itex] defined by:
    [itex] \zeta(i) = I [/itex]
    [itex] \zeta(a) = B [/itex]
    [itex] \zeta(b) = A [/itex]
    [itex] \zeta(c) = C [/itex]

    It satisfies [itex] \zeta((x)(y)) = \zeta(x) \zeta(y)[/itex]. For example:

    [itex] \zeta( (a)(c) ) = \zeta(b) = A [/itex] and [itex] \zeta(a)\zeta(c) = (B)(C) = A [/itex]
    [itex] \zeta( (a)(a)) = \zeta(i) = I [/itex] and [itex] \zeta(a) \zeta(a) = (B)(B) = I [/itex]

    You can try other cases yourself. You can also try the scenario about chocolate being spilled on the multiplication table for [itex]G[/itex] and see that [itex]\zeta[/itex] works as well as [itex]\phi[/itex] for finding products that you can't see in the table.

    I think about the situation this way: The definition of being isomorphic pleases me as long as two groups that I intuitively think are "essentially the same" satisfy the definition of being isomorphic and as long as two groups that I think are "not essentially the same" don't.

    In the example at hand, the groups [itex]G[/itex] and [itex]H[/itex] are, in my opinion, essentially the same. So the fact that there is more than one way to match-up there elements and satisfy the definition doesn't disturb me. ("The more, the merrier")

    The fact that there can more than one way does make me worry that there might be two groups that I think are "not essentially the same" that satisfy the definition of being isomorphic. The only way to investigate this mathematically would be to define precisely what I mean by "not essentially the same". That's a rather circular dilemma since the reason for creating the definition of being isomorphic was to make the ideas of "essentially the same" and "not essentially the same" precise.

    You can investigate the matter by examples. . For example you can try various mappings between the elements of [itex] G [/itex] and some group you feel is "not essentially the same". Try the group [itex]W[/itex] (which is called "a cyclic group of order 4") with multiplication table given by:

    A Multiplication Table For [itex]W[/itex]

    [tex] \begin{pmatrix} \ & \ & I & A & B & C \\ - & - & - & - & - & - \\ I & \ & I & A & B & C \\ A &\ & A & B & C & I \\ B & \ & B & C & I & A \\ C & \ & C & I & A & B \end{pmatrix} [/tex]

    [itex]W[/itex] contains the same elements as [itex]H[/itex] but its multiplication is fundamentally different. The element [itex]I[/itex] is the identity in both groups, but [itex]W[/itex] doesn't have as many elements that satisfy [itex](x)(x) = I [/itex] as [itex]H[/itex] does.
     
  5. Jun 23, 2013 #4

    Stephen Tashi

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    We need to define 'isomorphism". Did you think we already did? . We defined the relation of "being isomorphic". If the conventions of common speech apply, you know that "isomorphism" means "the property of being isomorphic". However, this is math, and that's not correct.

    The technical definition of "isomorphism" says it is a function. You'll recognize the idea behind it:

    If an author defines an "isomorphism" first, then he can define the relation of being isomorphic in a concise manner:

    To phrase a previously stated fact in fancier manner, if [itex] G [/itex] is isomorphic to [itex] H [/itex] then there exists at least one isomorphism from [itex] G [/itex] to [itex] H [/itex] and there may be more than one isomorphism between the two groups.

    Is a group [itex]G[/itex] isomorphic to itself? Yes. This can be proven. There is nothing in the definition of "[itex]G[/itex] is isomorphic to [itex]H[/itex]" that forbids [itex] G [/itex] and [itex] H [/itex] from being the same group. (Here "being the same" means being the same set of elements with the same binary operation.) If you want to find an isomorphism from [itex] G [/itex] to [itex] G [/itex], it's natural to pick the identity function, which maps each element to itself. You can prove the identity function is an isomorphism from [itex] G [/itex] to [itex] G [/itex].

    Here's where things get interesting (for people with mathematical tendencies) or scary (for people who like a flat intellectual landscape). There can be isomorphisms of a group to itself that are different than the identity function. In fact, for any finite group with more than 2 elements "it can be shown" that there are isomorphisms of the group to itself besides the identity function.

    In the previous example, where [itex] G [/itex] is a Klein 4-group, we can define an isomorphism of [itex] G [/itex] to itself that isn't the identity function. ( Take a hint from the function [itex] \zeta [/itex] in a previous example.)

    Define the function [itex]\psi[/itex] on the elments of [itex] G [/itex] by:
    [itex]\psi(i) = i[/itex]
    [itex] \psi(a) = b [/itex]
    [itex] \psi(b) = a [/itex]
    [itex]\psi(c) = c [/itex]

    You can work some examples to check that [itex]\psi [/itex] satisfies [itex]\psi((x)(y)) = \psi(x)\psi(y) [/itex]

    Isomorphisms of a group to itself are studied so often in group theory that a special name is given to them. They are called autmorphisms .

    You might find it disgusting that a group can have automorphisms that aren't the identity map. After all, if the definition of being isomorphic really captured the idea of "being essentially the same", shouldn't the only way to match-up "corresponding" elements of [itex]G[/itex] be to match them to themselves?

    In defense of the definition , it is only intended to express the idea that the results in two multiplication tables are essentially the same. For example, pretend the group [itex]G [/itex] in the example is "really" a set of 4 specific mathematical structures (like matrices or integers). When we assign symbols [itex]\{i,a,b,c\}[/itex] to represent these structures, we make arbitrary choices about assigning the names.

    Notice the function [itex] \psi [/itex] swaps [itex] a [/itex] and [itex] b [/itex] ( i.e. [itex] \psi(a) = b [/itex] and [itex] \psi(b) = a [/itex] ). One way to interpret [itex]\psi[/itex] is that is says "The way you assigned the names was OK, but if we swap the names [itex]a[/itex] and [itex]b[/itex] then nobody will be able to tell the difference by looking the multiplication table.

    You can test this by taking a multiplication table for [itex] G [/itex] (the example of a Klein 4-group) and swapping all the [itex] a[/itex]'s and [itex]b[/itex]'s in it (including those in the lists of elements, which are topmost row and the leftmost column). After you've done that, compute some products like [itex] (a)(c) [/itex]. You'll look it up at a different place in second table, but you'll get the same answer from both tables.

    Then try a similar experiment except swap the [itex]i[/itex]'s and [itex]a[/itex]'s. That will create a table that gives different answers. The effect in swapping the names [itex]a[/itex] and [itex]i[/itex] can be detected. The two tables don't give the same results.

    In a manner of speaking, the automoprphisms that are not the identity give you an indication of some freedom that you have to vary from your original assignment of names without changing the results given by the original table.

    Besides the relation of being isomorphic, there are other equivalence relations defined for groups. However, most only apply up when special kinds of groups are studied.. These other notions of being "essentially the same" add requirements to the definition of being isomorphic. For example, they might require that [itex] \phi [/itex] be a continuous function or a differential function. That type of requirement isn't meaningful when talking about abstract groups since their elements need not represent points in any sort of space.


    Now that you understand the definition for being isomorphic and the definition of isomorphism, you can anticipate studying topics in math where the instructor will tell you that they mean something different. This is because we have covered the concept of being isomorphic for groups. There are other mathematical structures (like rings, vector spaces, graphs) and they have different definitions for being isomorphic. You can console yourself with the fact that understanding the definition for groups is a significant help to understanding the other definitions.
     
    Last edited: Jun 23, 2013
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