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Dog leashes, braids, and groups
When we got our greyhounds, we were told to hold onto the leash using a particular method: lark's-head the leash loop around your wrist and wrap the leash around your fingers. This is quite secure: even if a dog lunges after a squirrel or something and pulls the leash out of your hand, the lark's-head will tighten around your wrist and you won't drop the leash. Of course, it also means that the leash, which is basically a long flat ribbon, gets twisted; if you're like me and your sense of the rightness of the universe requires untwisting the leash, it sometimes requires undoing this whole assembly, untwisting, and redoing the lark's-head and all that.
Of course, I have two dogs, whom I frequently walk together. Which means not only can each leash get twisted, but they can get twisted around one another, and you can't untwist either of them individually until you've undone the twist-around-one-another.
This sounds like a job for... Math! (Into a nearby phone booth, a quick clothes change, and out comes Algebraman!)
I'm sure that better mathematicians than myself have analyzed cord-twisting and braiding from a group-theory perspective, but (a) I want to try developing it from first principles myself, and (b) most of those mathematicians don't know about fingerloop braiding, so I might have something to add.
First, I should give a roadmap to what I'm doing and why. Much of what mathematicians do is "find a concrete phenomenon, erase as many details as possible, and prove things about the resulting abstract concept." If you can prove something about the abstract concept, then since the original concrete phenomenon was an example of the abstract concept, whatever you proved must be true of it. Furthermore, by erasing the details, you may find lots of other concrete phenomena, apparently completely unrelated, that turn out to be examples of the same abstract concept, so the same must be true of them too. So today's voyage is inspired by the concrete phenomena of twisting dog leashes and fingerloop braids, but it'll have things to say about lots of other concrete phenomena like piles of paper in my office and infinite checkerboards.
For the non-mathematicians in the audience: a "semigroup" is a set of things and an operation (which we'll call *) that combines any two of those things in a particular order, producing a third, with the associative property that (a*b)*c = a*(b*c) -- that is, if you combine a and b, and then combine the result with c, you get the same thing as if you combined a with the result of combining b and c. For example, addition and multiplication on ordinary numbers are both associative operations, so if you think of a, b, c as (let's say) integers, * could be the operation of addition, or it could be the operation of multiplication. Or, for that matter, * could be the "max" operation that gives the larger of its two operands. But not all operations are associative, e.g. subtraction: (a-b)-c ≠ a-(b-c). Or exponentiation: (ab)c ≠ a(bc).
Of course, the objects don't have to be numbers: they could be, for example, piles of papers, with the operation being "put this pile on top of that pile". Note that the operation is associative: putting pile a on top of pile b, and then putting that on top of pile c, produces the same result as putting pile a on top of the result of putting pile b on top of pile c. Or pieces of colored glass, where the operation is "put this one in front of that one": putting a red piece of glass in front of a blue piece of glass produces, for looking-through purposes, a purple piece of glass.
A semigroup may or may not be commutative, which means a*b = b*a -- that is, if you combine a with b (remember I said the operation combines them in a particular order), you get the same thing as if you combine b with a. Again, one can imagine operations that aren't commutative, such as subtraction on integers. And again, we're not restricted to numbers: the "put one pile of papers on top of another" operation is clearly not commutative (the papers end up in a different order), while "put one piece of colored glass in front of another" is, at least as far as light passing through them all is concerned.
Of course, if you combine an object with itself, that's always commutative -- a*a = a*a, and I bet you can't even tell that I wrote the second pair of a's in the opposite order from the first pair. Indeed, in any semigroup, any sequence of a single object combined together can be completely described by how many there are: associativity and commutativity mean it doesn't matter how they're grouped.
And a semigroup may or may not have an identity element, an element that can be combined with anything else without affecting the other thing. By convention, an identity element is written with the letter e, so this property amounts to e*a=a.
(Actually, that makes e a "left identity"; a "right identity" would have the property a*e=a. It turns out that if they both exist, they have to be the same, so there can be only one of them. Here's a proof: suppose e is a left identity, and e' is a right identity. Then what is e * e'? Because e is a left identity,e*e' must be e', but because e' is a right identity, e*e' must be e. Since this expression is equal to both e and e', those two must have been the same all along. That was your first example of proving a fact about an abstract concept that automatically applies to all the gazillions of concrete examples of it.)
For example, if we're working with integers and addition, 0 is an identity; if we're working with integers and multiplication, 1 is an identity; if we're piling papers, an empty pile of papers is an identity; and if we're working with pieces of colored glass, a perfectly clear piece of glass is an identity. A semigroup that does have an identity element is called a monoid.
A favorite kind of monoid among mathematicians is a set of objects that are themselves "actions", where the operation is "do this, then do that", which mathematicians call "composition". The identity element is "do nothing". The operation is often not commutative: my algebra teacher's favorite example was the difference between "open the window and put your head out" and "put your head out and open the window". In the world of Rubik's Cube, the action "twist the yellow face 90° clockwise" may not commute with "twist the blue face 90° clockwise" (if those faces are adjacent). Or (back to the dog leashes) "twist this leash clockwise and then twist the two leashes around one another clockwise" might be different from "twist the two leashes around one another clockwise and then twist this leash clockwise".
Once you've got a monoid, a natural question is whether, once you've left the identity element, you can get back to it: given a ≠ e, is there a b such that a * b = e? Such a thing is called an inverse of a. (Technically, that's a "right-inverse", but it turns out that if something has a right-inverse, it must also have a left-inverse, and they're the same. Here's a proof:
suppose a has right-inverse b, i.e. a*b=e. Then what is b*a*b? Since a*b=e, we get b*(a*b)=b*e=b. But that means (b*a)*b = b, which means b*a is acting like an identity, and since there can only be one identity, that means b*a=e, i.e. b is not only a's right-inverse but also its left-inverse.) It also turns out that the identity is always its own inverse; the proof is easy and left as an exercise for the reader.
The inverse of a is conventionally written a-1 (which does not mean anything about exponents; it's just a way of writing "the inverse of a"). If every element of a monoid has an inverse, then the monoid is called a group. It turns out that this bare-bones combination of properties -- a set with an associative operation, containing an identity element, and containing an inverse for every element -- is remarkably powerful, and leads to a whole field of mathematics called Group Theory which used to be considered utterly abstract and useless until people started developing public-key computer cryptosystems.
Anyway, some examples of groups. For addition of integers (what mathematicians call (Z,+)), 0 is the identity, and negatives provide the inverses, since a*(-a)=0 (remember that * means addition here), so the integers under addition form a group. For multiplication of integers (or (Z,*)), 1 is the identity but most things have no inverses: there's no integer you can multiply by 2 to get 1, so the integers under multiplication form a monoid but not a group. On the other hand, multiplication on positive rationals does have an inverse for every element, so the positive rationals under multiplication (aka (Q+,*)) do form a group.
Twisting cords, dog leashes, etc. form a classic example of a group. The objects are twisting actions, the operation is composition, i.e. "do this, then do that", the identity element is "don't twist anything", and any twisting action can be inverted by twisting the same amount in the opposite direction. Furthermore, any sequence of twists can be inverted by inverting each of them in reverse order: (a*b*c*d*e)-1 = e-1*d-1*c-1*b-1*a-1 (this is true in all groups, not just this example).
So let's consider some successively-more-interesting cord-twisting scenarios.
The simplest is a single featureless cord. Since it's featureless, you can't tell whether it's twisted, so we have a group with only an identity element. There's not much to say about it; it's called "the trivial group".
What about a cord with a stripe running along its length, or a flat-ribbon-shaped dog leash with one side light and the other side dark? Suppose we allow the action a="rotate the whole cord clockwise along its axis 180°". Obviously the inverse, a-1, is "rotate the whole cord counter-clockwise along its axis 180°". But what if we rotated the whole card clockwise 180°, and then another 180°? The cord or leash would be back to exactly where it started, as though nothing had happened to it, so a*a=e and a is its own inverse. The group has exactly two elements -- either the cord is flipped, or it's not -- and it behaves exactly like the integers under addition, if you treat all even numbers as equal and all odd numbers as equal (exercise: convince yourself that this makes sense), so mathematicians have given it the name Z2, pronounced "the integers mod 2". Likewise if we had started with an operation "rotate the whole cord clockwise along its axis 120°", we would have a group of 3 elements, Z3, in which a*a=a-1; 90° would give us Z4, in which a*a*a=a-1, and so on.
But what if, rather than rotating the whole cord, we left the far end fixed and twisted the near end 180°? Doing this twice brings the stripe back to the top at the near end, but the cord is still visibly twisted, as you can see the stripe angling around it. No (positive) number of repetitions of this operation will ever bring you back to the untwisted state; the only way to get back is to use the inverse operation, twisting counter-clockwise. Also note that any conceivable twist can be reached eventually by some number of repetitions of the basic action (this will be important in a moment).
This group behaves exactly like the integers under addition, which mathematicians call (Z,+).
The same can be said if you have two featureless cords with the far ends fixed and you allow the operation "cross the left cord over the right cord". Again, doing this twice puts the same cord on the left as before, but there's a visible twist, and no number of repetitions will untwist it unless you invert the operation and cross the right cord over the left cord. This scenario too is isomorphic to the group (Z,+).
What if you have two striped cords, or two ribbon-shaped dog leashes, one in each hand, with the far ends fixed? You can twist either one by 180°, and it doesn't affect the other one. So let action a be "twist the left cord clockwise 180°" and action b be "twist the right cord clockwise 180°. Note that if you twist the left cord and then the right cord, the result is the same as if you twist the right cord and then the left cord; in other words, a*b = b*a, "a and b commute". Since a already commutes with itself, and so does b, that means everything in the group commutes, just as in the integers (Z,+). But this isn't quite the same as (Z,+). Remember that for the integers, there's a single element 1 which, if you repeat it enough times forward or backwards, will get you to any desired element of the set (such an element is called a generator), while this group has no single element that generates everything; you need two distinct elements a and b, which are called a generating set for the group. (Intuitively, this group is 2-dimensional where Z is 1-dimensional.) This group is called the direct product of (Z,+) with itself, written (Z,+) x (Z,+). Another way of looking at it is a checkerboard, infinite to left, right, forward, and backward, in which moving right and then forward gets you to the same place as moving forward and then right.
Now let's suppose you have three featureless cords with the far ends fixed, and you allow the operations a="cross the left cord over the middle cord" and b="cross the right cord over the middle cord". Either operation alone would give us something equivalent to (Z,+), but the two operations do not commute: a*b produces a visibly different braid (indeed, a mirror image) from b*a. The classic three-cord braid we all learned at summer camp is the sequence ababababab.... and no positive number of repetitions of this will ever untwist the braid. Like (Z,+)x(Z,+), this group has a two-element generating set, but the two elements don't commute with one another. This is called the free group on 2 generators, F{a,b} or F2. You can think of its elements as finite-length "words" of a's and b's: if two words have different lengths, or if there's even one position where they disagree, they're different words.
What about the situation that originally inspired this post: two ribbon-shaped dog leashes that can be twisted around one another? We can think of this in terms of three generators: a="twist the left leash 180° clockwise", b="twist the right leash 180° clockwise", and c="cross the left leash over the right leash". At this point we have to consider whether the physical scenario we're modeling includes friction between the leashes. If so, and crossing one leash over another "blocks" a twist in one leash from sliding across the crossover, then a*c can't be rewritten as anything else (although a and b still commute with one another). I don't think this has a special name, but mathematicians would call it the free group on 3 generators modulo the equation a*b=b*a.
If not, then twisting a leash and then twisting it around another leash allows the former twist to "slide" up and down its length, except that the one that used to be "left" is now "right", so a*c=c*b and b*c=c*a. And of course a and b still commute with one another, since they're independent leashes. This could be called the free group on 3 generators modulo the equations a*b=b*a, a*c=c*b, and b*c=c*a.
And what about fingerloop braiding? We have some number of loops or "bowes", each with its far end fixed, and we have several operations we can perform: twist a single bow 180° clockwise (and its inverse), and take one bow through another. The medieval instruction books describe "taking a bow reversed" and "taking a bow unreversed", which correspond to whether or not there's a 180° twist coinciding with the action of passing one bow through another. (The instruction books also go into great length about which finger each bow is on, but this doesn't directly affect the braid; it's just a method of bookkeeping to make sure you do the other operations to the correct bows in the correct order.) Let's look at some successively more complex fingerloop braids from a group-theory perspective.
The simplest imaginable fingerloop braid consists of a single bow. Since there is no other bow to go through, the only operation is twisting it 180°, and this scenario is just like the single ribbon-shaped dog leash, isomorphic to (Z,+).
Next suppose we have two bowes, "left" and "right". There are four basic actions: reverse the left bow clockwise, reverse the right bow clockwise, take the left bow through the right bow, and take the right bow through the left bow. Let's name these operations Rl, Rr, Tl, and Tr. Obviously Rl and Rr commute with one another, since they're independent bowes. Note that when you take one bow through another, the one that was on the left is now on the right and vice versa. In particular, if you take the left bow through the right, and then the right bow (which used to be left) through the left (which used to be right), you're back where you started, so Tl*Tr = e = Tr*Tl -- in other words, Tr and Tl are inverses. Which means we don't really need both of them as generators. The instruction books routinely say "take this bowe through that bowe reversed" without specifying whether you should reverse the inner bow before or after taking it through, and in my own experience it really doesn't matter, except that after you've passed one bow through another, the one that used to be "left" is now "right", so
Rl*Tl= Tl*Rr and Rr*Tr=Tr*Rl. But if you're reversing the outer bow (the one through which the other is taken), it matters very much whether you do it before or after the T operation:
Rr*Tl≠Tl*Rl and
Rl*Tr≠Tr*Rr. So we have effectively the free group on 3 generators, mod the equations Rl*Rr=Rr*Rl, Rl*Tl= Tl*Rr, and Rr*Tl-1=Tr-1*Rl. This feels almost like the frictionless two-leash scenario described above, in that we have three generators, two of which commute, and one of the commuting ones turns into the other on crossing the third generator, but crossing in the opposite direction requires the inverse of the third generator. We've got the free group on 3 generators modulo the equations a*b=b*a, a*c=c*b, and b*c-1=c-1*a.
Perhaps calling the two bowes "left" and "right" isn't the clearest way to think about it. Alternatively, we could make one bow yellow and another blue, and define the operations in terms of which colored bow is doing what. There are again four basic operations: reverse the yellow bow (Ry), reverse the blue bow (Rb), take the yellow bow through the blue (Ty), and take the blue bow through the yellow (Tb). Now passing one bow through the other doesn't change its identity (blue is still blue and yellow yellow), so instead of Ty and Tb being one another's inverses, each is its own inverse. Ry and Rb still commute with one another (Ry*Rb=Rb*Ry), and each of them commutes with taking that-colored bow through the other (Ry*Ty=Ty*Ry and Rb*Tb=Tb*Rb). This doesn't have the same generators or equations as the previous paragraph, but it's actually the same group, because it describes the exact same phenomenon, only labeled differently.
That's not as surprising as it may seem. In the well-known example of "clock arithmetic", Z12 is the group of integers under addition, if you equate 12 with 0 (and thus 13 with 1, 14 with 2, etc.) It can be generated by 1: for any of the 12 elements of the group, you can eventually get to it by repeatedly adding 1. It can also be generated by 5: 5 -> 10 -> 15=3 -> 8 -> 13=1 -> 6 -> 11 -> 16=4 -> 9 -> 14=2 -> 7 -> 12=0. (Musicians will recognize this as the "circle of fourths"; going backwards, generating the set by -5=7, you get the "circle of fifths".) And it can also be generated by the two-element generating set {3,4}: neither one alone gets you all the elements, but by combining them you can get all the elements. So yes, it's possible for the same group to have different generating sets of different sizes.
Anyway, what can you do with two bowes? If all you do is reverse them, with no T operations, you get two two-strand twists (one all-blue, one all-yellow) each twisted some integer number of times, with no interaction between them; you're working in (Z,+) x (Z,+). If all you do is T operations (say, Tb*Ty*Tb*Ty...), you're working in a single copy of (Z,+), and the state of the system can be completely described by how many T operations you've done, in which direction. The resulting "braid" looks like two two-strand twists of blue and yellow, one above and one below, twisted the same number of times in opposite directions.
But if you combine T and R operations, things get more interesting. Following an Ry with a Tb causes the blue bow to "lock" the yellow reversal in place and prevents the blue and yellow bowes from getting separated. Following a Tb with an Ry causes the twist in the yellow bow to "lock" the blue threads together and prevent the upper and lower threads from getting separated. So by alternating T and R operations, you can form a single braid of four threads, not two two-strand twists as in the previous paragraph. If you stop alternating for a few passes, and do only T's or only R's, you'll get two separate two-strand twists briefly, rejoining when you return to alternating T's and R's.
Things get much more interesting when you have three or more bowes. But I'd better wrap this up and prepare for bed.
Of course, I have two dogs, whom I frequently walk together. Which means not only can each leash get twisted, but they can get twisted around one another, and you can't untwist either of them individually until you've undone the twist-around-one-another.
This sounds like a job for... Math! (Into a nearby phone booth, a quick clothes change, and out comes Algebraman!)
I'm sure that better mathematicians than myself have analyzed cord-twisting and braiding from a group-theory perspective, but (a) I want to try developing it from first principles myself, and (b) most of those mathematicians don't know about fingerloop braiding, so I might have something to add.
First, I should give a roadmap to what I'm doing and why. Much of what mathematicians do is "find a concrete phenomenon, erase as many details as possible, and prove things about the resulting abstract concept." If you can prove something about the abstract concept, then since the original concrete phenomenon was an example of the abstract concept, whatever you proved must be true of it. Furthermore, by erasing the details, you may find lots of other concrete phenomena, apparently completely unrelated, that turn out to be examples of the same abstract concept, so the same must be true of them too. So today's voyage is inspired by the concrete phenomena of twisting dog leashes and fingerloop braids, but it'll have things to say about lots of other concrete phenomena like piles of paper in my office and infinite checkerboards.
For the non-mathematicians in the audience: a "semigroup" is a set of things and an operation (which we'll call *) that combines any two of those things in a particular order, producing a third, with the associative property that (a*b)*c = a*(b*c) -- that is, if you combine a and b, and then combine the result with c, you get the same thing as if you combined a with the result of combining b and c. For example, addition and multiplication on ordinary numbers are both associative operations, so if you think of a, b, c as (let's say) integers, * could be the operation of addition, or it could be the operation of multiplication. Or, for that matter, * could be the "max" operation that gives the larger of its two operands. But not all operations are associative, e.g. subtraction: (a-b)-c ≠ a-(b-c). Or exponentiation: (ab)c ≠ a(bc).
Of course, the objects don't have to be numbers: they could be, for example, piles of papers, with the operation being "put this pile on top of that pile". Note that the operation is associative: putting pile a on top of pile b, and then putting that on top of pile c, produces the same result as putting pile a on top of the result of putting pile b on top of pile c. Or pieces of colored glass, where the operation is "put this one in front of that one": putting a red piece of glass in front of a blue piece of glass produces, for looking-through purposes, a purple piece of glass.
A semigroup may or may not be commutative, which means a*b = b*a -- that is, if you combine a with b (remember I said the operation combines them in a particular order), you get the same thing as if you combine b with a. Again, one can imagine operations that aren't commutative, such as subtraction on integers. And again, we're not restricted to numbers: the "put one pile of papers on top of another" operation is clearly not commutative (the papers end up in a different order), while "put one piece of colored glass in front of another" is, at least as far as light passing through them all is concerned.
Of course, if you combine an object with itself, that's always commutative -- a*a = a*a, and I bet you can't even tell that I wrote the second pair of a's in the opposite order from the first pair. Indeed, in any semigroup, any sequence of a single object combined together can be completely described by how many there are: associativity and commutativity mean it doesn't matter how they're grouped.
And a semigroup may or may not have an identity element, an element that can be combined with anything else without affecting the other thing. By convention, an identity element is written with the letter e, so this property amounts to e*a=a.
(Actually, that makes e a "left identity"; a "right identity" would have the property a*e=a. It turns out that if they both exist, they have to be the same, so there can be only one of them. Here's a proof: suppose e is a left identity, and e' is a right identity. Then what is e * e'? Because e is a left identity,e*e' must be e', but because e' is a right identity, e*e' must be e. Since this expression is equal to both e and e', those two must have been the same all along. That was your first example of proving a fact about an abstract concept that automatically applies to all the gazillions of concrete examples of it.)
For example, if we're working with integers and addition, 0 is an identity; if we're working with integers and multiplication, 1 is an identity; if we're piling papers, an empty pile of papers is an identity; and if we're working with pieces of colored glass, a perfectly clear piece of glass is an identity. A semigroup that does have an identity element is called a monoid.
A favorite kind of monoid among mathematicians is a set of objects that are themselves "actions", where the operation is "do this, then do that", which mathematicians call "composition". The identity element is "do nothing". The operation is often not commutative: my algebra teacher's favorite example was the difference between "open the window and put your head out" and "put your head out and open the window". In the world of Rubik's Cube, the action "twist the yellow face 90° clockwise" may not commute with "twist the blue face 90° clockwise" (if those faces are adjacent). Or (back to the dog leashes) "twist this leash clockwise and then twist the two leashes around one another clockwise" might be different from "twist the two leashes around one another clockwise and then twist this leash clockwise".
Once you've got a monoid, a natural question is whether, once you've left the identity element, you can get back to it: given a ≠ e, is there a b such that a * b = e? Such a thing is called an inverse of a. (Technically, that's a "right-inverse", but it turns out that if something has a right-inverse, it must also have a left-inverse, and they're the same. Here's a proof:
suppose a has right-inverse b, i.e. a*b=e. Then what is b*a*b? Since a*b=e, we get b*(a*b)=b*e=b. But that means (b*a)*b = b, which means b*a is acting like an identity, and since there can only be one identity, that means b*a=e, i.e. b is not only a's right-inverse but also its left-inverse.) It also turns out that the identity is always its own inverse; the proof is easy and left as an exercise for the reader.
The inverse of a is conventionally written a-1 (which does not mean anything about exponents; it's just a way of writing "the inverse of a"). If every element of a monoid has an inverse, then the monoid is called a group. It turns out that this bare-bones combination of properties -- a set with an associative operation, containing an identity element, and containing an inverse for every element -- is remarkably powerful, and leads to a whole field of mathematics called Group Theory which used to be considered utterly abstract and useless until people started developing public-key computer cryptosystems.
Anyway, some examples of groups. For addition of integers (what mathematicians call (Z,+)), 0 is the identity, and negatives provide the inverses, since a*(-a)=0 (remember that * means addition here), so the integers under addition form a group. For multiplication of integers (or (Z,*)), 1 is the identity but most things have no inverses: there's no integer you can multiply by 2 to get 1, so the integers under multiplication form a monoid but not a group. On the other hand, multiplication on positive rationals does have an inverse for every element, so the positive rationals under multiplication (aka (Q+,*)) do form a group.
Twisting cords, dog leashes, etc. form a classic example of a group. The objects are twisting actions, the operation is composition, i.e. "do this, then do that", the identity element is "don't twist anything", and any twisting action can be inverted by twisting the same amount in the opposite direction. Furthermore, any sequence of twists can be inverted by inverting each of them in reverse order: (a*b*c*d*e)-1 = e-1*d-1*c-1*b-1*a-1 (this is true in all groups, not just this example).
So let's consider some successively-more-interesting cord-twisting scenarios.
The simplest is a single featureless cord. Since it's featureless, you can't tell whether it's twisted, so we have a group with only an identity element. There's not much to say about it; it's called "the trivial group".
What about a cord with a stripe running along its length, or a flat-ribbon-shaped dog leash with one side light and the other side dark? Suppose we allow the action a="rotate the whole cord clockwise along its axis 180°". Obviously the inverse, a-1, is "rotate the whole cord counter-clockwise along its axis 180°". But what if we rotated the whole card clockwise 180°, and then another 180°? The cord or leash would be back to exactly where it started, as though nothing had happened to it, so a*a=e and a is its own inverse. The group has exactly two elements -- either the cord is flipped, or it's not -- and it behaves exactly like the integers under addition, if you treat all even numbers as equal and all odd numbers as equal (exercise: convince yourself that this makes sense), so mathematicians have given it the name Z2, pronounced "the integers mod 2". Likewise if we had started with an operation "rotate the whole cord clockwise along its axis 120°", we would have a group of 3 elements, Z3, in which a*a=a-1; 90° would give us Z4, in which a*a*a=a-1, and so on.
But what if, rather than rotating the whole cord, we left the far end fixed and twisted the near end 180°? Doing this twice brings the stripe back to the top at the near end, but the cord is still visibly twisted, as you can see the stripe angling around it. No (positive) number of repetitions of this operation will ever bring you back to the untwisted state; the only way to get back is to use the inverse operation, twisting counter-clockwise. Also note that any conceivable twist can be reached eventually by some number of repetitions of the basic action (this will be important in a moment).
This group behaves exactly like the integers under addition, which mathematicians call (Z,+).
The same can be said if you have two featureless cords with the far ends fixed and you allow the operation "cross the left cord over the right cord". Again, doing this twice puts the same cord on the left as before, but there's a visible twist, and no number of repetitions will untwist it unless you invert the operation and cross the right cord over the left cord. This scenario too is isomorphic to the group (Z,+).
What if you have two striped cords, or two ribbon-shaped dog leashes, one in each hand, with the far ends fixed? You can twist either one by 180°, and it doesn't affect the other one. So let action a be "twist the left cord clockwise 180°" and action b be "twist the right cord clockwise 180°. Note that if you twist the left cord and then the right cord, the result is the same as if you twist the right cord and then the left cord; in other words, a*b = b*a, "a and b commute". Since a already commutes with itself, and so does b, that means everything in the group commutes, just as in the integers (Z,+). But this isn't quite the same as (Z,+). Remember that for the integers, there's a single element 1 which, if you repeat it enough times forward or backwards, will get you to any desired element of the set (such an element is called a generator), while this group has no single element that generates everything; you need two distinct elements a and b, which are called a generating set for the group. (Intuitively, this group is 2-dimensional where Z is 1-dimensional.) This group is called the direct product of (Z,+) with itself, written (Z,+) x (Z,+). Another way of looking at it is a checkerboard, infinite to left, right, forward, and backward, in which moving right and then forward gets you to the same place as moving forward and then right.
Now let's suppose you have three featureless cords with the far ends fixed, and you allow the operations a="cross the left cord over the middle cord" and b="cross the right cord over the middle cord". Either operation alone would give us something equivalent to (Z,+), but the two operations do not commute: a*b produces a visibly different braid (indeed, a mirror image) from b*a. The classic three-cord braid we all learned at summer camp is the sequence ababababab.... and no positive number of repetitions of this will ever untwist the braid. Like (Z,+)x(Z,+), this group has a two-element generating set, but the two elements don't commute with one another. This is called the free group on 2 generators, F{a,b} or F2. You can think of its elements as finite-length "words" of a's and b's: if two words have different lengths, or if there's even one position where they disagree, they're different words.
What about the situation that originally inspired this post: two ribbon-shaped dog leashes that can be twisted around one another? We can think of this in terms of three generators: a="twist the left leash 180° clockwise", b="twist the right leash 180° clockwise", and c="cross the left leash over the right leash". At this point we have to consider whether the physical scenario we're modeling includes friction between the leashes. If so, and crossing one leash over another "blocks" a twist in one leash from sliding across the crossover, then a*c can't be rewritten as anything else (although a and b still commute with one another). I don't think this has a special name, but mathematicians would call it the free group on 3 generators modulo the equation a*b=b*a.
If not, then twisting a leash and then twisting it around another leash allows the former twist to "slide" up and down its length, except that the one that used to be "left" is now "right", so a*c=c*b and b*c=c*a. And of course a and b still commute with one another, since they're independent leashes. This could be called the free group on 3 generators modulo the equations a*b=b*a, a*c=c*b, and b*c=c*a.
And what about fingerloop braiding? We have some number of loops or "bowes", each with its far end fixed, and we have several operations we can perform: twist a single bow 180° clockwise (and its inverse), and take one bow through another. The medieval instruction books describe "taking a bow reversed" and "taking a bow unreversed", which correspond to whether or not there's a 180° twist coinciding with the action of passing one bow through another. (The instruction books also go into great length about which finger each bow is on, but this doesn't directly affect the braid; it's just a method of bookkeeping to make sure you do the other operations to the correct bows in the correct order.) Let's look at some successively more complex fingerloop braids from a group-theory perspective.
The simplest imaginable fingerloop braid consists of a single bow. Since there is no other bow to go through, the only operation is twisting it 180°, and this scenario is just like the single ribbon-shaped dog leash, isomorphic to (Z,+).
Next suppose we have two bowes, "left" and "right". There are four basic actions: reverse the left bow clockwise, reverse the right bow clockwise, take the left bow through the right bow, and take the right bow through the left bow. Let's name these operations Rl, Rr, Tl, and Tr. Obviously Rl and Rr commute with one another, since they're independent bowes. Note that when you take one bow through another, the one that was on the left is now on the right and vice versa. In particular, if you take the left bow through the right, and then the right bow (which used to be left) through the left (which used to be right), you're back where you started, so Tl*Tr = e = Tr*Tl -- in other words, Tr and Tl are inverses. Which means we don't really need both of them as generators. The instruction books routinely say "take this bowe through that bowe reversed" without specifying whether you should reverse the inner bow before or after taking it through, and in my own experience it really doesn't matter, except that after you've passed one bow through another, the one that used to be "left" is now "right", so
Rl*Tl= Tl*Rr and Rr*Tr=Tr*Rl. But if you're reversing the outer bow (the one through which the other is taken), it matters very much whether you do it before or after the T operation:
Rr*Tl≠Tl*Rl and
Rl*Tr≠Tr*Rr. So we have effectively the free group on 3 generators, mod the equations Rl*Rr=Rr*Rl, Rl*Tl= Tl*Rr, and Rr*Tl-1=Tr-1*Rl. This feels almost like the frictionless two-leash scenario described above, in that we have three generators, two of which commute, and one of the commuting ones turns into the other on crossing the third generator, but crossing in the opposite direction requires the inverse of the third generator. We've got the free group on 3 generators modulo the equations a*b=b*a, a*c=c*b, and b*c-1=c-1*a.
Perhaps calling the two bowes "left" and "right" isn't the clearest way to think about it. Alternatively, we could make one bow yellow and another blue, and define the operations in terms of which colored bow is doing what. There are again four basic operations: reverse the yellow bow (Ry), reverse the blue bow (Rb), take the yellow bow through the blue (Ty), and take the blue bow through the yellow (Tb). Now passing one bow through the other doesn't change its identity (blue is still blue and yellow yellow), so instead of Ty and Tb being one another's inverses, each is its own inverse. Ry and Rb still commute with one another (Ry*Rb=Rb*Ry), and each of them commutes with taking that-colored bow through the other (Ry*Ty=Ty*Ry and Rb*Tb=Tb*Rb). This doesn't have the same generators or equations as the previous paragraph, but it's actually the same group, because it describes the exact same phenomenon, only labeled differently.
That's not as surprising as it may seem. In the well-known example of "clock arithmetic", Z12 is the group of integers under addition, if you equate 12 with 0 (and thus 13 with 1, 14 with 2, etc.) It can be generated by 1: for any of the 12 elements of the group, you can eventually get to it by repeatedly adding 1. It can also be generated by 5: 5 -> 10 -> 15=3 -> 8 -> 13=1 -> 6 -> 11 -> 16=4 -> 9 -> 14=2 -> 7 -> 12=0. (Musicians will recognize this as the "circle of fourths"; going backwards, generating the set by -5=7, you get the "circle of fifths".) And it can also be generated by the two-element generating set {3,4}: neither one alone gets you all the elements, but by combining them you can get all the elements. So yes, it's possible for the same group to have different generating sets of different sizes.
Anyway, what can you do with two bowes? If all you do is reverse them, with no T operations, you get two two-strand twists (one all-blue, one all-yellow) each twisted some integer number of times, with no interaction between them; you're working in (Z,+) x (Z,+). If all you do is T operations (say, Tb*Ty*Tb*Ty...), you're working in a single copy of (Z,+), and the state of the system can be completely described by how many T operations you've done, in which direction. The resulting "braid" looks like two two-strand twists of blue and yellow, one above and one below, twisted the same number of times in opposite directions.
But if you combine T and R operations, things get more interesting. Following an Ry with a Tb causes the blue bow to "lock" the yellow reversal in place and prevents the blue and yellow bowes from getting separated. Following a Tb with an Ry causes the twist in the yellow bow to "lock" the blue threads together and prevent the upper and lower threads from getting separated. So by alternating T and R operations, you can form a single braid of four threads, not two two-strand twists as in the previous paragraph. If you stop alternating for a few passes, and do only T's or only R's, you'll get two separate two-strand twists briefly, rejoining when you return to alternating T's and R's.
Things get much more interesting when you have three or more bowes. But I'd better wrap this up and prepare for bed.