Suppose we have two "things": a and b. We don't what they are, but we have labels for them. Like any other algebraic variables, we can multiply them by putting them next to each other: aa, ab, ba, or bb. Since we don't know what these things are, we can't assume that their multiplication is commutative; i.e., ab does not necessarily equal ba. So, aa, ab, ba, and bb are four new unknown-but-labelled "things" that are generated by multiplying our original things.
We can also define multiplicative inverses, /a and /b, so we can do division. I general, multiplicative inverses aren't guaranteed to exist, but in this case, we can just assert that they do. Thus,
a/a = /aa = b/b = /bb =1
So, we can now generate additional new things named a/b, /ba, b/a, /ab, /a/a, and /b/b.
Additionally, since we have defined a thing times its inverse to equal 1, given a string of things like a/bba, we can simplify it to a1a = aa. Or, for something a little more complicated, we can simplify something like ba/bb/ab to aa1/ab = aa/ab = a1b = ab. Note, however, that because multiplication in general is not guaranteed to be commutative, we can't simplify a/bab or baa/b--the a's get in the way between the b and its inverse.
So, we can simplify things, but not very interesting things--we can only insert and remove sequences where items and their inverses are right next to each other. So let's add a third thing: c, and it's inverse /c. And, we'll assert some special properties of c:
ac = ca
a/c = /ca
c/a = /ac
/a/c = /c/a
In other words, c and its inverse multiply commutatively with a and its inverse. All other combinations are non-commutative.
Now, we can simplify things like a/bc/cba to a/b1ba = a1a = aa, as before; but we can also do things like baca/c/a/a. Note that nothing is adjacent to its inverse, but because of the commutativity rules, we can re-arrange things to put them next to their inverses: baca/c/a/a = baca/a/c/a = bcaa/a/a/c = bc/caa/a/a = ... = b.
Now, suppose we have two more complicated expressions that we want to multiply together; e.g., perhaps we want to square acb/a/c. By itself, it can't be simplified, because the b does not multiply commutatively, so it blocks re-arrangements of the a's and c's to allow cancellation. But when we square it, we get acb/a/cacb/a/c = acba/ac/cb/a/c = acbb/a/c.
Now, having introduced these unknown things and their algebraic properties... what's the point? Clearly, these things aren't regular numbers--do they correspond to anything familiar?
Yes!
It turns out that a and b behave exactly like (and thus, mathematically are) the generators of the braid group of order 3. In other words, If you have three parallel pieces of string, a is the name we give to twisting the first and second strings in a particular direction; b is the name we give to twisting the second and third strings in the same direction; /a and /b are the names we give to twisting those same pairs of strings in the opposite direction; and any three-strand braid can be algebraically described by some sequence of a's, b's, and their inverses. 1--the multiplicative identity--represents a section of straight, untwisted string. a/a = 1 because twisting a pair of strings and then untwisting it results in... untwisted, straight string. Same with /aa, b/b, and /bb. A typical braided hair ponytail is just a sequence a/ba/b, repeated over and over... or b/ab/a, to get the chevrons pointed in the other direction. Multiplying a's and b's together is not commutative because a twist in the first and second strands shares a strand with a twist in the second and third strands, so you can't slide them past each other to change their order.
Adding c gives you the three generators of the braid group of order 4--that is, a way to describe braids with 4 strings. So, why does c commute with a? Because a twist in strings one and two doesn't share any strands with a twist in strings three and four! So, you can slide them back and forth along their respective pairs of strings to change their linear order.
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