The symmetric group $S_n$ has presentation $$S_n = \{ s_1,...,s_{n-1} | s_is_{i+1}s_i=s_{i+1}s_is_{i+1}, \text{ } s_i^2=1 , \text{ and } s_is_j = s_js_i \text{ for } |i-j| \geq 2\}$$

If we take away the relation $s_i^2=1$ we get the braid group:

$$B_n = \{ s_1,...,s_{n-1} | s_is_{i+1}s_i=s_{i+1}s_is_{i+1},\ \text{ } s_is_j = s_js_i \text{ for } |i-j| \geq 2\}$$

Thus elements $b \in B_n$ are equivalence classes, with $b \cong c$ if the braid word that represents $b$ differs from the braid word representing $c$ by finite many applications of the relations.

I am trying to understand, given an element $b \in B_n$, what is its cardinality as an equivalence class as an element of $B_n$?

For example, let $b=(1,2,3,1,2,3)$. Then I'm pretty sure, by brute force, I've calculated that there are $7$ different braid words that can represent $b$:








Does anyone know how I could figure this out in a more sophisitcated way? I know word problems such as this have been well studied. Also, is there a systemic way to determine when two braid words are equivalent? Thank you.


The equivalence classes are all countably infinite.

In fact, from the presentations one reads off a surjective homomorphism $B_n \to S_n$ taking each generator of $B_n$ to the same-named generator of $S_n$. Your equivalence classes are simply the cosets of the kernel of this homomorphism. Since $B_n$ is countably infinite and $S_n$ is finite, and since the cosets all have the same cardinality (a fact true of every group homomorphism), it follows that the cosets are all countably infinite.

By the way, I'll add that the notation of your "brute force calculation" does not make much sense to me; I don't know what those sequences of natural numbers in that calculation are supposed to represent, either as elements of $S_n$ or of $B_n$. Elements of $S_n$ or of $B_n$ can, by definition of a generating set, each be written as a finite word in the generators. For example each element of $B_3$ or of $S_3$ can all be written as a word in the letters $s_1,s_2$; a specific example of such a word is $$s_1^2s_2^{-1}s_1^{4}s_2^3 = s_1 s_1 s_2^{-1} s_1 s_1 s_1 s_1 s_2 s_2 s_2 $$ Given two such words, to say they represent the same element of $B_3$ means, as you say, that you can get from one word to the other by applying the relations of $B_3$. On top of that, to say those two words represent the same element of $S_3$ means that you can get from one to the other by applying the larger set of relations of $S_3$.

  • 2
    $\begingroup$ It's still true that every coset contains infinitely many positive braids: for example you can stick $s_1^2$ into the middle of any word without changing the element of $S_n$, but definitely changing the element of $B_n$. $\endgroup$
    – Lee Mosher
    Jan 16 at 17:42
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    $\begingroup$ That's exactly my point: you cannot go from $s_2s_3$ to $s_2s_1^2s_3$ using the braid relations; but you can go from one to the other using the symmetric group relations. $\endgroup$
    – Lee Mosher
    Jan 16 at 18:02
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    $\begingroup$ So $s_2 s_3$ and $s_2 s_1^2 s_3$ are inequivalent element of the braid group, which nonetheless represent the same element of the symmetric group. $\endgroup$
    – Lee Mosher
    Jan 16 at 18:03
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    $\begingroup$ Let me see if I understand. First, the symmetric group paragraph is not relevant to your mathematical question, it's just there as a means of introducing the presentation for $B_n$. Second, you did not intend to talk about the braid group $B_n$, only the positive braid semigroup. Thus your question is: How many positive braid words are there are that can represent the same element of the positive braid semigroup? If that's correct then you should probably unaccept my answer (but also edit your post). $\endgroup$
    – Lee Mosher
    Jan 16 at 18:08
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    $\begingroup$ Haha, actually, you gave such a great answer to exactly what I asked in the OP. I'll ask the question that I intended to ask in another post later today. $\endgroup$
    – Nobel Cat
    Jan 16 at 18:10

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