Cardinality of the equivalence class of a given braid word? 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 $b$:
$(1,2,3,1,2,3)$
$(1,2,1,3,2,3)$
$(2,1,2,3,2,3)$
$(2,1,3,2,3,3)$
$(2,3,1,2,3,3)$
$(1,2,1,2,3,2)$
$(2,1,2,2,3,2)$
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.
 A: 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 $S_3$.
