Word Presentation of Fundamental Group of Trefoil Knot Complement I'm trying to understand Slide 35 of Joan Birman's presentation on Lorenz knots, available here: https://www.math.columbia.edu/~jb/Lorenz-general-audience.pdf
I'm struggling with this particular bit. Let $G=\pi_1(S^3\smallsetminus \mathcal{T}) = \langle U,V: U^2=V^3\rangle$ be the fundamental group of the trefoil knot complement. She says that every free homotopy class of $G$ is represented by the cyclic word in $U$ and $V$ of the form $W=C^kUV^{\epsilon_1}....UV^{\epsilon_r}$ where $\epsilon_i=\pm1$ and $C=U^2=V^3$ generates the center of $G$. 
My questions:
1) I don't see where she gets this particular presentation of the free homotopy class from. My guess is that this might have something to do with the fact that the fundamental group of the trefoil knot complement is the same as the braid group on 3 strands, $B_3$, and braid groups have a solvable word problem. So maybe that's where the presentation comes from...? I can't find the explicit justification for it though.
2) Just a clarification: isn't the trefoil knot complement a connected space? If so, isn't every element of the fundamental group a free homotopy class since all fundamental groups of a connected space is isomorphic, regardless of the base-point? If so,then Birman's presentation is true for all elements of the fundamental group and not just a special class of them, right?
 A: This is what I have. It's just a sketch, but I think it's right. If it's wrong, feel free to point it out!
Note that the generic element $W\in G$ is of the form $W=U^{a_1}V^{b_1}U^{a_2}V^{b_1}...U^{a_n}V^{b_n}$, where $a_i,b_i\in\mathbb{N}$, for all $1\leq i\leq n$. By using the fact that $C=U^2=V^3$, we can reduce it to the following form:
$W=C^{k}U^{a'_1}V^{b'_1}UV^{\pm1}UV^{\pm1}...U^{a'_m}V^{b'_m}$
where $a'_1,a'_m \in \{0,1\}$, where $b'_1,b'_m \in \{0,1,-1\}$. 
Suppose that none of the following exponents are zero: $a'_1,a'_m,b'_1, b'_m$. Then we are done because the reduced form of $W$ is exactly in the form we want. Now, suppose that $a'_1=0,b'_1=\pm1$ and that $a'_m=1$, $b'_m=\pm1$. Then, $W= C^{k}V^{\pm1}...UV^{\pm1}$ is free homotopic to $(UV^{\pm1})W(UV^{\pm1})^{-1}=C^{k}U(V^{\pm1})^2...UV^{\pm1}(UV^{\pm1})^{-1} = C^{k}UV^{\mp1}...UV^{\pm1}$, 
which is exactly in the form we want. 
The other cases (i.e. when at least one of the exponents $a'_1,a'_m,b'_1, b'_m$ are zero) can be solved through similar methods and exploiting the fact that $C$ commutes with every element $g\in G$.
