From this document exercise 2.13 states:
Show that the trefoil knot group is isomorphic to the group $\langle a,b \space | \space a^3 = b^2 \rangle$.
From Fact 2.9 (and also the fact that Example 2.12 mentions this) it can be shown that the trefoil group is equivalent to $B_3 = \langle x_1, x_2 \space | \space x_2x_1x_2 = x_1x_2x_1 \rangle$
I would guess then to show there is an isomorphism we need to find $\phi((x_1, x_2))$ such that $\phi : T \rightarrow \pi$, (where $T$ is the Trefoil Group), where $\phi$ is bijective and a homomorphism.
I tried to see if there was any sort of connection in the relation:
$$x_1x_2x_1 = x_2x_1x_2$$ $$x_1x_2x_1x_2 = x_2x_1x_2x_2$$
So if you let $a = x_1x_2$ then $a^2 = x_1x_2x_1x_2$ but no relationship to the notion of a $b^3$...
Not much success. It really isn't obvious to me why there should be an isomorphism between the two structures.
I am pretty lost and don't know exactly where to start. Any hints or ideas would be appreciated.
If $a = x_1 x_2$, then
$$a^3 = x_1 x_2 x_1 x_2 x_1 x_2 = x_1 x_1 x_2 x_1 x_1 x_2 = (x_1^2 x_2)^2.$$
So this suggests letting $b = x_1^2 x_2$. Can you finish from here?
One way you could've figured out that you had to do something like this is that the abelianization of both of these groups is isomorphic to $\mathbb{Z}$: in the case of $B_3$, the abelianization sends both generators to $1$, and in the case of the other group, it sends $a$ to $2$ and $b$ to $3$.