Wirtinger Presentation for the Figure eight knot, Rolfsen exercise

Question

I have been working through Rolfsen's "Knots and Links" and have found myself frustrated by exercise 4 on page 58. It concerns the Wirtinger Presentation of the figure eight knot, where the (unsimplified) knot group can be found with the four generators $x_1,\ldots,x_4$ and relations: $$ x_1 x_3 = x_3 x_2 $$ $$ x_4 x_2 = x_3 x_4 $$ $$ x_3 x_1 = x_1 x_4 $$ $$ x_2 x_4 = x_1 x_2. $$ The exercise says "Show combinatorially that the fourth relation is a consequence of the other three." Originally, I assumed this was just a simple algebraic manipulation, but after a while of no success, I am assuming that I am going about this the wrong way. If I just am missing the simple algebra solution, I would be satisfied and move on. Otherwise, what kind of "combinatorial" argue is intended here? I know that you are always suppose to be able to eliminate one of your relations, but cannot see how in this case. Thank you in advance for any response.

Answer 1

We do algebraic manipulation on the first three equations to derive the fourth one. equation (1) is equivalent to the following $x_{1} = x_{3}x_{2}x_{3}^{-1}$.\ equation (2) is equivalent to $x_{4}x_{2}x_{4}^{-1} = x_{3}$\

Now put eqn(1) into eqn(3) to get: \begin{align*} & x_{4}x_{2}x_{2}x_{4}^{-1}x_{2}x_{4}x_{2}^{-1}x_{4}^{-1} = x_{4}x_{2}x_{4}^{-1}x_{2}x_{4}x_{2}^{-1}\\ \Longleftrightarrow\;\;\;\; & x_{2}x_{4}^{-1}x_{2}x_{4}x_{2}^{-1}x_{4}^{-1} = x_{4}^{-1}x_{2}x_{4}x_{2}^{-1}\\ \Longleftrightarrow\;\;\;\; & x_{4}x_{2}x_{4}^{-1}x_{2}x_{4}x_{2}^{-1}x_{4}^{-1} = x_{2}x_{4}x_{2}^{-1}\\ \end{align*} But, $x_{4}x_{2}x_{4}^{-1} = x_{3}, x_{4}x_{2}^{-1}x_{4}^{-1} = x_{3}^{-1}$, so we have

\begin{align*} & x_{3}x_{2}x_{3}^{-1} = x_{2}x_{4}x_{2}^{-1}\\ \Longleftrightarrow \;\;\;& x_{1} = x_{2}x_{4}x_{2}^{-1}\\ \Longleftrightarrow \;\;\;& x_{1}x_{2} = x_{2}x_{4} \end{align*}