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.
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