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.
migrated from mathoverflow.net Aug 23 '13 at 19:09
This question came from our site for professional mathematicians.