I'm currently working through N.D. Gilbert and T. Porter's Knots and Surfaces. In it the idea of a Wirtinger presentation and a Dehn presentation for a group associated with a given knot is introduced. It then says that:

Both presentations are a presentation of the same group.

I'm trying to prove this statement but I'm not sure how exactly I would go about it. Is it enough to show that generators for one both generate and satisfy the relations for the other?

I'm only looking at tame knots, and so we can assume that for a knot $K$ there are a finite number of generators for the Wirtinger presentation, say $n$ many, and from this we know that there are $n+2$ many generators for the Dehn presentation, via Euler's formula (for planar graphs). (Is this much correct?)

Also a Wirtinger presentation gives relations of the form $xyx^{-1}z^{-1}$ or $x^{-1}yxz^{-1}$, whereas a Dehn relation is always of the form $ab^{-1}cd^{-1}$, as well as the one relation of setting one generator equal to the identity.

Am I thinking about it the right way? And are my preliminary assumptions correct? If so, I would really appreciate a nudge in the right direction as I haven't been able to connect the generators of one to the relations of the other.


I have not read this book, so maybe my answer is not appropriate for the approach taken there. But both of these presentations are presentations of the fundamental group of the knot complement and this is probably the best way to see they are presentations of the same group.

  • 2
    $\begingroup$ I realise this is the case, and it is something I'm working towards, however this fact is treated quite a few chapters on and is not the method sought to prove the above, I believe. Thanks for your comment though! $\endgroup$ – WakeUpDonnie Nov 27 '13 at 1:22

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