How would you go about proving the $(p, q, r)$ -pretzel knot is equivalent to the $(p, r, q)$ -pretzel knot?

By "equivalent" I mean you can change one knot into the other by elementary deformations.

I've found this question/example in several books and papers on knot theory where they state the proof as obvious.

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    $\begingroup$ Have you tried looking at a sequence of Reidemeister moves to get the $(p,r,q)$ pretzel knot from the $(p,q,r)$ pretzel knot? In particular you are trying to find an ambient isotopy. I think one can just use the definition of elementary deformation (e.g. look at the top of the second tangle and choose a point $P$ not collinear and try to deform it to the same section in the third tangle, etc...) So we keep moving down the second tangle. $\endgroup$ – PrimeNumber Feb 17 '11 at 3:37

You can change the $(p,q,r)$ pretzel link into the $(r,q,p)$ pretzel link by a 180 degree rotation around the midpoint of the central twist box (the one containing $q$ twists).

There is also a way to change the $(p,q,r)$ link into the $(q,r,p)$ link. This is a bit three-dimensional, so takes a bit more work to see. Draw the diagram on a sphere $S$ in $R^3$ instead of on a plane in $R^3$. Now you can drag the twist box containing $r$ twists around the back of $S$ ("past infinity") until it becomes the first twist box. (The arcs connecting the twist boxes rearrange themselves in the most convenient way!)

Composing these two operations correctly will move you between the two diagrams. To get a sequence of Reidemeister moves, project all of these movements to a fixed plane.


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