Haven't you people been paying attention? Cauchy doesn't make errors.

Random832 wrote:Now, the question remains - is there a way to do it while having everyone move the same way?

Basically you're asking, is there some permutation [imath]\pi \in S_{2k}[/imath] so that some 2k-1 powers of [imath]\pi[/imath], acting on the set of people, represent an appropriate set of seating arrangements (yes: see Cauchy's),

and so that all 2k people are in the same cycle?*

Well, no. If they're all in the same cycle, then you can straighten it out so they're moving around a circle. You could represent such a scheme as k pairs of chords between 2k evenly spaced points on a circle, which you then rotate a bunch of times to get the different seating arrangements. At least one of these chords must be a diameter, or else people across from each other never date. But there are only k distinct diameters spanning these 2k points, so you can't have 2k-1 different rotations of this diagram.

Unless, of course, 2k-1=k. That's the trivial case where two people date each other, and nobody moves at all.

*EDIT: The notation of this paragraph is obviously begging for a slick algebraic proof, but the method I was trying (okay, so pi has order 2k, and 2k-1 of its powers comprise the complete set....) kept getting bogged down in special cases, which is why I gave up and appealed to geometry. Anyone got an elegant algebraic solution?