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Do rotation permutations form a group, as it seems to suggest here?
Re rotation permutation: yes, they form a group because it’s just transporting the structure of the group ℤn↪Aut({1,…,n}) across the given bijection i on which the definition depends. But the language “rotation permutation of X” could easily invite confusion since it suppresses mention of this i which is actually crucial. Clearly rotations relative to different i’s do not compose.
At first the article had the title cyclic permutation which could be defined as a transitive ℤ-set structure on X: σ:ℤ→Aut(X) (normally considered given as σ(1); also this definition allows the case where X to be countably infinite, although usually one would constrain to X finite). That would be the notion of rotation permutation but untethered to a particular i.
Yes, the cyclic group.
[edit: ah, overlapped with Todd]
Isn’t the wording misleading at rotation permutation?
A rotation permutation is, roughly speaking, a permutation in which, if we view the elements of a finite set as people standing in a circle, everybody shifts one step to the right, or everybody shifts one step to the left.
This sounds to me like a generator.
Thanks David, fixed now.
Also attempted to improve the rotation permutations example according to Todd’s remarks (thanks!).
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