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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 $\mathbb{Z}_n \hookrightarrow Aut(\{1, \ldots, n\})$ 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 $\mathbb{Z}$-set structure on $X$: $\sigma: \mathbb{Z} \to Aut(X)$ (normally considered given as $\sigma(1)$; 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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