Sophie, it might be a combination of factors. The most important one is that we're requiring \\(f^{(k)} = \mathrm{id}\\) for some \\(k\\), which can only happen if \\(f\\) is a bijection. The fact that \\(f\\) is a loop on a single object is what allows us to iterate \\(f\\) like this, but I wouldn't say that's the most important part -- just a necessary detail.

Once you're working with bijections on finite sets, you quickly fall into the usual patterns of group theory.

(I _just_ got a new copy of Dummit & Foote in the mail. Perhaps this is my cue to crack it open and refresh myself on group theory.)

Once you're working with bijections on finite sets, you quickly fall into the usual patterns of group theory.

(I _just_ got a new copy of Dummit & Foote in the mail. Perhaps this is my cue to crack it open and refresh myself on group theory.)