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.)