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.)