What's a proof that a set of disjoint cycles is a bijection? Consider a function $f : D \to D$ (where $D$ is a finite set) so that for every $d \in D$, there is an integer $n$ so that $f(f(...(n\text{ times})...f(d)...) = d$.


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*Prove that $f$ is a bijection.

*Prove that if $f : D \to D$ (where $D$ is a finite set) is a bijection, then for every $d \in D$, there is an integer $n$ so that $f(f(...(n\text{ times})...f(d)...) = d$.


I know this proposition is true (because a bijection over a finite set is just a permutation which is always representable using cycles), I'm just having a hard time formulating a proof for it that a first-year algebra student could understand.

For (1), it's easier to prove that if $f$ is not injective then neither is $f^{n}$ and if $f$ is not surjective then neither is $f^n$, for any $n \ge 1$. You know that $f^n$ is bijective, and hence so must $f$ be.
For (2), note that the bijections on a set form a group under composition, and that if $D$ is finite then this group is finite, so all its elements have finite order.
