/sci/ - Science & Math

>>11621810
Hi, I was just thinking about you(r progress).
(30) If you have any finite set S and a function f from S to S, then f(S) is contained in S, yes. Now, if f is injective, then the number of elements in f(S) is the same as the number of elements in S, and so f is onto. If it is not injective, then it is not onto either. Since the problem assumes the function f is 1-1, it is bijective by 29. This result is related to the automorphism groups, where (for a finite set) the group itself is finite, and so you can compose an automorphism with itself a few times to get the identity function. Do you want a hint for this?