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Cycle of Permutations

Source: 2023 Taiwan Round 1 Mock Exam P4

March 18, 2023
combinatoricsPermutation cyclesTaiwanfunction

Problem Statement

Let kk be a positive integer, and set n=2kn=2^k, N={1,2,,n}N=\{1, 2, \cdots, n\}. For any bijective function f:NNf:N\rightarrow N, if a set ANA\subset N contains an element aAa\in A such that {a,f(a),f(f(a)),}=A\{a, f(a), f(f(a)), \cdots\} = A, then we call AA as a cycle of ff. Prove that: among all bijective functions f:NNf:N\rightarrow N, at least n!2\frac{n!}{2} of them have number of cycles less than or equal to 2k12k-1. Note: A function is bijective if and only if it is injective and surjective; in other words, it is 1-1 and onto.
Proposed by CSJL