Find the smallest natural number n ∈ ℕ such that $S_n$ contains a cyclic subgroup of order 101.
Proof: We seek the smallest n such that Sn contains a permutation of order 101. Permutation can be written in terms of a collection of disjoint cycles. Let σ be permutation of order 101 consisting of q disjoint cycle, and let the length be ki,…,kq. The order of the element will be equal to the least common multiple of the length cycle. By the fundamental theorem of arithmetic, any number can be uniquely decomposed into a product of powers of prime numbers. It is easy to see that 101 =1 • 101, so each of the ki's must have a prime decomposition consisting of 1 & 101. Hence smallest natural number is 1
Is this correct or I am missing? If missing what will be the correct way?
As you note, every permutation can be expressed as a product of disjoint cycles. The order of such a product is given by the least common multiple of the cycle lengths. The smallest $n$ such that $S_n$ contains a permutation of that form is the sum of the cycle lengths.
Since 101 is prime, the only disjoint cycle expression of a permutation of order 101 is a single cycle of length 101. So $n$ = 101.
A more interesting problem is given by replacing 101 with a composite number, such as 12.