Find the smallest \(n\) for which the symmetric group \(S_n\) contains a cyclic subgroup of order \(111\).
Solution
Let the partition \(n = n _ { 1 } + n _ { 2 } + \ldots + n _ { k }\) represent the cycle structure of an element \(g \in S _ { n }\) , i.e. g is a product of disjoint cycles of the lengths \(n _ { 1 } \le n _ { 2 } \le \dots \le n _ { k }\) . The order of the cyclic subgroup generated by \(g\) is equal to the least common multiple of \(n _ { 1 } , . . . , n _ { k }\) . We want this least common multiple to be \(1 1 1 = 3 \cdot 3 7\) One of the possibilities is \(( n _ { 1 } , n _ { 2 } , . . . , n _ { k } ) = ( 3 , 3 7 )\) in which case \(n = 3 + 3 7 = 4 0\) . We claim that this value of n is the minimal possible. Indeed, if 111 is the least common multiple of \(n _ { 1 } , . . . , n _ { k }\) then each of the prime factors 3, 37 divides at least one of the numbers \(n _ { i }\) and moreover, the sum of such factors dividing \(n _ { i }\) does not exceed their product and thus does not exceed \(n _ { i }\) . This implies \(n = n _ { 1 } + . . . + n _ { k } \geq 3 + 3 7 = 4 0\) .