Question

Which one of the following is TRUE for the symmetric group S₁₃ ?

• A. S₁₃ has an element of order 42
• B. S₁₃ has an element of order 35
• C. S₁₃ has an element of order 27
• D. S₁₃ has an element of order 60

Answer 1

The Correct Option is A

The symmetric group \(S_{13}\) is the group of all permutations of 13 elements. The order of a permutation is the least common multiple (LCM) of the lengths of the cycles in its cycle decomposition.

Let's evaluate each given option to determine which one is true: 

1. To find an element of order 42:
   • 42 can be obtained as the LCM of 6 and 7.
   • A possible permutation in \(S_{13}\) can consist of a 6-cycle and a 7-cycle because 6 + 7 = 13, which uses all elements.
   • This combination can exist, giving us a permutation of order 42.
2. To find an element of order 35:
   • 35 can be obtained as the LCM of 5 and 7.
   • A possible permutation would need a 5-cycle and a 7-cycle, the sum of which gives 12, and thus uses only 12 elements, not all 13.
   • This combination does not exist in \(S_{13}\), so no such element exists.
3. To find an element of order 27:
   • 27 is a power of 3 (3³).
   • Any cycle whose length is a power of 3 needs to be smaller than the total number of elements, but use all 13 elements within the restrictions of \(S_{13}\). This is not possible.
   • No permutation of order 27 can be formed in \(S_{13}\).
4. To find an element of order 60:
   • 60 can be factored as a combination of 3ys, 4s, and 5s (e.g., LCM of 3, 4, and 5).
   • The sum of these cycle lengths can be any of their configuration permutations which do not sum up to 13. This proves infeasible within \(S_{13}\).
   • No suitable cycles can form a single permutation of order 60.

Since finding permutations of such cycle lengths that sum up to n are needed, the correct choice for existing permutation is only for option 1.

Therefore, the correct answer is S₁₃ has an element of order 42.