技术咨询：求对称群S₁₁中阶最大的元素α

Hey there! Let's break down how to find the element with the largest order in (S_{11}). First, a quick refresher: the order of any permutation in (S_n) is equal to the least common multiple (LCM) of the lengths of its disjoint cycles. So our goal boils down to finding a partition of 11 into positive integers where the LCM of those integers is as large as possible.

Testing Possible Partitions of 11 #

Let's go through the most promising partitions and calculate their LCMs to find the maximum:

• Single 11-cycle: LCM(11) = 11. Not great, since we can do better.
• 10-cycle + 1-cycle: LCM(10,1) =10. Worse than 11.
• 9-cycle + 2-cycle: LCM(9,2)=18. Better, but still not the peak.
• 8-cycle +3-cycle: LCM(8,3)=24. Getting warmer.
• 7-cycle +4-cycle: LCM(7,4)=28. Nice jump, but we can go higher.
• 6-cycle +5-cycle: LCM(6,5)=30. This is a big improvement.
• 5-cycle +3-cycle +2-cycle +1-cycle: LCM(5,3,2)=30. Same LCM as the 6+5 partition (the 1-cycle doesn't affect the order, since it's just the identity on that element).

What about other partitions? Let's rule them out quickly:

• Partitions like 4+3+2+2: LCM(4,3,2)=12, way lower than 30.
• 3+3+3+2: LCM(3,2)=6, no contest.
• Any partition including 7 and another large number (like 7+3+1) gives LCM(7,3)=21, which is less than 30.

We can't get an LCM higher than 30 because:

• To get a multiple of 7 and 5 (which would give LCM 35), we'd need 7+5=12 elements—more than 11.
• To get LCM 42 (2×3×7), we'd need 2+3+7=12 elements, again over 11.
• All other combinations either sum to more than 11 or give a lower LCM.

Conclusion #

The maximum order of any element in (S_{11}) is 30. Examples of such elements include:

• A disjoint product of a 6-cycle and a 5-cycle (written as ((1\ 2\ 3\ 4\ 5\ 6)(7\ 8\ 9\ 10\ 11)) in cycle notation), or
• A disjoint product of a 5-cycle, a 3-cycle, and a 2-cycle (like ((1\ 2\ 3\ 4\ 5)(6\ 7\ 8)(9\ 10))—the 11th element is fixed, so we don't write it in the cycle decomposition).

内容的提问来源于stack exchange，提问作者Charlie Etheridge