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How does one find the element with the highest order in $S_9$ ?

I can only guess that the element may be (1 2 3 4)(5 6 7 8 9) whose order is 20.

Rescy_
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    The idea is very good. Express the permutation as a product of disjoint cycles. Then the order is the lcm of their lengths. There are a number of possibilities to track down, but not too many. – André Nicolas Sep 19 '15 at 23:22
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    Relevant: http://oeis.org/A000793 – Arthur Sep 19 '15 at 23:23

1 Answers1

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The conjugacy classes of $S_n$ are in one-to-one correspondence with the partitions of $n$, that is, a partition of $n$ is an ordered $k$-tuple ($k$ is arbitrary) of positive integers ordered increasingly whose sum is $n$. The partitions of $9$ are $$ (1,1,1,1,1,1,1,1,1) \\ (1,1,1,1,1,1,1,2) \\ (1,1,1,1,1,1,3) \\ \vdots \\ (1,8), \\ (1,1,1,1,1,2,2) \\ (1,1,1,1,2,3) \\ (1,1,1,2,4) \\ (1,1,2,5) \\ (1,2,6) \\ (2,7), \\ (1,1,1,3,3) \\ (1,1,3,4) \\ (1,3,5) \\ (3,6) \\ (1,4,4) \\ (4,5) \\ (1,1,1,2,2,2) \\ (1,1,2,2,3) \\ (1,2,2,4) \\ (2,2,5) \\ (1,2,3,3) \\ (2,3,4) \\ (3,3,3)\\ (1,1,1,2,2,2)\\ (1,2,3,3) \\ (1,2,2,2,2) \\ (2,2,2,3) \\ (9) $$ So by checking all their LCMs, you quickly notice that your 20 is indeed the correct one. I don't know if there is a general method for $S_n$ though.

Hope that helps,

Patrick Da Silva
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