3
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I understand that you have to write out all the disjoint cycles and then take the least common multiple which yields the highest order.

But my question is, do I have to write all elements of $S_5$, write them as disjoint cycles, and then find the largest least common multiple, or is there a shortcut?

$S_5$ has $5!$ elements and I would not like writing all of these permutations out...

I have read the other answers on here but I have not seen anything to help me with this question.

For example, here (https://math.stackexchange.com/a/231893/133156) the answer lists six disjoint cycles of $S_5$, how did he get there without writing them all out?

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You only need to write out all possible structures for disjoint cycles. They correspond to the (additive) partitions for $5$:

$5 = 5 $ corresponds to one $5$-cycle.

$5 = 4 + 1 $ corresponds to one $4$-cycle and one $1$-cycle, but $1$-cycles can be ignored.

$5= 3 + 2 $ corresponds to one $3$-cycle and one $2$-cycle.

etc...

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    $\begingroup$ So it would follow as such: $$5=5+0$$ $$5=4+1$$ $$5=3+2$$ $$5=3+1+1$$ $$5=2+2+1$$ $$5=2+1+1+1$$. So the largest order would be $$lcm(3,2)=6$$. $\endgroup$ – H5159 Apr 11 '14 at 2:41
  • $\begingroup$ How have I missed any? These are the only additions to yield 5 for disjoint cycles. I'm confused how I missed four. Also, the author of the answer in the link only lists these 6, not 10. $\endgroup$ – H5159 Apr 11 '14 at 2:48
  • $\begingroup$ I still do not understand, can you elaborate? The largest least common multiple is with the product of two disjoint cycles of length 2 and 3. I don't see why you are mentioning 4 as a least common multiple. $\endgroup$ – H5159 Apr 11 '14 at 2:57
  • $\begingroup$ @Frumpy, you're right, I misread. $\endgroup$ – lhf Apr 11 '14 at 2:58
  • $\begingroup$ I'm going to try $S_6$ now :) $\endgroup$ – H5159 Apr 11 '14 at 3:01

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