Finding element of order in the symmetric group

Question

What is the proper and technical way to find if there exists an element in the symmetric group $S_n$ (Wikipedia article on symmetric groups.). I do know how to disprove if there does not exists such elements, but how should I find the element which does exists of this order?

For example, in the $S_{12}$, there is not such element of order $13$ because The only elements of order $13$ in $S_n$ are unions of disjoint $13$-cycles, since $13$ is prime. This would require $S_{12}$ to contain at least $13$ symbols, which it does not. But I think that there is an element of order $35$ in $S_{12}$. How should I find it?

EDIT: I do understand that I have to find a $7$-cycle and a $5$-cycle, but how?

First, prove that the order of an element in $S_n$ equals the least common multiple of the lengths of the cycles in its cycle decomposition. Hence, by the answer provided by @Ross, you can prove your claim. — toric_actions, Jan 01 '19 at 15:16

score 4 · Accepted Answer · answered Jan 01 '19 at 15:04

4

If you want an element of order $k$, you can factor $k$ and make some cycles whose $\operatorname {LCM}$ is $k$. In your example of $35$ we note that $35=5\cdot 7$, so if you make a $5-$cycle and a $7-$cycle you will have an element of order $35$. As $5+7=12$ you have enough room to do this in $S_{12}$

answered Jan 01 '19 at 15:04

Ross Millikan

374,822

Thanks I get it now. How can I find the maximum order possible? – vesii Jan 01 '19 at 15:15
1

$35$ is the best you can do with just two cycles, but once $n$ gets large enough you can do better with three separate cycles. $12$ is large enough that three cycles gives a higher LCM. Values of the maximum are given in http://oeis.org/A000793 – Ross Millikan Jan 01 '19 at 15:26

score 0 · Answer 2 · answered Jan 01 '19 at 15:03

0

Hint: $35 = 5 \cdot 7$ and $5+7=12$.

answered Jan 01 '19 at 15:03

lhf

216,483

Finding element of order in the symmetric group

2 Answers2