# For the following permutation in $S_9$, compute the sign, order and give cycle decomposition into disjoint cycles

For the following permutation in $S_9$, compute the sign, order and give cycle decomposition into disjoint cycles, justifying your steps.

$\sigma \in S_9$ with

$$\sigma(1)=6, \sigma(2)=4, \sigma(3)=2, \sigma(4)=5, \sigma(5)=3, \sigma(6)=8, \sigma(7)=7, \sigma(8)=9, \sigma(9)=1$$

My current solution

$(1, 6, 8, 9)(2, 4, 5, 3)$

Therefore $sgn(\sigma)=+1$

Not sure how do find the order, thanks in advance my knowledge on this topic is poor.

• The order is the smallest $n$ such that $\sigma^n$ is the identity. – Lord Shark the Unknown Aug 3 '18 at 15:00
• Find $\sigma^n$ for small $n=1,2,3,4,\dots$ – drhab Aug 3 '18 at 15:00
• Could you expand a bit on "Therefore sgn$(\sigma)=+1$? Just to make sure you have a correct reasoning because there are lots of correct ways but also several ways to get the correct result with a fallacious argument. – Arnaud Mortier Aug 3 '18 at 15:02
• @ArnaudMortier I have used $(-1)^m$ where $m$ is the number of transpositions in the permutation – Ben Jones Aug 3 '18 at 15:07
• That's correct, and how many transpositions are there? – Arnaud Mortier Aug 3 '18 at 15:16

In general, the order of a permutation equals to the $lcm$ of the lengths of its disjoint cycles. (the cycles must have no common elements of course)

• Okay, so in this case the $lcm=4$ – Ben Jones Aug 3 '18 at 15:02
• Looking at another example, $(1, 8, 9)(2, 7, 5, 3)(4, 6)$ the $lcm$ would be equal to 12? – Ben Jones Aug 3 '18 at 15:03
• Yes, exactly. If you are not sure why the statement that I wrote is true then you can try to prove it, it's really easy. – Mark Aug 3 '18 at 15:05