-2
$\begingroup$

Let $\sigma$ be the permutation given by enter image description here

Is their a short way to do this.Thanks

$\endgroup$
  • $\begingroup$ Write $\sigma$ as a product of disjoint cycles. $\endgroup$ – user296602 Dec 31 '15 at 7:42
  • $\begingroup$ I guess this was downvoted because it's a pasted image with multiple choice answers. I don't see a problem with the question. $\endgroup$ – Matt Samuel Dec 31 '15 at 7:43
1
$\begingroup$

Hint: Every permutation can be written as a product of disjoint cycles. The order of the permutation is the least common multiple of the order of each cycle.

$\endgroup$
  • $\begingroup$ So answer is 12 $\endgroup$ – Sophie Clad Dec 31 '15 at 7:54
  • $\begingroup$ Yes ma'm that is correct. $\endgroup$ – Brandon Thomas Van Over Dec 31 '15 at 8:16
2
$\begingroup$

We have a cycle of length 4, a cycle of length 3, and a cycle of length 2. Raising to a power that is a multiple of 4 makes the length 4 cycle go away, as well as the length 2. Raising to a multiple of 3 makes the 3-cycle go away. But we want them all to go away. What can we do?

$\endgroup$
  • $\begingroup$ "Raising to a power that is a multiple of 4 makes the length 4 cycle go away"..Why is that so? $\endgroup$ – Sophie Clad Dec 31 '15 at 7:54
  • $\begingroup$ @Sophie try explicitly raising a 4 cycle to the fourth power and you'll see. In general an $n$-cycle has order $n$. The reason the method in the other answer works is because the disjoint cycles commute, so raising the permutation to a power is the same as raising all of the cycles to the same power. $\endgroup$ – Matt Samuel Dec 31 '15 at 7:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.