1

Could someone explain the process of simplifying the following permutation in $S_6$

(1,3,5)(2,4,5)(2,3,6)

An explanation on how you arrived at the simplified form would also be greatly appreciated.

Thanks

Caitlin Y
  • 13
  • The order of the cycle multiplication is important. A lot of texts go right to left, some go left to right (including wolfram alpha) – David P Dec 14 '14 at 00:58
  • I know, but no matter what I do I never arrive at the correct simplified form. So I'm assuming I have the wrong method – Caitlin Y Dec 14 '14 at 01:03
  • Assuming it is right to left, start with 1. 1 isn't sent anywhere by the first cycle or second cycle, the third cycle sends $1\to 3$. So begin your answer as $(1,3...).$ Next look at $3$. $3\to 6$ by the first, the others fix $6$, so your answer is so far $(1,3,6,...)$. – David P Dec 14 '14 at 01:06

1 Answers1

0

In the book I read this from they started at the right. Having this in mind we will now simplify $(1,3,5)(2,4,5)(2,3,6)$. To do this we start with $1$. Where does $1$ end up after this permutation? to see this look for the first appearence of $1$ (starting from the right), it only appears once and it goes to $3$, so write $(1,3$ . We now wish to know where $3$ goes to. This time $3$ appears in the first cycle (right to left), so $3$ goes to $6$. And the other cycles don't move $6$, so we write $(1,3,6$ we know want to go where $6$ goes to. Notice $6$ goes to $2$, and then $2$ goes to $4$ in the next cycle, and two is not moved by the last cycle, so $6$ goes to $4$, we proceed to write $(1,3,6,4$. We know wish to know where $4$ goes. $4$ is not moved by the first cycle, but it is moved by the second cycle, notice $4$ goes to $5$, and in the last cycle $5$ is sent to $1$, so we write $(1,3,6,4)$. We have closed the first cycle, however notice we have not inspected what happens to elements $2$ and $5$ yet. since $2$ is the smallest numbers missing we write $(1,3,6,4)(2$ and proceed to see where $2$ is mapped under the permutation. Notice $2$ is sent to $3$ by the first cycle and $3$ not moved by the second cycle, it is however moved by the last cycle, it is sent to $5$, so $2$ goes to $5$, we proceed to write $(1,3,6,4)(2,5$ Where does $5$ go to? well all the other elements are taken so $5$ must go to $2$ and we conclude:

$(1,3,5)(2,4,5)(2,3,6)=(1,3,6,4)(2,5)$

Asinomás
  • 105,651