Problem with cycle multiplication:

56 Views Asked by At

From the rules of the cycle multiplication I know I can do this: $(123)(145)=(231)(145)=(23145)$. However, if I try to solve the following I get repeated elements (the $1$):

$$(126)(123)(145)=(126)(23145)=(612)(23145)=(6123145)$$

Is there a rule I am forgetting here? (I already know how to do it manually, but I want to do it using rules of cycle multiplication). The answer should be $(1456)(23)$

2

There are 2 best solutions below

1
On BEST ANSWER

You are right to question your final answer. I have no idea why you ended up with repeated $1$s.

You are not missing a rule. You may not understand the rules, or you may have made a mistake.

You are multiplying right to left (which is one convention).

In the product $$ (612)(23145) $$ $$ 3 \to 1 \to 2 $$ where your calculation has $3 \to 1$. So the correct answer should have the sequence $32$ in one of the cycles, and that should be the only cycle containing a $2$ or a $3$.

What you say the answer should be is not in final form since the cycles are not disjoint.

There is no need to rearrange $(126)$ as $(621)$. All that does is introduce another place where you might make a copying error. Just track what happens to each number, one at a time, right to left. You can even do that with the original product of three permutations - no need to group two of them first.

0
On

Rotate the $5$-cycle again to align the $1$.

$$(612)(23145)\to(6123145)\to(6114523)\to(614523)\to(6145)(23)$$