Computing the product of disjoint cycles

433 Views Asked by At

I'm trying to compute the product of disjoint cycles:

$$(2 6)(3 4)(5 9)(7 8)(1 2 3 4 5 6 7 8 9 10)$$

I know the answer is $(1 6 8 5 2 4 9 10)$, but why? I was thinking $1$ maps to $2$ in the right most term then $2$ maps to $6$ in the rightmost term so I get that. Then $2$ maps to $3$ which maps to $4$ but the third entry isn't $4$. Can someone please explain? This isn't homework, I am just learning by myself.

Also why is the order equal to $8$? Because I thought it's equal to the lcm of all the lengths, which equals lcm of 2 and 10, which is 20. I don't understand this either

2

There are 2 best solutions below

0
On BEST ANSWER

You got $1$ goes to $6$, that is correct. So to compute the cycle the next entry you should check is $6$, not $2$. We have $6\to 7\to 8$, so the third entry is $8$, and you can really see that in the answer. So now we check where does $8$ map. $8\to 9\to 5$, so the fourth entry is really $5$. And continue that way.

As for your second question-the order is the lcm of the lengths of the cycles only when the permutation is written as a composition of disjoint cycles, which means every element might appear only in one cycle, not more. For example, the permutation that you showed in the answer has just $1$ cycle of length $8$, so the order is $8$. And by the way, $lcm(2,10)$ is $10$ and not $20$.

0
On

Think of it like this:

  • $1\mapsto2\mapsto6$
  • $6\mapsto7\mapsto8$
  • $8\mapsto9\mapsto5$
  • $5\mapsto6\mapsto2$
  • $2\mapsto3\mapsto4$

et cetera, agreeing with $(168524\dots)$