Question

Permutations as a product of disjoint cycles...
I do not care for the answer just the process; I stumped on these two and my brain hurts :/

b). (13256)(23)(46512)=

C). (12)(13)(23)(142)=

Answer 1

I know the answers, but I cannot for the life of me figure out how...
B)= (124)(35)
C)= (1423)

Answer 2

Let's go with b, for now...
Pick a number from 1 - 6
But I'm guessing you'll pick 1, anyway, shall we proceed? :)

Answer 3

Well, when i tried i started with 4, just because it seemed to be standard. I know we work right to left, but i am not getting the same answer

Answer 4

I got the first part anyways, but not so much the second...of part b

Answer 5

Okay, let's go with 4 :)
(1 3 2 5 6)(2 3)(4 6 5 1 2)
So... the rightmost permutation sends 4 to 6, the middle, fixes 6, and the leftmost, sends 6 to 1.

So this permutation sends 4... to 1 right?

Answer 6

I concur

Answer 7

Now, let's see what it does to 1.
The rightmost sends 1 to 2, the middle sends 2 to 3, and the leftmost, sends 3 to 2
So this permutation sends 4 to 1, and sends 1 to 2. Still with me?

Answer 8

Yes, i am still following. That's as much as I have, then my confusion sets in.

Answer 9

Okay, so now we start with 2.
The rightmost sends 2 to 4. The middle, fixes 4. The leftmost... also fixes 4.
So, this permutation sends 4 to 1, 1 to 2, and 2 BACK to 4.
This is important, catch me so far?

Answer 10

Yes, this makes sense. Mine looked like (412) and I know that is the same as (124). So, i am still with ya

Answer 11

okay, so now we can account for what in the blazes (^.^) this permutation does to 1, 2, and 4, but not what it does to the other three numbers from 1 to 6. Pick another number, from 1 to 6, except 4, 1, or 2.

Answer 12

Okie dokie then, let us go with 3, please. :D

Answer 13

Okay. Start with 3.
The rightmost fixes 3. The middle, sends 3 to 2. The leftmost sends 2 to 5.
So this permutation sends 3 to 5. So far so good?

Answer 14

Yes, making perfect sense so far. Not sure what I was doing wrong, but let's keep on keeping on

Answer 15

So, let's see what it does to 5.
The rightmost sends 5 to 1. The middle fixes 5. The leftmost sends 1 to 3.
So this permutation sends 3 to 5, and 5, BACK to 3.
And again, we've 'closed the loop', so to speak.

Answer 16

oh okay sweet, awesome opossum. And how/when do we know if we are done, if not all the numbers are used?

Answer 17

We're 'technically' not yet done, we don't know what it does to 6 yet :D
From instinct, you probably guess it fixes 6, but let's be formal and sure :)
The rightmost sends 6 to 5. The middle fixes 5. The leftmost sends 5 to 6.
So this permutation sends 6 BACK to 6, so to speak...

So this permutation, as a product of disjoint cycles, is just
(412)(35)(6)
And (6) is just an identity permutation, it can be removed from this notation...
(412)(35)
And now, just rearrange the numbers within the cycles as you wish :)

Answer 18

Ohhhh right!!! DUh!!! Goodness! So you just exhaust all the permutations of each number and that's basically how you know you are done. Thanks a lot! I believe I can confidently do part c now.

Answer 19

No problem :)