OWhat is the product of these permutations?

[Firepanda]

I don't have any example of this in my notes:

I have the permutation p1 = (1 5)(2 4 6 3)

and the permutation p2 = (1 3 7 4)(2)(5 8 6)

and I have to find p2 * p1

i THINK you take it in the form p1 then p2, like:


(1 5)(2 4 6 3)(1 3 7 4)(2)(5 8 6) = ...

But them I'm stuck.

I tried googling some info, but it was all too vague.

I had an attempt which ended up in:

1 -> 5 in p1, then 5 -> 8 in p2 so 1 -> 8

etc

so p2*p1 = (1 8)(2 1 8)(3 2 1 8)(4 5 3 2 1 8)(6 7)

= (4 5 3 2 1 8)(6 7)

Looks right? :P

 

[HallsofIvy]

I don't have any example of this in my notes:

I have the permutation p1 = (1 5)(2 4 6 3)

and the permutation p2 = (1 3 7 4)(2)(5 8 6)

and I have to find p2 * p1

i THINK you take it in the form p1 then p2, like:


(1 5)(2 4 6 3)(1 3 7 4)(2)(5 8 6) = ...

But them I'm stuck.

I tried googling some info, but it was all too vague.

I had an attempt which ended up in:

1 -> 5 in p1, then 5 -> 8 in p2 so 1 -> 8

etc

so p2*p1 = (1 8)(2 1 8)(3 2 1 8)(4 5 3 2 1 8)(6 7)

= (4 5 3 2 1 8)(6 7)

Looks right? :P

Take it "one step at a time". p2 is (1 3 7 4)(2)(5 8 6) and p1 is (1 5)(2 4 6 3).
p1 takes 1 to 5 and then p2 takes 5 to 8: together p1p2 takes 1 to 8.
p1 takes 2 to 4 and then p2 takes 4 to 1: p1p2 takes 2 to 1.
p1 takes 3 to 2 and then p2 takes does not change 2: p1p2 takes 3 to 2.
p1 takes 4 to 6 and then p2 takes 6 to 5: p1p2 takes 4 to 5.
p1 takes 5 to 1 and then p1 takes 1 to 3: p1p2 takes 5 to 3.
p1 takes 6 to 3 and then p2 takes 2 to 7: p1p2 take 6 to 7
p1 does not change 7 but p2 takes 7 to 4: p1p2 takes 7 to 4
p1 does not change 8 but p2 takes 8 to 6: p1p2 takes 8 to 6
(That's how I am interpreting the fact that neither 7 nor 8 appear in the definition of p1.)

Now look for cycles: 1 changes to 8 which changes to 6 which changes to 7 which changes to 4 which changes to 5 which changes to 3 which changes to 2 which changes to 1: a cycle is (18674532). Since that includes every number from 1 to 8, p1p2 is that cycle: p1p2= (18674532).

 

[Firepanda]

I just tried to comnpute p2*p2

(1 3 7 4)(2)(5 8 6)(1 3 7 4)(2)(5 8 6)

To show where i went wrong ill miss out some steps and jum straight to it.

I got 1 -> 3 then 3 -> 7 : so 1->7

I also got 7 -> 4 then 4 -> 1 : so 7->1

when starting off my cycle: 1 goes to 7, then 7 goes to 1...

so my cycle would look like (17...)

but it goes back to 1, does the cycle have to be in one set of brackets?

otherwise I can do it as (17)(...)

thanks!

 

[CompuChip]

You are right, so p2 * p2 starts with (17)
Then you open a new pair of brackets and start with a number you haven't looked at yet. So take for example 2, then p2 * p2 will become
(17)(2...)
If you complete the cycle while still not having had all the numbers 1 up to 8, open a new bracket again, etc.

 

[Firepanda]

I now how to write p2 as a product of tranbspositions :P

so (1 3 7 4)(2)(5 8 6)

= (1 3) (3 7) (7 4) ( ? ) (5 8) (8 6)

what goes in the bracket of ( ? ), i believe it needs to have 2 values, so (2 2)?

 

[happyg1]

What happens to the 4? What happens to the 6?
2 goes to 2, so it's just (2)

 

[Firepanda]

Well i have an example in my book, and it says:

( a1, a2, a3, ... , a(k-1), a(k)) = (a1 a2) (a2 a3) (a3 a4) ... (a(k-1) a(k))

Which suggests I would leave the final values inside the bracket alone, as there is no (a(k) a1) term.

And the example I have is:

(2 4 6 7 3)(2 5 1 3)(7 4 8)

= (1 2 5)(3 4 8)(6 7)

= (1 2)(2 5)(3 4)(4 8)(6 7)

Which also leaves out the 5 and 8 (final values of the brackets), is this wrong or?

 

[happyg1]

Our Professor had us putting in the last thing as well. I think it's just a difference in the method...My professor wanted it written out like (1 3) (3 7) (7 4) ( 4 1)(2) (5 8) (8 6)(6 5)
I looked in my book and your form of the transpositions isn't in there.

I think you're good.
CC