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Group theory question - permutation groups watch

1. I'm stuck on how to do these kind of questions and was hoping someone might be able to explain:

Peform the following permutation calculations, writing the answer as a product of disjoint cycles (remember to start from the right):

(i) (1 8 2 4)(3 5 9 7)(2 7)(1 9 3)
(ii) (2 5 3 7)(1 3 7)(2 6 4)(2 7 3 5)
2. The idea is that you have to chase the numbers around until you can find the disjoint cycles. I won't give you the answers to your problems, but I'll illustrate with another example.

Suppose your permutation is (1 4 2 3)(2 6)(3 4 5)(1 6 2 3) on the set {1,2,3,4,5,6}.

We start by looking where 1 goes:
(1 4 2 3)(2 6)(3 4 5)(1 6 2 3) 1 --> (1 6 2 3) sends 1 to 6, so this is equal to:
(1 4 2 3)(2 6)(3 4 5) 6 --> (3 4 5) has no effect on 6, so this is equal to:
(1 4 2 3)(2 6) 6 --> (2 6) sends 6 to 2, so this is equal to:
(1 4 2 3) 2 --> (1 4 2 3) sends 2 to 3, so this is equal to 3

So 1 goes to 3, and our first disjoint cycle starts (1 3..., so now we look where 3 goes. As before:
(1 4 2 3)(2 6)(3 4 5)(1 6 2 3) 3
= (1 4 2 3)(2 6)(3 4 5) 1
= (1 4 2 3)(2 6) 1
= (1 4 2 3) 1
= 4

Now we see where 4 goes:
(1 4 2 3)(2 6)(3 4 5)(1 6 2 3) 4
= (1 4 2 3)(2 6)(3 4 5) 4
= (1 4 2 3)(2 6) 5
= (1 4 2 3) 5
= 5

Now we see where 5 goes:
(1 4 2 3)(2 6)(3 4 5)(1 6 2 3) 5
= (1 4 2 3)(2 6)(3 4 5) 5
= (1 4 2 3)(2 6) 3
= (1 4 2 3) 3
= 1

...and we end up back at 1, so one of our disjoint cycles is (1 3 4 5). We still don't know what happens to 2 or 6, and following those through as above reveals that 2 and 6 are fixed, and so (1 4 2 3)(2 6)(3 4 5)(1 6 2 3) = (1 3 4 5)(2)(6), or we can just write (1 3 4 5).

Apply this method to your cycles.
3. (Original post by nuodai)
The idea is that you have to chase the numbers around until you can find the disjoint cycles. I won't give you the answers to your problems, but I'll illustrate with another example.

Suppose your permutation is (1 4 2 3)(2 6)(3 4 5)(1 6 2 3) on the set {1,2,3,4,5,6}.

We start by looking where 1 goes:
(1 4 2 3)(2 6)(3 4 5)(1 6 2 3) 1 --> (1 6 2 3) sends 1 to 6, so this is equal to:
(1 4 2 3)(2 6)(3 4 5) 6 --> (3 4 5) has no effect on 6, so this is equal to:
(1 4 2 3)(2 6) 6 --> (2 6) sends 6 to 2, so this is equal to:
(1 4 2 3) 2 --> (1 4 2 3) sends 2 to 3, so this is equal to 3

So 1 goes to 3, and our first disjoint cycle starts (1 3..., so now we look where 3 goes. As before:
(1 4 2 3)(2 6)(3 4 5)(1 6 2 3) 3
= (1 4 2 3)(2 6)(3 4 5) 1
= (1 4 2 3)(2 6) 1
= (1 4 2 3) 1
= 4

Now we see where 4 goes:
(1 4 2 3)(2 6)(3 4 5)(1 6 2 3) 4
= (1 4 2 3)(2 6)(3 4 5) 4
= (1 4 2 3)(2 6) 5
= (1 4 2 3) 5
= 5

Now we see where 5 goes:
(1 4 2 3)(2 6)(3 4 5)(1 6 2 3) 5
= (1 4 2 3)(2 6)(3 4 5) 5
= (1 4 2 3)(2 6) 3
= (1 4 2 3) 3
= 1

...and we end up back at 1, so one of our disjoint cycles is (1 3 4 5). We still don't know what happens to 2 or 6, and following those through as above reveals that 2 and 6 are fixed, and so (1 4 2 3)(2 6)(3 4 5)(1 6 2 3) = (1 3 4 5)(2)(6), or we can just write (1 3 4 5).

Apply this method to your cycles.
Sorry, I'm still getting really confused! You explained so eloquently as well ! Hmm...

So far I got (1 8 2 4)(3 5 9 7)(2 7)(1 9 3)1-->(1 9 3) sends 1 to 9 so
=(1 8 2 4)(3 5 9 7)(2 7)9 has no effect on 9 so
=(1 8 2 4)(3 5 9 7)9-->(3 5 9 7) sends 9 to 7
=(1 8 2 4)7
=7
is that right? I don't understand how I go on from there
4. (Original post by kellywellydoodle)
Sorry, I'm still getting really confused! You explained so eloquently as well ! Hmm...

So far I got (1 8 2 4)(3 5 9 7)(2 7)(1 9 3)1-->(1 9 3) sends 1 to 9 so
=(1 8 2 4)(3 5 9 7)(2 7)9 has no effect on 9 so
=(1 8 2 4)(3 5 9 7)9-->(3 5 9 7) sends 9 to 7
=(1 8 2 4)7
=7
is that right? I don't understand how I go on from there
You have to keep chasing the numbers until you get back to 1. You know 1 goes to 7. But where does 7 go? And where does the number after go? And so on. When you get back to 1, you've found one of your disjoint cycles. So for example if 1 went to 4, 4 went to 5, 5 went to 2, 2 went to 9 and 9 went back to 1, then your first cycle would be (1 4 5 2 9). [Note that this doesn't happen here.] When you've found the first disjoint cycle, you pick a number not in that cycle and chase that round until you get back to it, to give you another disjoint cycle... and so on.

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