Given , find a permutation such that .
Is there an easy way to do this? I came to an answer but I dont think I used an efficient way to get there.
Right excuse the quick paint job but here's an idea.
This is what i normally do given these probs (our lecturer advised us to do them this way!)
In the first diagram the red lines are y's perm, the black lines are x's perm.
Then just fill in lines so that the everthing goes to its right place. (example shown in the second diagram).
So my z would be (1)(2 3)(4)(5)(6). But there could be more than one way of doing it.
How did you get that as my z?
Pretty sure its just (1)(2 3)(4)(5)(6)
Since 1 goes to 1 -> (1)
2 goes to 3 which goes to 2 -> (2 3)
4 goes to 4 -> (4)
5 goes to 5 -> (5)
6 goes to 6 -> (6)
Your one would be something like this...
1 goes to 1 -> (1)
2 goes to 5 which goes to 3 which goes back to 2 -> (252)
4 goes to 6 which goes to 4 -> (46)