Permutation

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.

this is awesome! thank you

how do you put the part where 5 and 6 change in your z?
nevermind i managed to see it. your z should be (1)(253)(46) or something like that.
but your method is brilliant!

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)

in the diagram you drew you have (56) as part of x but this is not how x was defined. by definition of x 5->5 and 6->6 which means that that the change in y (5->6 and 6->5) must be included in z.

Oh yeah! My x would've been (12)(34)(56) and my y (13)(24)(56)... I just sketched another one and you're right. (1)(253)(46) is what i got as well. Used the same method though, think its quite a good wee way of doing it.