Hey, I am trying to work out what y is (y is gamma) where
y^{1}(14)(256)y=(145)(23)
Everywhere I look including wikipedia keep on using poor examples which leaves me not knowing what's going on, so any help would be great!
How to work out conjugating elements for group permutations?
Announcements  Posted on 


 Follow
 1
 23042012 01:11

 Follow
 2
 23042012 02:25
(Original post by Lewk)
Hey, I am trying to work out what y is (y is gamma) where
y^{1}(14)(256)y=(145)(23)
Everywhere I look including wikipedia keep on using poor examples which leaves me not knowing what's going on, so any help would be great!
g = (a_1, a_2, ..., a_k1)(b_1, b_2, ..., b_k2)...(z_1, z_2, ..., z_kn)
and
h = (A_1, A_2, ..., A_k1)(B_1, B_2, ..., B_k2)...(Z_1, Z_2, ..., Z_kn)
The conjugating permutation x (so that x^(1)gx = h) is the one taking a_1 to A_1, a_2 to A_2, ..., z_kn to Z_kn (assuming you compose permutations left to right).
So for example, in S6 take g = (12)(34)(5)(6) and h = (23)(16)(4)(5). They have the same cycle structure. Let x be the permutation taking 1>2, 2>3, 3>1, 4>6, 5>4 and 6>5. So x = (123)(465). You can work out that x^(1) is (132)(456). Then
x^(1)gx = (132)(456)(12)(34)(5)(6)(123)(46 5) = (16)(23)(4)(5) = h
Note, there are options here. I could have written h as (23)(16)(5)(4). Then x would be (123)(46)(5) which also works. So to find x you just line up the cycles eg. by putting them in order of length (including cycles of length 1) and read off what x does to each number.Post rating:1 
 Follow
 3
 23042012 03:31
(Original post by SsEe)
There's a "method" for this. You perhaps know that two elements are conjugate if and only if they have the same cycle structure. So you can write them something like:
g = (a_1, a_2, ..., a_k1)(b_1, b_2, ..., b_k2)...(z_1, z_2, ..., z_kn)
and
h = (A_1, A_2, ..., A_k1)(B_1, B_2, ..., B_k2)...(Z_1, Z_2, ..., Z_kn)
The conjugating permutation x (so that x^(1)gx = h) is the one taking a_1 to A_1, a_2 to A_2, ..., z_kn to Z_kn (assuming you compose permutations left to right).
So for example, in S6 take g = (12)(34)(5)(6) and h = (23)(16)(4)(5). They have the same cycle structure. Let x be the permutation taking 1>2, 2>3, 3>1, 4>6, 5>4 and 6>5. So x = (123)(465). You can work out that x^(1) is (132)(456). Then
x^(1)gx = (132)(456)(12)(34)(5)(6)(123)(46 5) = (16)(23)(4)(5) = h
Note, there are options here. I could have written h as (23)(16)(5)(4). Then x would be (123)(46)(5) which also works. So to find x you just line up the cycles eg. by putting them in order of length (including cycles of length 1) and read off what x does to each number.
(132)(46)(5)(12)(34)(5)(6)(123)( 46)(5) = (13)(26)(4)(5) =/= h
Have I done something wrong or am i just confused still? :/Last edited by Lewk; 23042012 at 03:33. 
 Follow
 4
 23042012 03:55
(Original post by SsEe)
There's a "method" for this. You perhaps know that two elements are conjugate if and only if they have the same cycle structure. So you can write them something like:
g = (a_1, a_2, ..., a_k1)(b_1, b_2, ..., b_k2)...(z_1, z_2, ..., z_kn)
and
h = (A_1, A_2, ..., A_k1)(B_1, B_2, ..., B_k2)...(Z_1, Z_2, ..., Z_kn)
The conjugating permutation x (so that x^(1)gx = h) is the one taking a_1 to A_1, a_2 to A_2, ..., z_kn to Z_kn (assuming you compose permutations left to right).
So for example, in S6 take g = (12)(34)(5)(6) and h = (23)(16)(4)(5). They have the same cycle structure. Let x be the permutation taking 1>2, 2>3, 3>1, 4>6, 5>4 and 6>5. So x = (123)(465). You can work out that x^(1) is (132)(456). Then
x^(1)gx = (132)(456)(12)(34)(5)(6)(123)(46 5) = (16)(23)(4)(5) = h
Note, there are options here. I could have written h as (23)(16)(5)(4). Then x would be (123)(46)(5) which also works. So to find x you just line up the cycles eg. by putting them in order of length (including cycles of length 1) and read off what x does to each number.
I see how the first part works (sort of visually) very nice. I can also see some other ways of doing this, but not sure if I can see the method to construct all possible x. 
 Follow
 5
 23042012 11:35
(Original post by Lewk)
I tried to see if your second rearrangement worked, where you said you could have written h as (23)(16)(5)(4), then x = (123)(46)(5) and x^{1}=(132)(46)(5), but x^{1}gx =/= h:
(132)(46)(5)(12)(34)(5)(6)(123)( 46)(5) = (13)(26)(4)(5) =/= h
Have I done something wrong or am i just confused still? :/
Also, make sure you understand why it works. Then you don't have to memorize which way round it goes. If you're composing right to left then x takes A_1 to a_1, A_2 to a_2 etc. So x^(1)gx(A_1) = x^(1)g(a_1) = x^(1)(a_2) = A_2 = h(A_1). etc. So x^(1)gx = h. 
 Follow
 6
 23042012 12:04
(Original post by Blutooth)
Incidentally, I was just working on a very similar question.
I see how the first part works (sort of visually) very nice. I can also see some other ways of doing this, but not sure if I can see the method to construct all possible x. 
 Follow
 7
 23042012 22:35
(Original post by SsEe)
You need to permute all cycles of the same length and then "cycle" each cycle. eg (123) = (312) = (231). So I think that if you could write with cycle lengths as n1, ..., n1, n2, ..., n2, ..., nr, ..., nr where ni is repeated ki times, you'd be able to construct k1! k2! ... kr! n1^k1 n2^k2 ... nr^kr values of x. There is the question of whether or not all possible x are covered by this method and whether each one is different.
Write a reply…
Reply
Submit reply
Register
Thanks for posting! You just need to create an account in order to submit the post Already a member? Sign in
Oops, something wasn't right
please check the following:
Sign in
Not got an account? Sign up now
Updated: April 23, 2012
Share this discussion:
Tweet
Related discussions:
TSR Support Team
We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.