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# How to work out conjugating elements for group permutations? Watch

1. 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!
2. (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!
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.
3. (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 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-1gx =/= 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? :/
4. (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.
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.
5. (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-1gx =/= 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? :/
I think you're composing permutations right to left. I do it left to right. In that case, take x to be the map taking A_1 to a_1 etc (ie, x is the inverse of what you'd get if you did it as I originally wrote it).

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.
6. (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.
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.
7. (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.
I got the right answer with the question in the OP when I did xgx-1=h instead of x-1gx .. no idea why that works but i'm just going to stick with it.

Updated: April 23, 2012
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