Yet another small problem for this new and interesting topic
Let http://dientuvietnam...mimetex.cgi?S_n be the symmetric group of degree http://dientuvietnam...n/mimetex.cgi?n and let
http://dientuvietnam.net/cgi-bin/mimetex.cgi?K_n
ii) Prove that, for any http://dientuvietnam.net/cgi-bin/mimetex.cgi?\pi is equal to http://dientuvietnam...mimetex.cgi?K_n
iii) http://dientuvietnam...mimetex.cgi?K_n
Conjugacy class
Bắt đầu bởi RongChoi, 18-10-2005 - 22:43
#1
Đã gửi 18-10-2005 - 22:43
#2
Đã gửi 24-10-2005 - 14:56
I have solved this "small" problem in the case http://dientuvietnam...n/mimetex.cgi?n is odd . In this case, http://dientuvietnam...mimetex.cgi?K_n is empty and all statements are followed.
(I think Rong choi had a typing mistake in the formula defining http://dientuvietnam...imetex.cgi?K_n:
.)
(I think Rong choi had a typing mistake in the formula defining http://dientuvietnam...imetex.cgi?K_n:
.)
#3
Đã gửi 24-10-2005 - 20:11
Can I try?
Prop. 1: an element x in S_n is an involution, i.e. x^2=id, if and only if x is a product of disjoint transpositions.
Prop. 2: 2 elements in S_n are in the same conjugacy class if and only if they have the same cycle structure.
For the case when n is even (n=2m):
(i) The elements in K_n are precisely the elements in S_n which are the products of m disjoint transpositions. A combinatorial counting will give us the size of K_n as follows: |K_n| = (2m)! / (m!*2^m).
(ii) Straightforward from Prop. 2.
(iii) I don't understand what "uniformly distributed" means in this case, so I guess it must be one of the following meanings:
a. "uniformly distributed" means: txt^(-1) runs through K_n when x runs through K_n. This can be seen from the fact that o(txt^(-1))=2 and txt^(-1) (i) <> i for all i in {1,2,...,n}.
b. "uniformly distributed" means: txt^(-1) runs through K_n "uniformly" when t runs through S_n. This can be seen from the fact that txt^(-1)=axa^(-1) iff t and a make the same (left) cosets of C(x) in S_n (the centralizer of x in S_n), i.e. tC(x)=aC(x). The number of such cosets is exactly the cardinality of the conjugacy class of x, which is |K_n|. Thus txt^(-1) runs through K_n "uniformly" (i.e. runs through K_n for |C(x)| times) when t runs through S_n.
Prop. 1: an element x in S_n is an involution, i.e. x^2=id, if and only if x is a product of disjoint transpositions.
Prop. 2: 2 elements in S_n are in the same conjugacy class if and only if they have the same cycle structure.
For the case when n is even (n=2m):
(i) The elements in K_n are precisely the elements in S_n which are the products of m disjoint transpositions. A combinatorial counting will give us the size of K_n as follows: |K_n| = (2m)! / (m!*2^m).
(ii) Straightforward from Prop. 2.
(iii) I don't understand what "uniformly distributed" means in this case, so I guess it must be one of the following meanings:
a. "uniformly distributed" means: txt^(-1) runs through K_n when x runs through K_n. This can be seen from the fact that o(txt^(-1))=2 and txt^(-1) (i) <> i for all i in {1,2,...,n}.
b. "uniformly distributed" means: txt^(-1) runs through K_n "uniformly" when t runs through S_n. This can be seen from the fact that txt^(-1)=axa^(-1) iff t and a make the same (left) cosets of C(x) in S_n (the centralizer of x in S_n), i.e. tC(x)=aC(x). The number of such cosets is exactly the cardinality of the conjugacy class of x, which is |K_n|. Thus txt^(-1) runs through K_n "uniformly" (i.e. runs through K_n for |C(x)| times) when t runs through S_n.
#4
Đã gửi 25-10-2005 - 05:43
@ noproof: thanks to fix the typo.
@ madness:
i) and ii) : your proposed solution is very clear
iii) : ìweak (but general) uniformed distribution” means that, .
ìstrong uniformed distribution” combines your a) and b) definitions.
Your explanation in b) seems correct but I have not completely seen it yet…
@ madness:
i) and ii) : your proposed solution is very clear
iii) : ìweak (but general) uniformed distribution” means that, .
ìstrong uniformed distribution” combines your a) and b) definitions.
Your explanation in b) seems correct but I have not completely seen it yet…
#5
Đã gửi 26-10-2005 - 07:34
From your interpretation of "weak uniformly distributed", this can be seen from the fact stated in a) alone.
My explanation in b) is not clear Let me try to make it clearer.
b. "uniformly distributed" means: txt^(-1) runs through K_n "uniformly" when x \in S_n is fixed t runs through S_n. Now we try to understand the behavior of txt^(-1) when t runs and x is fixed.
Note that txt^(-1)=axa^(-1) iff a^(-1)t.x.t^(-1)a=x iff a^(-1)t.x=x.a^(-1)t iff a^(-1)t \in C(x) (the centralizer of x in S_n), i.e. tC(x)=aC(x).
The number of left cosets of C(x) in S_n is [S_n:C(x)], this is exactly the cardinality of the conjugacy class of x, which is |K_n|. Thus txt^(-1) runs through K_n "uniformly" (i.e. runs through K_n for |C(x)| times) when t runs through S_n.
My explanation in b) is not clear Let me try to make it clearer.
b. "uniformly distributed" means: txt^(-1) runs through K_n "uniformly" when x \in S_n is fixed t runs through S_n. Now we try to understand the behavior of txt^(-1) when t runs and x is fixed.
Note that txt^(-1)=axa^(-1) iff a^(-1)t.x.t^(-1)a=x iff a^(-1)t.x=x.a^(-1)t iff a^(-1)t \in C(x) (the centralizer of x in S_n), i.e. tC(x)=aC(x).
The number of left cosets of C(x) in S_n is [S_n:C(x)], this is exactly the cardinality of the conjugacy class of x, which is |K_n|. Thus txt^(-1) runs through K_n "uniformly" (i.e. runs through K_n for |C(x)| times) when t runs through S_n.
#6
Đã gửi 26-10-2005 - 11:25
Very well! :clap
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