有一題證明想了很久還是沒有方向希望代數強者教我一下Suppose G is a group of order 6 Prove:There is exactly one subgroup of order 3. 拜託了.....感謝!! ... <看更多>
subgroup 代數 在 Re: [代數] normal subgroup - Math | PTT Web 的推薦與評價
Re:[代數]normalsubgroup@math,共有0則留言,0人參與討論,0推0噓0→, ※ 引述《rich1119 (We)》之銘言:: 請問: 有人能解釋normal subgroup 嗎: 我覺得只懂他的定義: ... ... <看更多>
subgroup 代數 在 Normal Subgroups and Quotient Groups (aka Factor Groups) 的推薦與評價
Normal subgroups are a powerful tool for creating factor groups (also called quotient groups). In this video we introduce the concept of a ... ... <看更多>
subgroup 代數 在 Re: [代數] subgroup - 看板Math - 批踢踢實業坊 的推薦與評價
※ 引述《s620555 (小毛)》之銘言:
: 麻煩幫解
: 1) show that {A屬於GLn(R)∣det A = ±1 } is a subgroup of GLn(R) un-der
: multiplication
The definition of subgroup (by I.N. Herstein) (可以請教這裡的hersteinXD)
A subset H of the group G is a subgroup of G if and only if it is
nonempty and closed under products and inverses.
(The closure conditions mean the following: whenever a and b are in H, then
-1
ab and a are also in H)
GLn(R)其實可以寫成很多形式
https://en.wikipedia.org/wiki/General_linear_group
在Lie group 裡面 我們就是把 GLn(R) 當做是一種n*n維空間中的一般線性變換群
所以可以寫成一般的矩陣轉換形式且係數是實數且 det(A)≠0
A'_i = Σ a_ij A_j .where A and A'are n*n matrix 且 det(a_ij)= ±1
j
pf:
let A'_i and B'_i belong to H
n
A'_i = Σ a_ij A_j ; det(a_ij)= ±1 .where A and A'are n*n matrix
j
with matrix element a_ij;i,j=1...n
n
B'_i = Σ b_ik B_k ; det(b_jk)= ±1 .where B and B'are n*n matrix
k
with matrix element b_jk;j,k=1...n
First
n n
C'i ≡ A'_i*B'_i = Σ a_ij b_jk C_k =Σ c_ik C_k (這步實在讓我很掙扎想好久)
jk k BBS要寫整個矩陣很麻煩
應該純寫矩陣相乘就好了??
and det(a_ij b_jk) = det(a_ij) det(b_jk)= ±1 ,
so the multiplication is closed.
Second
-1 n -1 -1
A'_i =( Σ a_ij) A_j det (a_ij)= ±1
j
because det(a_ij)= ±1 ≠ 0 ,so the inverse n*n matrix exists
hence
{ A is belong to GLn(R)∣det A = ±1 } is a subgroup of GLn(R)
PS:
A屬於GLn(R)∣det A = ±1 } is a subgroup of GLn(R)
是有物理意義的,且det A = 1 是 SLn(R)
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◆ From: 140.113.28.113
※ 編輯: Lindemann 來自: 140.113.28.113 (10/06 03:04)
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