[離散]證明subgroup的問題

2007-11-22 10:35 pm
Let G be a group and let a 屬於 G.
The set C(a)={x 屬於 G | xa=ax} for all
elements of G that commute with a.
Show C(a) is a subgroup of G.

ie.證
1.C(a) 不為空集
2.C(a) is closed
3.C(a) has the identity e
4.C(a) has the inverse property
5.C(a) is associative

我已經知道

1.
let a 屬於 G ,
因為aa = aa => a 屬於 C(a) 不為空集

2.closed:
(先證b = a之inverse 屬於 C(a) )
因為 b = a之inverse 屬於 G and
ba = e = ab
=> b 屬於 C(a)
(再證closed)
任意 x , y 屬於 C(a)
xa = ax , ya = ay
=>xab = axb , yab = ayb => x = axb , y = ayb
=>xy = axbayb
=>xy = axyb
=>xya = axy => (xy)a = a(xy)
=>xy 屬於 C(a)

3.
因為 ea = ae => e 屬於 C(a)

4.
inverse不會證

5.
C(a) inherits the associative property from G
=>G is associative


請問怎麼證有inverse property?

回答 (2)

2007-11-23 2:27 am
✔ 最佳答案
4.
For x in C(a), xa=ax.
Let y=x^{-1},
ya = y(ae) = y(a(xy)) = y((ax)y) = y((xa)y) = ((yx)a)y = (ea)y = ay
所以 y=x^{-1} in C(a).

2. (我看不懂你的證明 orz)
Assume x, y in C(a), then
(xy)a = x(ya) = x(ay) = (xa)y = (ax)y = a(xy)
這證明了 C(a) 中運算的封閉性.
2007-11-23 9:00 am
4.的另外作法(意思一樣)
For x in C(a), xa=ax
let y = x^{-1}
因為 xa = ax
=>(xa)y = (ax)y
=>(xa)y = a(xy)
=> a = (xa)y
=> ya = y(xa)y => ya = (yx)ay
=>ya = ay


收錄日期: 2021-05-04 01:42:47
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20071122000016KK04460

檢視 Wayback Machine 備份