Re: [代數] 兩個有關group的問題
※ 引述《ssss50201 (ssss)》之銘言：
: 我覺得學group好痛苦阿
: 遇到判斷題幾乎都是死翹翹>"<
: 1. Let * be the binary operation on the rational numbers is given by
: a*b=a+b+2ab. Which of the following are true?
: (A) * is commutative
: (B) There is a rational number that is a *-identity
: (C) Every rational number has a *-inverse.
: Ans: A,B
: 我知道A是對的（因為Q本身就commutative）
: 關於B，我試著用a*e=a解identity e
: a*e=a+e+2ae=a+e+2a=2a+e = a (by assumption e is identity, a*e=a)
↑這個e不是普通乘法的identity,是(Q,*)中*的
在(Q,*)當中:
Given a belongs to Q
a*e = a+e+2ae = a (by def.)
=> e(1+2a)=0 => e = 0 (這就是*的identity)
: 所以e=-a
: 但當我想檢查我找到的e是不是能讓a*e=e時，卻不對了
: 關於C, 沒有頭緒怎麼找inverse在沒有找到identity之前...
根據題目 找到一個有理數a讓他在(Q,*)中不可逆就好了
(i.e. a*x ≠0 for all x belongs to Q)
↑identity
Let a*x= a + x + 2ax = 0 (in order to find counter example)
-a
=> x = --------- (if 1+2a≠0 )
1+2a
所以取 a = -1/2，它沒有inverse
: 2. A group G in which (ab)^2=(a^2)(b^2), for all a, b in G is necessarily
: (A) finite
no,consider the cyclic group with infinite order
: (B) cyclic
no,consider K4 ={1,a,b,ab}, a^2 = b^2 = 1
K4 is abelian, note that If G is abelian
then (ab)^2=(a^2)(b^2) must hold
: (C) of order 2
no,same as (B),K4 with order 4
: (D) abelian
Because G is a group
we can apply cancellation laws
: (E) none of the above
: Ans:D
: 我有嘗試看看是不是abilian
: (ba)^2=(b^2)(a^2) =? (ab)^2 不知道怎麼從=?左邊推到右邊
: 謝謝幫忙~~
大概這樣
不過寫完後覺得2(C)怪怪的 它是問a&b必定order2嗎?
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