[代數] 兩個有關group的問題

看板Math作者 (ssss)時間11年前 (2012/11/06 01:33), 編輯推噓0(003)
留言3則, 2人參與, 最新討論串1/4 (看更多)
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) 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... 2. A group G in which (ab)^2=(a^2)(b^2), for all a, b in G is necessarily (A) finite (B) cyclic (C) of order 2 (D) abelian (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~~ -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 108.3.154.49

11/06 01:50, , 1F
a*b=a+b+ab 注意後面是實數運算..所以ae不等於a
11/06 01:50, 1F

11/06 01:54, , 2F
上面少寫2倍..a=a*e=a+e+2ae => 0=(1+2a)e 剩下自己
11/06 01:54, 2F

11/06 04:03, , 3F
我覺得(ab)^2 = (ab)(ab)耶,似乎abelian是必要的
11/06 04:03, 3F
文章代碼(AID): #1Gb_XM9s (Math)
文章代碼(AID): #1Gb_XM9s (Math)