Anonymous
Anonymous asked in 科學數學 · 1 decade ago

# 英文版的代數學喔!有興趣的話...就看看唄!!

Question4

Let H be a non-empty subset of the

group G. Show that H is a subgroup of G if and only

if ab(-1在右上角小寫)屬於H (倒A)a,b屬於H

Question 5

Show that if H and K are subgroups

of an abelian group G, then

{hk|h屬於H and k屬於K}

is a subgroup of G.

Question 6

(a) Give a detailed definition for a set G with a binary

operation ¤ to be a group.

(b) A binary operation ¤ on (立體的Z) is defined by

x ¤ y = x + y + 1 for x,y 屬於(立體的Z)

Show that (立體的Z) is a group under ¤.

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Question4

Let H be a non-empty subset of the

group G. Show that H is a subgroup of G if and only

if ab(-1在右上角小寫)屬於H (倒A)a,b屬於H

------

=>)∀ a,b ∈H,since H is a subgroup of G,϶c∈H,s.t.bc=cb=e(identity)

=>b^-1bc=b^-1e =>c=b^-1 (也就是再證了subgroup的反元素即是原group的反元素)

c∈H i.e.b^-1∈H,hence ab^-1∈H (closed)

(其實如果之前已學過子集合的單位元素與反元素即是原群中之單位元素與反元素

則此方向只須用運算的封閉性來說即可)

<=)(1)choosea,a∈H(nonempty),then aa^-1∈H, i.e. e∈H,

(2)∀a∈H,ea^-1∈H, i.e. a^-1∈H

Hence H is a subgroup of G

-----------------------------------------

Question 5

Show that if H and K are subgroups of an abelian group G, then

{hk|h屬於H and k屬於K} is a subgroup of G.

Let G*={hk|h∈H and k∈K}

(1)Given a=h1k1∈G*,b=h2k2∈G*

then ab=h1k1h2k2=(h1h2)(k1k2)(因abelian)∈G*,i.e. G* is closed.

(2)e∈H,e∈K, then e=e*e∈G*

(3)∀a∈G*,϶h∈H,k∈K,s.t.a=hk ,since h^-1∈H,k^-1∈K,we obtain h^-1k^-1∈G*

let b=h^-1k^-1, then ac=hkh^-1k^-1=hh^-1kk^-1(abelian)=ee=e,similarly ca=e

By (1)(2)(3) we conclude that G*={hk|h∈H and k∈K} is a subgroup of G.

------------------------------------

Question 6

(a) Give a detailed definition for a set G with a binary operation to be a group.

(b) A binary operation on (立體的Z) is defined by x y = x + y + 1 for x,y 屬於(立體的Z)

Show that (立體的Z) is a group under .

(a)請抄課本,因為這是定義

(b)我檢驗幾個重要的

(1)closed封閉性: trivially x+y+1 ∈ Z

(2)結合律: (a b )c=(a+b+1)c=a+b+1+c+1=a+b+c+2

a (b c)=a(b+c+1)=a+(b+c+1)+1=a+b+c+2, hence (a b )c=a (b c)

(3)identity單位元素: e=-1,since x(-1)=x+(-1)+1=x,(-1)x=(-1)+x+1=x

(4)inverse反元素:given x,then x(-x-2)=x+(-x-2)+1=-1(e),(-x-2)x=(-x-2)+x+1=-1(e)

i.e. each x has an inverse=(-x-2)

By above , we conclude that Z is a group under

2008-11-02 13:29:31 補充：

還好他問的是很基本的,再難些,我應該也是要投降

Source(s): 國中數學老師的腦袋
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