让其从组(G,o)中定义了二进制运算o的集合G。如果满足以下三个属性,则G是一个组:
- 关联性
- 身份
- 逆
组的属性:
属性1 :
如果a,b,c∈G那么aob = aoc⇒b = c
证明: –
Given a o b = a o c, for every a, b, c ∈ G
Operating on the left with a-1, where a-1 ∈ G we have
a-1 o (a o b) = a-1 o (a o c)
or (a-1 o a) o b = (a-1 o a) o c [using associative property]
or e o b = e o c, [using inverse property]
or b = c, [using identity property]
请注意,aob也写为ab。
这被称为左抵消法。
属性2:
对于每个a∈G,eoa = a = aoe,其中e是恒等元素。即,左标识元素也是右标识元素。
证明: –
If a-1 be the left inverse of a, then
a-1 o (a o e) = (a-1 o a) o e [using associative property]
or a-1 o (a o e) = e o e [using inverse property]
= e [using identity property]
or a-1 o (a o e) = a-1 o a [using inverse property]
i.e. a-1 o (a o e) = a-1 o a
因此,aoe = a by property-1,即左抵消定律。因此,我们发现e也是正确的标识元素,因此仅称为标识元素。
属性3:
对于每个a∈G,-1 oa = e = aoa -1,即元素的左逆也是其右逆。
证明: –
a-1 o (a o a-1) = (a-1 o a) o a-1 [using identity property]
= e o a-1 [using inverse property]
= a-1 o e [by property 2]
i.e. a-1 o (a o a-1)= a-1 o e
Hence, a o a-1 = e, by left cancellation law.
因此,我们发现元素a的左逆a -1也是其右逆,因此a -1仅被称为a的逆。
属性4:
如果a,b,c∈G,则是boa = coa⇒b = c
证明: –
Given a o b = a o c, for every a, b, c ∈ G
Operating on the left with a-1, where a-1 ∈ G we have
(b o a) o a-1 = (c o a) o a-1
or b o (a-1 o a) = c o (a-1 o a) [using associative property]
or b o e = c o e, [using inverse property]
or b = c, [using identity property]
这就是所谓的权利取消法。
属性5:
对于每个a,b∈G,我们有(aob) -1 = b -1 oa -1,即G组两个元素a,b的乘积(或复合物)的倒数是a的乘积(或复合物)。以相反顺序获取的两个元素的逆。
证明: –
Let a-1 and b-1 be the inverses of a and b.
Now,(a o b) o (b-1 o a-1) = a o (b o b-1) o a-1 [using associative property]
= a o e o a-1 [using inverse property]
= a o a-1 [using identity property]
= e [using inverse property]
(a o b) o (b-1 o a-1) = e
Similarly, (b-1 o a-1) o ( a o b)= e
因此,根据逆b -1的定义oa -1是ao bie(aob) -1 = b -1 oa -1的逆
这称为逆转规则。