令集合 G 从一个群 (G , o) 定义二元运算 o 。如果 G 满足以下 3 个性质,则 G 是一个群:
- 关联性
- 身份
- 逆
组的属性:
属性-1 :
如果 a , b, c ∈ G 那么,是 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]
注意 aob 也写成 ab。
这被称为左抵消定律。
属性-2:
对于每个 a ∈ G ,eoa = a = aoe,其中 e 是单位元素。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即左消消法。因此我们发现 e 也是正确的单位元,所以它只被称为单位元。
属性 3:
对于每个 a ∈ G , a -1 oa = e = aoa -1即一个元素的左逆也是它的右逆。
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也是它的右逆,因此 a -1仅称为 a 的逆。
属性 4:
如果 a , b, c ∈ G 那么,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]
这被称为权利取消法。
属性-5:
对于每个 a , b ∈ G 我们有 (aob) -1 = b -1 oa -1即 G 组的两个元素 a、b 的乘积(或复合)的逆是取相反顺序的两个元素的倒数。
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,由逆 b -1 oa -1的定义是 ao bie (aob) -1 =b -1 oa -1 的逆
这被称为反转规则。