1202 ex1 Q3:

1/20/2007

Q3:

Because G is a finite set,

So we can say {g1,g2,g3...gn}=G

For each gi belongs to G, define a new set F={f1,f2,f3...fn}=G

S.T.

gi * g1 = f1

gi * g2 = f2

gi * gn = fn

because * is an associative binary operation so * is a map G*G-->G

so f1, f2, ... fn belongs to G

If fp = fq, fp,fq belongs to F

i.e. gi*gp = gi*gq

==> gp = gq by cancellation law.

so F=G

so there is a fj = gi I call it "fj", which also known as "gi" in G

i.e. gi*gj=gi=gi*e

so gj = e

so there is a fk = gj I call it "fk", which also known as "gj" in G

i.e. gi*gk=gj=e

so gi has an inverse element, which is gk.

so G is a group.

QED

4:06 PM | Blog it | maths

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	shadi wangwrote: 樊兄高见，愚不敢妄评。 ==我把题目找出来。 Feb. 19
	费斯坦但提勒斯wrote: Q3: 因为G 是有限集所以令 {g1,g2,g3...gn}=G 因为每个gi 属于 G, 定义新集 F={f1,f2,f3...fn}=G 使得 gi * g1 = f1 gi * g2 = f2 : : : gi * gn = fn 因为 * 是一个相关二元运算所以 * 是一个 GG-->G的集合所以 f1, f2, ... fn 属于 G 若 fp = fq, fp,fq 属于F 即由相消率gigp = gigq ==> gp = gq 所以 F=G 所以有一个 fj = gi 称之为"fj", 在G中被认为是"gi" 即 gigj=gi=gie 所以 gj = e 所以这有一个 fk = gj 称之为 "fk",在G中被认为是"gj" 即 gigk=gj=e 所以 gi 有一个逆元素gk 所以G是一个群证毕翻的不对之处，还望沙地兄指出。另外能不能把题写出来？ Feb. 19
	zhang00000wrote: 我英语不好这是问题还是解答如果是解答问题在哪 Feb. 16
	yaoyao zhouwrote: 大家都是学数学的，为什么我看这道题晕成这样，你们都住得很远吗？要用这种　方式～～这可能就是bath这个小小campus的优点了吧~~ Jan. 31
	靓李wrote: 哦买糕... 博客真是多种功效呀... Jan. 27
	Ellawrote: haha, 沙地，你还真绝！ Jan. 20
	昕钱wrote: 弟弟啊，姐姐真是感激涕零啊。真是public resource...... 谢了，周一请你吃糖哦。 Jan. 20

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