The proof that the permutation is clearly defined, Ishian can exchange all elements in the group

1.

How to understand the permutation it says.
Qinhuai Family 22:06:34
Is that the elements of each column in each row are not the same,
Jing Weiren's Sorrow 22:07:28
So can I swap the third and second columns for a position?
Jing Weiren's Sorrow 22:07:58
The exchange is over, the new table is still the G replacement?

Qinhuai Family 22:08:47
It doesn't matter how it's changed. As long as columns and columns are changed, rows and rows

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What's the one by one transformation?
Qinhuai Family 21:30:59
Double shot.
Jing Weiren's Sorrow 21:31:14
Oh
Jing Weiren's Sorrow 21:31:42
It's a one by one map.
Qinhuai Family 21:32:24
That's right. But this one-to-all mapping is set to itself, not to another collection

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The permutation in the congruence relationship, how do you understand
Qinhuai Family 22:18:41
For an operation, if you replace the operand with an element that is equivalent to it, then the result is also equivalent to the original result, a~ B, c~ D, ac~ BD, which is the permutation. ac~ BD This is the replacement which is needed to prove.

Qinhuai Family 22:21:49
Not for every one of them.

Jing Weiren's Sorrow 22:34:23
Displacement occurs in 2 places: one is congruence, and the other is the permutation group of the group. Is there a similar place?
Qinhuai Family 22:37:53
are permutations, except that the objects are different.
Qinhuai Family 22:38:36
Permutation of operands
Jing Weiren's Sorrow 22:39:19
a~ B, c~ D, ac~ BD It's a replacement.

Qinhuai Family 22:40:28
That is, the equivalent of two elements as operands, you can use one permutation of the other, the result remains an equivalence relationship, so called to meet the permutation, for example, A and B can be exchanged, C and D can be exchanged.

Modify here, Date Modified 2010.09.18,08:01am:

Add: Last night came back to discuss this problem with Zheng Zhigao classmate, below add my humble opinion:

If I give you an alphabet of a={a,b,c,d}, then the algebraic system v=<a,o> form an operation table, then now tell you that the table satisfies a relationship R, the relationship R, is the equivalence relationship on a, and R for all operations in V (where there is only one operation O) has a permutation nature, then you have to prove that R is the congruence relationship on V.

Obviously this satisfies the basic definition of the congruence relationship, so you have to prove that you have to use a more basic definition, that is, I want to prove that the basic nature of the congruence, that is, the R has reflexivity, symmetry, transitivity, and finally it to all operations in V (here only one o) has a permutation nature, then I only care about the proof of We combine the old willow to explain, I understand, that is, by a~b,c~d I want to go to the AC~BD, that is to prove that AC in the calculation table about the operation o corresponds to the value, to be equivalent to the BD in the table of Operations o the corresponding value, if the relationship is satisfied, Then we can say that the equivalence relation on A is a permutation of all operations in V (there is only one operation O here), plus the first proof of R is reflexive, symmetric, transitive, that is, R is the congruence relationship on V.

2010.11.16,09:15AM:

In the morning, I received a short message from Xiao Zheng, very surprised and happy. He said: "Since I have not been clear about the concept of" permutation ", the contents of PKU's discrete textbook algebra system can read Wengfeng's" abstract algebra ". In the book, "permutation" is defined as:

The reversible mapping of a non-empty finite set A to a itself becomes a permutation on a, which is a permutation of a.

Thank Xiaozheng very much, remember this matter, all the past 2 months, very touched.

I replied to the definition of "permutation" that I understood and summed up:

The permutation on a non-empty finite set A is the substitution of all operations (that is, operations) that are acting on the set a as a function of any element in set a (i.e., operations).

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If there is only 2 order in Group G, then this element can be exchanged with all elements in G.
Please give proof.
The answers are as follows:
Proof: Set Ishian to X, to any y belongs to G have: |yxy^-1|=|x|=2
Launched Yxy^-1=x
Launched Yx=xy

Jing Weiren's Sorrow 21:42:41
|yxy^-1| because I have the following textbook content as the basis, so it can be understood as a subgroup
The elements in

Qinhuai Family 21:46:04
Note that X is Ishian
Jing Weiren's Sorrow 21:46:07
Only E in the group has a power-equal law
Qinhuai Family 21:46:55
Who used the power of the law.
Jing Weiren's Sorrow 21:46:55
Yxy^-1 is a few orders.
Qinhuai Family 21:47:00
2 steps.
Jing Weiren's Sorrow 21:47:05
Why
Qinhuai Family 21:47:47
Yxy^-1yxy^-1=yxxy^-1=yy^-1=e
Jing Weiren's Sorrow 21:48:04
Yxy^-1, how did you get here?
Qinhuai Family 21:48:37
What's the matter, I've been fooled by you.
Jing Weiren's Sorrow 21:48:54
Why write Yxy^-1.
Qinhuai Family 21:49:43
You're not trying to prove he's a Ishian. I'll take two of them and get E.
Jing Weiren's Sorrow 21:50:17
I don't want to be second-order.
Jing Weiren's Sorrow 21:50:25
I need a license to exchange.
Jing Weiren's Sorrow 21:50:38
You read the question.
Qinhuai Family 21:51:13
Two orders, and said there is only a second order, so yxy^-1=x
Qinhuai Family 21:51:19
No, just come out?
Jing Weiren's Sorrow 21:53:14
So how do you start thinking about using Yxy^-1 to testify? How to think, how to conceive.
Qinhuai Family 21:53:29
I see your answer.

Qinhuai Family 22:03:42
But look at the answer after the idea can be pushed ah. You have to testify against any y Yx=xy, which is yxy-1=x.