Proving Isomorphism: Aut n(K) and Symmetric Group Sq^n

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In summary: An endomorphism is a map from a set to itself. An automorphism is an endomorphism that is a bijection. So an automorphism is a special kind of endomorphism.In summary, the conversation is discussing how to prove that the group of polynomial mappings over a finite field of q elements, denoted as Aut n(K), is isomorphic to the symmetric group Sq^n. N is the number of variables/indeterminants and q is the number of elements in the field. The conversation also explores the definition of polynomial mappings and the difference between endomorphisms and automorphisms. There is confusion over the terminology and the problem statement, and the conversation concludes with a plan to clarify
  • #1
grimster
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how do i prove that Aut n(K) is isomorphic to the symmetric group Sq^n.

K is a finite field of q elements. Aut n(K) is the group of polynomial automorphisms over K.

n i just the number of variables/indeterminants.

so i guess i have to somehow show that Aut n(K) are the permutations of the elements of K. so i guess i have to do something with the coeficients of a polynomial...
 
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  • #2
Any function from a finite field to itself is (equivalent to) a polynomial...
 
  • #3
Hurkyl said:
Any function from a finite field to itself is (equivalent to) a polynomial...

i'm sorry? how does this help me?


i rewrite the question, becuase it was stated a bit sloppy:

i'm supposed to prove that Aut n(K) is isomorphic to the symmetric group Sq^n.

K is a finite field of q elements. Aut n(K) is the group of polynomial mappings over K.

n i just the number of variables/indeterminants.

Aut n(K) is the so called cremona group. the set of polynomial mappings over K.

here is a definition of a polynomial map:
http://mathworld.wolfram.com/PolynomialMap.html

every mapping from K^n -> k^n is polynomial(i have already proven this).


i guess i should have mentioned that i work with the automorphisms as "mappings". this
is because i can subsitute all X^q by X.
 
  • #4
i'm sorry? how does this help me?

It means you can just say "automorphisms" instead of "polynomial automorphisms".


The more I look at the problem statement, the more it confuses me, and I think I've identified the problem:

Aut n(K) is the group of polynomial automorphisms over K.

Automorphisms of what? I don't recognize the terminology n(K).

If it wasn't for the fact you haven't seemed to be able to connect it to what you're trying to prove, I would have thought you're trying to describe (in a roundabout way) a permutation of the elements of k^n.
 
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  • #5
Hurkyl said:
It means you can just say "automorphisms" instead of "polynomial automorphisms".


The more I look at the problem statement, the more it confuses me, and I think I've identified the problem:



Automorphisms of what? I don't recognize the terminology n(K).

If it wasn't for the fact you haven't seemed to be able to connect it to what you're trying to prove, I would have thought you're trying to describe (in a roundabout way) a permutation of the elements of k^n.

just forget about the automorphisms. use the term map instead.

n i just the number of variables/indeterminants.

aut n(K) the group of polynomial mappings over K.

here is a definition of a polynomial map:
http://mathworld.wolfram.com/PolynomialMap.html

any mapping from K^n -> k^n is polynomial(i have already proven this).
 
  • #6
I repeat -- you are doing a very poor job describing what you mean by "Aut n(k)". If I don't know what that means, I can't help you. More importantly, if you don't know what that means, you can't solve the problem.

In this latest post, you told me that "Aut n(K)" means the group of all functions from k^n --> k^n. That is clearly not isomorphic to a symmetric group.
 
  • #7
Hurkyl said:
I repeat -- you are doing a very poor job describing what you mean by "Aut n(k)". If I don't know what that means, I can't help you. More importantly, if you don't know what that means, you can't solve the problem.

In this latest post, you told me that "Aut n(K)" means the group of all functions from k^n --> k^n. That is clearly not isomorphic to a symmetric group.

i don't get it. what is the problem? aut n(k) are the polynomial mappings over k. I've defined what a polynomial mapping is(the link):
http://mathworld.wolfram.com/PolynomialMap.html

i'm supposed to show that this group is isomorphic to Sn^q. n is the number of variables/indeterminants. q is the number of elements in k.


what is it you don't get? any of the terms? because aut n(K) IS isomorphic to the symmetric group Sq^n. i just have to prove it...
 
  • #8
The group of polynomial mappings from K^n --> K^n is not isomorphic to a symmetric group.

As we've noted, the terms "polynomial map from K^n --> K^n" and "function from K^n --> K^n" are synonymous, because K is finite.

And, we know that the number of functions from a set P to a set Q is [itex]|Q|^{|P|}[/itex]. In this case, that would be [itex]q^{nq^n}[/itex].

However, the symmetric group on [itex]q^n[/itex] elements consists of [itex](q^n)![/itex] permutations.

The groups have different sizes, so they're clearly not equal.


An example of a polynomial map from K^n --> K^n is the map that sends every point of K^n to the zero vector. Is that really something that's supposed to be in your Aut n(K)? Aut n(K), then, wouldn't even be a group!
 
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  • #9
it has to be isomorphic, otherwise my professor is just f*cking with me. he gave us this exercise last week and I'm still struggling with it.

we were supposed to show that aut n(K) is isomorphic to Sq^n and then that
|aut n(k)| = (q^n)!.

i guess i have to ask him in class on monday. this is really confusing...

i think it has something to do with the fact that one can subsitute all apperances of X^q with X.

what you are saying makes sense, because i have already proven that
end n(K) consists of q^nq^q elements. where end n(K) is the set of polynomial mappings k^n -> k^n.
 
  • #10
I think I'm convinced that by n(k), you mean the set of all n-tuples with elements in k... whether you know you mean that is a different question...


Anyways, do you know the difference between an endomorphism and an automorphism?
 
  • #11
Hurkyl said:
I think I'm convinced that by n(k), you mean the set of all n-tuples with elements in k... whether you know you mean that is a different question...


Anyways, do you know the difference between an endomorphism and an automorphism?

auto - iso to itself
endo - homo to itself.
the group end n(K) is obviously bigger than aut n(K).

aut n(K) is the cremona group. the group of polynomial mappings over a finite field K. K has q elements. in this group one can subsitute X^Q by X. the link i provided shows what a polynomial map is.
 
  • #12
What is n(K)? Is it the set of all n-tuples with elements in K?


Now, as you've stated, an automorphism of n(K) is an isomorphism. And as I've stated, not just any old map K^n -> K^n is isomorphic... aut n(K) cannot consist of all maps K^n -> K^n.
 
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  • #13
Hurkyl said:
What is n(K)? Is it the set of all n-tuples with elements in K?


Now, as you've stated, an automorphism of n(K) is an isomorphism. And as I've stated, not just any old map K^n -> K^n is isomorphic... aut n(K) cannot consist of all maps K^n -> K^n.


aut n(K) is the group of invertible polynomial mappings from K^n -> K^n.
 

Related to Proving Isomorphism: Aut n(K) and Symmetric Group Sq^n

1. What is isomorphism in mathematics?

Isomorphism is a concept in mathematics that describes a structural similarity between two mathematical objects. It means that two objects may look different, but they have the same underlying structure. In other words, they can be mapped onto each other while preserving their essential characteristics.

2. What is Aut n(K) and Symmetric Group Sq^n?

Aut n(K) is the automorphism group of a finite group K. It consists of all the bijective mappings from K to itself that preserve the group operation. Symmetric Group Sq^n is a group consisting of all the permutations of n symbols. It is a fundamental group in group theory, often used to describe the symmetry of objects with n distinguishable parts.

3. How do you prove isomorphism between Aut n(K) and Symmetric Group Sq^n?

To prove isomorphism between Aut n(K) and Symmetric Group Sq^n, we need to show that there exists a bijective mapping between the two groups that preserves their group operations. This can be done by constructing an explicit mapping and showing that it is both one-to-one and onto. Additionally, we need to demonstrate that the mapping preserves the group operation, meaning that the product of two elements in one group maps to the product of their images in the other group.

4. Why is proving isomorphism important?

Proving isomorphism between two mathematical objects is important because it allows us to understand and analyze their underlying structures. It also enables us to translate results and properties from one object to another, making it a powerful tool in mathematical research. In the case of Aut n(K) and Symmetric Group Sq^n, proving isomorphism allows us to study the symmetry of finite groups in terms of permutations, providing a deeper understanding of their structural properties.

5. What are some applications of isomorphism?

Isomorphism has many applications in various fields of mathematics, including abstract algebra, group theory, and graph theory. It is also used in computer science to study data structures and algorithms. In real-world applications, isomorphism is used to analyze and compare complex systems, such as chemical compounds, protein structures, and social networks. It is also used in cryptography to create secure encryption algorithms.

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