Which equalities makes phi to be automorphism?

[Sauraj]

Hello,
I have this structure \(\displaystyle \mathcal{G} = \{ \mathbb{R}, +^ \mathcal{G}, 0^ \mathcal{G} \}\) and I used this automorphism \(\displaystyle
\psi\,\colon = x \mapsto -x\) to prove something (that "<" cannot be expressed with this structure). Which equalities makes psi to be an automorphism?
Thx

 

[Sauraj]

is it enough to say that \(\displaystyle \psi\) is a automorphism because for every \(\displaystyle a, b \in \mathbb{R}\) it holds \(\displaystyle \psi(a+b) = \psi(a)+\psi(b)\) without proving this equation?

 

[Sauraj]

Well in general saying that \(\displaystyle \psi\) goes from \(\displaystyle \mathbb{R}\) to \(\displaystyle \mathbb{R}\), is bijective and is homomorphism would be enough as explanation that \(\displaystyle \psi\) is automorphism?

 

[Dr.Peterson]

Which equalities makes psi to be an automorphism?

Are you saying you are supposed to PROVE that it is, or just to ASSERT it?

It will be helpful if you tell us the exact wording of the problem you are working on. There is a lot of context missing here.

 

[Sauraj]

Are you saying you are supposed to PROVE that it is, or just to ASSERT it?

It will be helpful if you tell us the exact wording of the problem you are working on. There is a lot of context missing here.

I just want to mention the required properties. I don't want to prove them.
So just properties that makes \(\displaystyle \psi\) automorphism.

 

[Dr.Peterson]

I just want to mention the required properties. I don't want to prove them.
So just properties that makes \(\displaystyle \psi\) automorphism.

I presume you have looked up the definition of automorphism in your textbook. What did you find?

Here is one source: https://en.wikipedia.org/wiki/Automorphism

Notice this statement:

An automorphism is simply a bijective homomorphism of an object with itself. (The definition of a homomorphism depends on the type of algebraic structure; see, for example, group homomorphism, ring homomorphism, and linear operator).​

So, what kind of structure do you have? Then look up the definition for homomorphism in that context.

 

[Sauraj]

Ok I have the definition now.
How should I say, the function is a isomorphism or the function is isomorph (isomorphic)?

 

[Dr.Peterson]

Ok I have the definition now.
How should I say, the function is a isomorphism or the function is isomorph (isomorphic)?

The function is an isomorphism (noun); the structures are isomorphic (adjective). I'm not sure isomorph is a word; if it is, it would be a noun, and you might say one structure is an isomorph of another.

I looked it up, and it's a word, but apparently not in math ...

 

[Sauraj]

I presume you have looked up the definition of automorphism in your textbook. What did you find?

Here is one source: https://en.wikipedia.org/wiki/Automorphism

Notice this statement:

An automorphism is simply a bijective homomorphism of an object with itself. (The definition of a homomorphism depends on the type of algebraic structure; see, for example, group homomorphism, ring homomorphism, and linear operator).​

So, what kind of structure do you have? Then look up the definition for homomorphism in that context.

(unfortunately) they want a prove that \(\displaystyle \psi\) is automorphism. I can prove it by saying that it is a bijection and it is an isomorphism.
This is obvious but I can't write it in mathematical language.
My try:
The function \(\displaystyle \psi\ : \mathbb{R} \mapsto \mathbb{R}, \psi(x)=-x\) is bijective, since for each \(\displaystyle y\) there is a unique \(\displaystyle x = -y \) such that \(\displaystyle \psi(x) = y\)

How to show isomorphism? Is it now enough to say that \(\displaystyle \psi\) is Homomorphism, since
\(\displaystyle \psi(x+y) = \psi(x)+\psi(y) \) and for the neutral element in the structure it holds \(\displaystyle \psi(0)=0\) and bijective+homomorphism=automorphism.
Its too many things here for my, I cannot handle with them and come to the conclusion that psi is automorphism.

 

[Dr.Peterson]

(unfortunately) they want a prove that \(\displaystyle \psi\) is automorphism. I can prove it by saying that it is a bijection and it is an isomorphism.
This is obvious but I can't write it in mathematical language.
My try:
The function \(\displaystyle \psi\ : \mathbb{R} \mapsto \mathbb{R}, \psi(x)=-x\) is bijective, since for each \(\displaystyle y\) there is a unique \(\displaystyle x = -y \) such that \(\displaystyle \psi(x) = y\)

How to show isomorphism? Is it now enough to say that \(\displaystyle \psi\) is Homomorphism, since
\(\displaystyle \psi(x+y) = \psi(x)+\psi(y) \) and for the neutral element in the structure it holds \(\displaystyle \psi(0)=0\) and bijective+homomorphism=automorphism.
Its too many things here for my, I cannot handle with them and come to the conclusion that psi is automorphism.

That's probably enough; but as I pointed out, it depends on what your "structure" is, so I can't say anything positively without knowing the context.

 

[Sauraj]

That's probably enough; but as I pointed out, it depends on what your "structure" is, so I can't say anything positively without knowing the context.

structure is: \(\displaystyle \mathcal{G} = \{ \mathbb{R}, +^ \mathcal{G}, 0^ \mathcal{G} \}\)

 

[Sauraj]

I have now everything