MHB Groups of Automporphisms - Aut(C) .... Anderson and Feil Ch. 24 ....

[Math Amateur]

I am reading Anderson and Feil - A First Course in Abstract Algebra.

I am currently focused on Ch. 24: Abstract Groups ... ...

I need some help in understanding some claims in Chapter 24 by Anderson and Feil ... ...Anderson and Feil claim that $$\text{Aut} ( \mathbb{C} ) $$is a group with only two elements $$\{i, f \}$$ ... ... where $$i$$ is the identity automorphism and $$f$$ is the complex conjugation map defined by $$f(a + bi) = a - bi$$ ... ...

... can someone please help me to prove the assertion that $$\text{Aut} ( \mathbb{C} ) $$is a group with only two elements $$\{i, f \}$$ ...

Peter

 

[Euge]

Hi Peter,

Could you check back Anderson and Feil to see if they meant the automorphisms of $\Bbb C$ fixing $\Bbb R$?

 

[Math Amateur]

Hi Peter,

Could you check back Anderson and Feil to see if they meant the automorphisms of $\Bbb C$ fixing $\Bbb R$?

I am beginning to understand that whether the automorphisms of $\Bbb C$ fix $\Bbb R$ may be an issue ... but from my reading of Anderson and Feil they seem to imply that the the automorphisms of $\Bbb C$ actually fix $$\mathbb{R}$$ anyway ... but you seem to be implying that this is not necessarily the case ...

The relevant text from Anderson and Feil is as follows:First the example that started me thinking:
https://www.physicsforums.com/attachments/6855
Now the above example involves a Galois Group ... and in this group $$\mathbb{R}$$ is certainly fixed under Anderson and Feil's definition of a Galois Group ... which reads as follows:

https://www.physicsforums.com/attachments/6856
https://www.physicsforums.com/attachments/6857

Now, Example 47.3 refers to Exercise 24.14 which asserts that $$\text{Aut} ( \mathbb{C} )$$ has only two elements ... note the exercise does not mandate the fixing of $$\mathbb{R}$$ ... Exercise 24.14 reads as follows:View attachment 6858Hope that clarifies what Anderson and Feil have to say about the issue ... ...

Peter

 

[Euge]

If Anderson and Feil reject the axiom of choice, then it would make sense for them to say $\operatorname{Aut}(\Bbb C)$ has two elements. However, if they do accept the axiom of choice, their statement about $\operatorname{Aut}(\Bbb C)$ would be incorrect, for in that case, $\operatorname{Aut}(\Bbb C)$ would be uncountably infinite.

 

[Math Amateur]

If Anderson and Feil reject the axiom of choice, then it would make sense for them to say $\operatorname{Aut}(\Bbb C)$ has two elements. However, if they do accept the axiom of choice, their statement about $\operatorname{Aut}(\Bbb C)$ would be incorrect, for in that case, $\operatorname{Aut}(\Bbb C)$ would be uncountably infinite.

Thanks Euge ...

Well ... I am surprised by that ... can you please explain how the Axiom of Choice makes such a difference in this matter ...

... it is not obvious to me how the Axiom of Choice makes such a difference ...

Peter

 

[Euge]

With the axiom of choice, one can use the concept of a transcendence base to construct field automorphisms of $\Bbb C$ different from the identity and complex conjugation. More generally, it can be shown that the cardinal number of the field automorphisms of an algebraically closed field $\Bbb k$ is $2^{\text{card}(\Bbb k)}$. Since $\Bbb C$ has cardinal number $2^{\aleph_0}$, then with axiom of choice, the cardinal number of $\operatorname{Aut}(\Bbb C)$ is $2^{2^{\aleph_0}}$.

A natural question arises: how much choice is required for $\text{card}(\operatorname{Aut}(\Bbb C)) > 2$? I'm pretty certain you don't need the full AC for it to hold, but I don't know what the weakest choice assumption needs to be. Perhaps a logic expert like Evgeny.Makarov knows.

 

[Math Amateur]

With the axiom of choice, one can use the concept of a transcendence base to construct field automorphisms of $\Bbb C$ different from the identity and complex conjugation. More generally, it can be shown that the cardinal number of the field automorphisms of an algebraically closed field $\Bbb k$ is $2^{\text{card}(\Bbb k)}$. Since $\Bbb C$ has cardinal number $2^{\aleph_0}$, then with axiom of choice, the cardinal number of $\operatorname{Aut}(\Bbb C)$ is $2^{2^{\aleph_0}}$.

A natural question arises: how much choice is required for $\text{card}(\operatorname{Aut}(\Bbb C)) > 2$? I'm pretty certain you don't need the full AC for it to hold, but I don't know what the weakest choice assumption needs to be. Perhaps a logic expert like Evgeny.Makarov knows.

Thanks Euge ...

Still reflecting on this post ... need to read and think around this matter ...

Peter

 

[Euge]

I would advise you not to think too deeply about this, for these issues are difficult to deal with, even for professional mathematicians. Just ignore the exercise as it is a faulty one.

 

[Math Amateur]

I would advise you not to think too deeply about this, for these issues are difficult to deal with, even for professional mathematicians. Just ignore the exercise as it is a faulty one.

Thanks for the advice/warning Euge ...

I will now leave this matter and move on to more Galois theory ...

Thanks for your help on this issue/problem ...

Peter