Proving Isomorphism of Complex Number Conjugate: General Algebraic Systems

In summary, This conversation is discussing the problem of proving that the mapping a+bi->a-bi of each complex number onto its conjugate is an automorphism of the additive group of C. The solution presented seems to be correct, though it may benefit from showing surjectivity as well. The conversation also discusses the difficulty of the problem and the pace of learning mathematics as a self-taught student. It is suggested to take more time to fully understand the material if it involves more advanced concepts such as proofs for abstract algebra.
  • #1
complexPHILOSOPHY
365
2
I am working through this algebra book and some of the problems. The chapter this comes out of is General Algebraic Systems and the section is Isomorphisms. I am new to proofs and maths higher than calculus I so I am not sure if I am following the text or not. There aren't any solutions and this book is really old, out of print and not on the internet (temporary until I can purchase one of the texts already suggested to me).

Here is the problem:

Prove that the mapping [tex] a+bi\rightarrow a-bi [/tex] of each complex number onto its conjugate is an automorphism of the additive group of C.

[tex]
(a+bi)\alpha=a-bi
[/tex]
[tex]
(c+di)\alpha=c-di
[/tex]

[tex]
[(a+bi)+(c+di)]\alpha=[/tex]
[tex]
[(a+c)+(b+d)i]\alpha=[/tex]
[tex]
(a+c)-(b+d)i=[/tex]
[tex]
(a-bi)+(c-di)=[/tex]

[tex]
(a+bi)\alpha+(c+di)\alpha
[/tex]

Since [tex] a-bi=c-di [/tex]
Then [tex] (a-c)+(d-b)i=0 [/tex]

[tex]a=c [/tex] and [tex] b=d [/tex]

[tex] a+bi=c+di [/tex]

This is one-to-one right?

I think this is correct but I am not sure if it is even complete or if I am on the right track. If someone could help me, that would be great.
 
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  • #2
Everything looks fine.

You've shown that the mapping preserves the binary operation under each respective group.

You've shown injectivity with

[tex]a=c [/tex] and [tex] b=d [/tex]

Technically, you only need to show surjectivity, though it's a trivial exercise.

Other than that, everything looks perfectly fine.
 
  • #3
Kreizhn said:
Everything looks fine.

You've shown that the mapping preserves the binary operation under each respective group.

You've shown injectivity with

[tex]a=c [/tex] and [tex] b=d [/tex]

Technically, you only need to show surjectivity, though it's a trivial exercise.

Other than that, everything looks perfectly fine.

An automorphism is an isomorphism from a set of elements onto itself, right? So, by demonstrating surjection as well, would I be illustrating a bijection? Do I need to show surjection for proof or were you suggesting it to help improve my understanding?

I appreciate your response. Also, since my book seems to be the only copy that was printed on the earth, I don't know how difficult the problems are considered. This problem seemed almost too simple (considering I have no idea what I am doing) so I am wondering if my book is very easy? Does this problem look pretty average or would you suggest more difficult problems to get a better intuitive feel for groups? I am a complete algebra newbie. This is my third day reading through the text, so bear with me.

Also, one further question. How long does the average person spend on each chapter when working through maths? I am completely self-taught all the way up to Calculus 1 and I have no experience learning math from anyone except books, so I am not sure what the pace of average students is compared to me.

I am curious because I will be taking Calculus II this semester and I am wondering if I will be slower than them.
 
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  • #4
Well, from what I recall, the definition for a group isomorphism is

Definition The map [tex]\phi : G \rightarrow H [/tex] is an isomorphism if

1) [tex]\phi[/tex] is a homomorphism

2) [tex]\phi[/tex] is a bijection

where the group homomorphism preserves the binary operation mapped from G to H, and a bijective function is both injective and surjective. With that in mind, it's necessary, although relatively trivial to show that the complex conjugate is a one-to-one correspondence (i.e. Injective and Surjective).

It isn't usually until somewhat higher level courses that one is introduced to this kind of algebra (except possibly in an advanced classical algebra class), and so it's outstanding that you'd be learning this on your own.

As for this question in particular, it does indeed seem fairly rudimentary, but the difficulty of question you might want depends greatly on your background. Have you been introduced to the formal idea of proofs for abstract algebra? Is that the kind of thing you'd be interested in?

It's also somewhat hard to gauge what exactly can be meant by a chapter, and I suppose that also depends on the text that you're using. I too learned Calculus I and II via the textbook, and I found that doing a "chapter" daily was very sufficient. However, note that a university student doing Calc II typically only has this lecture 3 times a week (at least where I come from) and so a chapter daily would actually be an advanced pace.

I would imagine that this also depends on the amount of theoretical material presented in the course. When you studied Calc I were you introduced to the [tex]\epsilon\delta[/tex] notion of continuity? Cauchy Sequences? The notion of compact sets? Banach Spaces? Lipschitz, Topological, or metric definitions of continuity? I don't believe these are taught outside of advanced Calculus classes, but if this is the kind of thing you are learning, it wouldn't hurt to take a bit more time to ensure that you fully understand the material.
 

What is an isomorphism?

An isomorphism is a mathematical concept that describes a one-to-one correspondence between two structures or systems, such that they have the same underlying structure and can be transformed into one another without losing any essential properties.

How is isomorphism related to complex number conjugates?

In the context of complex numbers, isomorphism can be used to prove that the conjugate of a complex number is unique. This means that for any given complex number, there is only one complex number that when multiplied together, result in a real number.

What is the process for proving isomorphism of complex number conjugates?

The process for proving isomorphism of complex number conjugates involves showing that the mapping between the complex number and its conjugate is a bijective homomorphism. This means that the mapping is both one-to-one and onto, and it preserves the algebraic structure of the complex number system.

Why is proving isomorphism of complex number conjugates important?

Proving isomorphism of complex number conjugates is important because it allows us to understand the fundamental properties and relationships within the complex number system. It also provides a basis for solving complex algebraic equations and performing other mathematical operations involving complex numbers.

What are some practical applications of understanding isomorphism of complex number conjugates?

Understanding isomorphism of complex number conjugates has practical applications in various fields such as engineering, physics, and computer science. It is used in signal processing, control systems, and cryptography, among others. It also has implications in the study of symmetry and group theory.

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