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## Main Question or Discussion Point

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:

[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.

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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