Show mapping is an automorphism

[math_nerd]

Show that the mapping Phi(a+bi)=a-bi is an automorphism of the group of complex numbers under addition.

I have this as of now:

Let (c+di) and (a+bi) be elements in group G.

1) phi(a+bi) is function phi from G to G, by assumption. Therefore the function is a mapping.

2) 1-1: assume that phi(a+bi) = phi (c+di), so a-bi = c-di.
Then, how can I show that this somehow equals a+bi = c+di?

3) onto: phi is onto if for every element g' in G, we can find an element g in G, s.t. Phi(g)=g'. So phi(a+bi) = c-di. Where c-di is an element in G(?). With some arithmetic, I got phi(a+bi) = a-bi= c+di.

4) addition is preserved, by showing phi(a+bi+c+di) = phi(a+bi) + phi(c+di) is true.

 

[Dick]

For part 2, a+bi=c+di only if the real parts are equal and the imaginary parts are also equal, yes? a,b,c and d are real. Doesn't that mean a=c and b=d?

 

[math_nerd]

So for part 2, I can say that a-bi=c-di, so a=c and b=d must be true for a-bi=c-di. This then implies a+bi=c+di is true.

Also, is rest of the proof right?

 

[Dick]

So for part 2, I can say that a-bi=c-di, so a=c and b=d must be true for a-bi=c-di. This then implies a+bi=c+di is true.

Also, is rest of the proof right?

The phrasing on the rest of it is pretty vague. I just responded to the only part you had a '?' on. Why don't you detail a little more exactly what your question is? Focus on one part at a time.

 

[math_nerd]

So I need to show that the mapping Phi(a+bi)=a-bi is an automorphism of the group of complex numbers under addition. The way I have proved it is to show that it is a mapping (we are given this piece of information), the mapping is 1-1 and onto, and the operation is preserved. Also, I assumed that g=a+bi and g'=c+di, where g,g' are some elements in the group, G. What looks vague about it?

 

[math_nerd]

So I need to show that the mapping Phi(a+bi)=a-bi is an automorphism of the group of complex numbers under addition. The way I have proved it is to show that it is a mapping (we are given this piece of information), the mapping is 1-1 and onto, and the operation is preserved. Also, I assumed that g=a+bi and g'=c+di, where g,g' are some elements in the group, G. What looks vague about it?

So I have to show group G maps to itself.

 

[Dick]

So I need to show that the mapping Phi(a+bi)=a-bi is an automorphism of the group of complex numbers under addition. The way I have proved it is to show that it is a mapping (we are given this piece of information), the mapping is 1-1 and onto, and the operation is preserved. Also, I assumed that g=a+bi and g'=c+di, where g,g' are some elements in the group, G. What looks vague about it?

Nothing if the parts you aren't showing are ok. Like "So phi(a+bi) = c-di. Where c-di is an element in G(?). With some arithmetic, I got phi(a+bi) = a-bi= c+di.". I'm not sure what that means and I'm not sure what part of this you are having problem with.

 

[math_nerd]

I want to know if the way I have choose g'=c-di and g=a+bi to be two elements in G is written with the correct logic?

For the part I said I did the arithmetic, I did incorrectly.
I want to prove the mapping from G to G is onto.
So I need to show that phi(a+bi)= c+di. How?

 

[Dick]

I want to know if the way I have choose g'=c-di and g=a+bi to be two elements in G is written with the correct logic?

For the part I said I did the arithmetic, I did incorrectly.
I want to prove the mapping from G to G is onto.
So I need to show that phi(a+bi)= c+di. How?

Pick any element of G and call it a+bi. a-bi is also in G. phi(a-bi)=a+bi. So a+bi is in the range of phi and it's onto. There's no need to introduce a 'c' and 'd'.

 

[math_nerd]

Okay thanks!