math_nerd
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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.
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.