# Prove Beta is an isomorphism of groups

1. Nov 22, 2004

### SqrachMasda

i cant grasp these concepts, 1-to-1 and onto have always annoyed me.

here's 1 question, (i dont know how to post symbols so Beta ..)
(C is Complex numbers)
Let Beta:<C,+> -> <C,+> by Beta(a+bi)=a-bi (that is, the image is a +(-b)i).
Prove Beta is an isomorphism of groups.

i have alot of these problems but maybe a one may help me understand the rest

2. Nov 22, 2004

### nnnnnnnn

1-1: f(a+bi)=y
f(c+di)=y ->
f(a+bi)=f(c+di)->a-bi=c-di
->a=c & -bi=-di
->-b=-d->b=d as inverse is unique in integers under addition
so if f(x)=f(n)->x=n

onto: if y is an element of f(x) then y=a-bi for some a,b element of integers. But, c+di is an element of <C,+> for all c, d elements of integers. a, b are integers so a+bi is an element of <C,+> so the function is onto.

f(a+bi + c+di) = a-bi + c-di = f(a+bi) + f(c+di)...

3. Nov 22, 2004

### SqrachMasda

thanks nnnnnnnn

4. Nov 23, 2004

### matt grime

don't forget that since f has an obvious inverse map , namely itself, then it must be a bijection, so it suffices to prove it is a homomorphism of additive groups.