Prove Beta is an isomorphism of groups

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Discussion Overview

The discussion centers on proving that the function Beta, defined from the group of complex numbers under addition to itself, is an isomorphism of groups. The focus is on understanding the properties of the function, specifically its injectivity (1-to-1), surjectivity (onto), and whether it preserves the group operation.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant expresses difficulty with the concepts of injectivity and surjectivity, specifically in the context of the function Beta defined as Beta(a+bi) = a-bi.
  • Another participant attempts to demonstrate that Beta is injective by showing that if Beta(a+bi) = Beta(c+di), then it follows that a = c and b = d.
  • A further point is made regarding surjectivity, suggesting that for any element y in the codomain, there exists an element a+bi in the domain such that Beta(a+bi) = y.
  • One participant notes that since Beta has an obvious inverse (itself), it must be a bijection, implying that it suffices to prove that it is a homomorphism of additive groups.

Areas of Agreement / Disagreement

Participants appear to be exploring the properties of the function Beta, with some agreement on the necessity of proving it is a homomorphism. However, the discussion does not reach a consensus on the overall proof of isomorphism.

Contextual Notes

Some assumptions about the nature of the integers and the properties of addition in the context of complex numbers are implied but not explicitly stated. The discussion also does not resolve all mathematical steps necessary for a complete proof.

SqrachMasda
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i can't grasp these concepts, 1-to-1 and onto have always annoyed me.

here's 1 question, (i don't 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 a lot of these problems but maybe a one may help me understand the rest
 
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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)...
 
thanks nnnnnnnn
 
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.
 

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