Homomorphism question

  • Thread starter Juanriq
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Homework Statement

Consider the mapping [itex] \phi \colon (\mathbb{Z} , + ) \to (\mathbb{Z}, *) [/itex] such that [itex] \phi (a) = a+2[/itex]. Define * such that [itex] \phi [/itex] is a homomorphism. For [itex] (\mathbb{Z}, *) [/itex], define the identity element.

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The Attempt at a Solution

Well, first I thought I was oversimplifying this problem. I know that since we have a homomorphism, the identity of [itex] (\mathbb{Z} , \plus ) [/itex] will map to the identity element of our image. The identity of [itex] (\mathbb{Z} , \plus ) [/itex] will be 0, so [itex] \phi (0) = 1 = e^{\prime} [/itex]. But that seemed too easy.


So my question, do I have to break out properties of the homomorphism, namely that [itex] \phi (a+ b) = \phi (a) * \phi (b) = (a+2) * (b+2) [/itex] somehow.

I appreciate any help, thanks!
 

Answers and Replies

  • #2
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So for z,z', we need to know what z*z' is. Lets presume first that [tex]\phi[/tex] is indeed an homomorphism.

We know that [tex] z=\phi(z-2) [/tex] and [tex] z^\prime=\phi(z^\prime-2) [/tex]. Since [tex]\phi[/tex] is a homomorphism:

[tex] z*z^\prime= \phi(z-2)*\phi(z^\prime-2)=\phi(z+z^\prime-4)=z+z^\prime-2 [/tex]

So it is natural to define z*z'=z+z'-2. So now you just need to check that * indeed defines a group structure on Z and that [tex]\phi[/tex] indeed defines a homomorphism...
 

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