Homomorphisms and kernels,images

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Homework Help Overview

The discussion revolves around the function defined from Z12 to Z12, specifically examining whether it is a group homomorphism and exploring its kernel and image.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are computing the function's output and questioning the validity of their calculations. There is confusion regarding the operation used in Z12, leading to a reevaluation of the kernel and image.

Discussion Status

Participants have identified potential elements of the kernel and are actively discussing which elements belong to it. There is a recognition of the need to clarify the operation in Z12, and some guidance has been offered regarding the kernel's elements.

Contextual Notes

There is an ongoing discussion about the definitions and properties of group operations in Z12, particularly focusing on addition versus multiplication, which affects the interpretation of the homomorphism.

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Homework Statement


Show that the function i : Z12 → Z12 defined by i([a]) = 3[a] for all [a] ∈ Z12 is a
group homomorphism and determine the kernel and image.


Homework Equations





The Attempt at a Solution


Well, I started by computing i([a]i()
=3[a]3
=9[ab]
It should equal i[ab], but that equals 3[ab]
 
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Z12 are group on operation addition, not multiplication
 
Oh that makes a lot more sense now.
kernel you want i(a)=e
kernel= [9]
I'm not quite sure about image.
 
[0] is in the kernel too, also [4] and [8] and maybe even more
 
[0],[4],[[8]
Now that I think about [9] isn't in the kernel 3[9]=[27]=[3], not e
 
Well, it seems that we found the kernel as [10] and [11] are NOT in the kernel.
 
yeah [10] and [11] are not in the kernel.
3[10]=[30]=6
3[11]=33=11
 

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