Proving Group Homomorphism Between c and dc:G1--->G3

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Homework Help Overview

The problem involves proving that the composition of two group homomorphisms, dc:G1--->G3, is itself a homomorphism, and that the kernel of c is a subset of the kernel of dc. The subject area pertains to group theory and homomorphisms.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the properties of homomorphisms and the definitions of kernels. There is an exploration of the implications of composing homomorphisms and how this affects elements in the kernel.

Discussion Status

Some participants have provided insights into the relationship between the kernels and the composition of homomorphisms. There appears to be a productive exchange regarding the implications of the definitions involved, though no explicit consensus has been reached.

Contextual Notes

Participants are navigating through the definitions and properties of group homomorphisms and kernels, with some uncertainty about the implications of their reasoning.

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Homework Statement


Let c:G1--->G2 and d:G2--->G3 be group homomorphisms. Prove that dc:G1--->G3 is a homomorphism. Prove that ker(c) is a subset of ker(dc).


Homework Equations





The Attempt at a Solution


If a,b are in G1, then c(ab)=c(a)c(b) in G2 and so d(c(ab))=d(c(a)c(b))=d(c(a))d(c(b)) in G3

ker c is defined as x in G1 such that c(x)=e
ker dc is defined as x in G1 such that dc(x)=e
Then I get stuck
 
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Take x in ker(c), then c(x)=e. What happens if you compose both sides with d?
 


c(x)=e
dc(x)=de
e=de
e=d
 


I guess I do that and am unsure where that leads me
 


c(x)=e
d(c(x))=d(e)
d(e)=d(e)
Ahhh, so that shows it
 


Yes, It looks like you've got it!
 

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