Proving Homomorphism of Groups: Showing f(e)=e' and f(g^-1)=f(g)^-1

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SUMMARY

The discussion focuses on proving two properties of group homomorphisms: that the image of the identity element under a homomorphism is the identity element of the target group, expressed as f(e) = e', and that the image of the inverse of an element is the inverse of the image, expressed as f(g^-1) = f(g)^-1. The first property is established through the definition of the neutral element, while the second property remains unresolved, prompting further inquiry into the relationship between f(gg^-1) and the homomorphic property f(x·y) = f(x)·f(y).

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  • Understanding of group theory and the definition of homomorphisms.
  • Familiarity with the properties of identity elements in groups.
  • Knowledge of inverse elements in group structures.
  • Basic algebraic manipulation of functions and equations.
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  • Research the properties of group homomorphisms in more depth.
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  • Explore the concept of kernel and image in the context of homomorphisms.
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LagrangeEuler
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Homework Statement


Show that
##f(e)=e'## and ##f(g^-1)=f(g)^-1##

Homework Equations


Homomorphism
f(x\cdot y)=f(x)\cdot f(y)

The Attempt at a Solution


I show the first one. Neutral element is element which satisfied
##e\cdot e=e##.
So
##f(e)=f(e\cdot e=e)=f(e)\cdot f(e)##
So ##f(e)=e'##.
But how to show
##f(g^{-1})=f(g)^{-1}##?
 
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LagrangeEuler said:

Homework Statement


Show that
##f(e)=e'## and ##f(g^-1)=f(g)^-1##



Homework Equations


Homomorphism
f(x\cdot y)=f(x)\cdot f(y)



The Attempt at a Solution


I show the first one. Neutral element is element which satisfied
##e\cdot e=e##.
So
##f(e)=f(e\cdot e=e)=f(e)\cdot f(e)##
So ##f(e)=e'##.

But how to show
##f(g^{-1})=f(g)^{-1}##?

What is [itex]f(gg^{-1})[/itex]?
 
Tnx.
 

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