Order of Homomorphisms and Finite Elements

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Homework Help Overview

The problem involves a homomorphism ψ from group G to group H, focusing on the implications of finite order elements within these groups. The original poster is tasked with demonstrating that the order of the image of an element under the homomorphism divides the order of the original element.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to relate the order of the element g in G to its image ψ(g) in H, expressing uncertainty about the initial steps. Some participants suggest starting with the property that ψ(g) raised to the order of g equals the identity element in H. Others question whether the identity element in H should be denoted as eH and explore the implications of isomorphisms on the orders of the groups.

Discussion Status

The discussion is active, with participants providing guidance on how to approach the proof. There is an exploration of different interpretations regarding the identity element and the nature of isomorphisms, indicating a productive exchange of ideas without reaching a consensus.

Contextual Notes

Participants are navigating the definitions and properties of group homomorphisms and finite order elements, with some expressing confusion about notation and assumptions related to isomorphisms.

Locoism
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Homework Statement


Let ψ: G→H be a homomorphism and let g ε G have finite order.
a) Show that the order of ψ(g) divides the order of g

The Attempt at a Solution


I'm really lost here, but I'm guessing we can use the fact |ψ(g)| = {e,g...,g|g|-1}
and ψ(g|g|-1) = ψ(g)ψ(g)ψ(g)ψ(g)ψ(g)... (|g|-1 times)
I still have no idea where to start.
 
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First show that

[tex]\psi(g)^{|g|}=e[/tex]

Then use the general fact that if [itex]a^n=e[/itex], then |a| divides n.
 
Ah ok but I would use eH?
Also, if it is an isomorphism, could I show |ψ(g)| = |g|?
 
Locoism said:
Ah ok but I would use eH?
Also, if it is an isomorphism, could I show |ψ(g)| = |g|?

Yes to both.
 

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