Abstract Algebra Homomorphism Proof

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Homework Help Overview

The discussion revolves around a proof in abstract algebra concerning homomorphisms between groups. The original poster is tasked with proving that if \( f: G \rightarrow H \) is a homomorphism and \( a \in G \) with order \( n \) and \( b = f(a) \) with order \( m \), then \( n \) is a multiple of \( m \).

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to relate the orders of elements \( a \) and \( b \) through their definitions, questioning how to formally prove the relationship between \( n \) and \( m \). Some participants inquire about the properties of homomorphisms and suggest considering the implications of the orders of the elements involved.

Discussion Status

The discussion is ongoing, with participants exploring various ideas related to the proof. Some guidance has been offered regarding the relationship between \( n \) and \( m \), particularly through the establishment of inequalities and potential equations involving \( n \) and \( m \).

Contextual Notes

Participants are considering the implications of the definitions of order and the properties of homomorphisms, with some questioning the assumptions underlying the proof. There is an acknowledgment of the need for further clarification on these concepts.

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Homework Statement



Let G and H be two groups. If f: G [tex]\rightarrow[/tex] H is a homomorphism, a [tex]\in[/tex] G and b = f(a). If ord(a) = n, ord(b) = m, then n is a multiple of m. (Let [tex]e_{1}[/tex] be the identity of G and [tex]e_{2}[/tex] be the identity of H)

I have to prove that n is a multiple of m.

Homework Equations


N/A

The Attempt at a Solution



I know an = ?
and that bn= ?
then I conclude from what I get that n and m are multiples.

Can I have some help with this proof? Thanks.
 
Physics news on Phys.org
a^n = e (in G)
b^m = e (in H)...
 
A homomorphism has what property?
 
Consider these ideas:

(1) Why is n >= m ?
(2) With n >= m established, you can write n = q m + r, where 0 <= r < m and q > 0.
 

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