Abstract Algebra Homomorphism Proof

In summary, we are trying to prove that if f: G → H is a homomorphism, and a and b are elements in G and H respectively, with ord(a) = n and ord(b) = m, then n is a multiple of m. To do this, we use the properties that a^n = e (the identity in G) and b^m = e (the identity in H). From this, we can conclude that n and m are multiples. To prove this, we consider the following ideas: (1) Why is n >= m? (2) With n >= m established, we can write n = q m + r, where 0 <= r < m and q > 0.
  • #1
The_Iceflash
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0

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.
 
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  • #2
a^n = e (in G)
b^m = e (in H)...
 
  • #3
A homomorphism has what property?
 
  • #4
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.
 

What is abstract algebra homomorphism?

Abstract algebra homomorphism is a mathematical concept that describes a mapping between two algebraic structures that preserves their operations. In simpler terms, it is a function that preserves the structure of algebraic objects.

What is a proof in abstract algebra homomorphism?

A proof in abstract algebra homomorphism is a logical argument that demonstrates the validity of a statement or theorem related to homomorphisms. Proofs in abstract algebra often involve using axioms, definitions, and previously proven theorems.

How do you prove a homomorphism in abstract algebra?

To prove a homomorphism in abstract algebra, you must show that the function in question preserves the algebraic structure of the two objects. This can be done by showing that the function satisfies the definition of a homomorphism, which states that f(a⋅b) = f(a)⋅f(b) for all elements a and b in the algebraic structures.

Can a homomorphism be bijective?

Yes, a homomorphism can be bijective, meaning it is both injective (one-to-one) and surjective (onto). In other words, it is a one-to-one correspondence between the elements of the two algebraic structures, preserving the operations.

What are some common applications of abstract algebra homomorphism?

Abstract algebra homomorphisms have various applications in mathematics and other fields, including cryptography, coding theory, and computer science. They are also used in the study of group theory, ring theory, and field theory.

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