Homomorphisms and Cyclic Groups

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In summary, the author is trying to solve problem (ii) of a homework statement. They state that there is a homomorphism from G to G' if and only if the order of b' is an integral divisor of the order of a. They are not sure how to go about proving this.
  • #1
sutupidmath
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Homework Statement

Well, first i appologize for posting problems so often, but i have an exam comming up soon, so i am just working some problems on my own.

Problem:

Let G be a cyclic group [a](generated by a). Let b' be any element of a grou p G'.
(i)Show that ther eis at most one homomorphism from G to G' with [tex] \theta(a)=b'[/tex]
(ii)Show that there is a homomorphism [tex]\theta[/tex] from G to G' with [tex]\theta(a)=b'[/tex] if and only if the order of b' is an integral divisor of the order of a.
(iii) state a condition on the orders of a adn b' fro the homomorphism (ii) to be injective.



Homework Equations





The Attempt at a Solution


(i) [tex]\theta:G->G'[/tex] [tex] \theta(a)=b'[/tex] I am not sure whether i am getting the question right. I am assuming that in this case b' would be a fixed element of G' right. ??Because with this in mind, will my reasoning/proof follow below.

Well, i think that if b'=e' then we will certainly have a homomorphism. SInce

for any two elements x,y in G, where [tex]x=a^m,y=a^m[/tex] we would have

[tex]\theta(a^ma^n)=\theta(a^{m+n})=e'=e'e'=\theta(a^m)\theta(a^n)[/tex] so such a mapping would be a group homomorphism between these two groups.
Now,as in one of my previous questions(which i haven't received any answers yet) i am having trouble how to go about proving that we cannot have any other hommomorphism defined by this theta. So how would i prove this?

(ii)=> let [tex]\theta:G->G'[/tex] [tex] \theta(a)=b'[/tex] be a homomorphism. Let o(a)=p and o(b')=q.WE want to show that q|p??

Ok, let [tex]e'=\theta(e)=\theta(a^p)=[\theta(a)]^p=(b')^p=>q|p[/tex]
<= Let q|p. Now we want to show that [tex]\theta:G->G'[/tex] [tex] \theta(a)=b'[/tex] is a homomorphism.?

That is we want to show that for any two elements x,y in G, where [tex]x=a^m,y=a^n[/tex] [tex]\theta(xy)=\theta(a^ma^n)=\theta(a^m)\theta(a^n)??[/tex]

I'm not sure how to go about this one either?
 
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  • #2
Instead of taking b to be identity, just construct the only possible homomorphism. Powers of a must be mapped to powers of b. If you change one of the powers of a to map to a different power of b you no longer have a homomorphism.
 
  • #3
AHA! So, you are saying to construct my isomorphism for part (i) something like this:

[tex]\theta(a^i)=b^i[/tex] for i in Z. Well, yeah, i easily showed that this is a homomorphism.

Now, to prove that this is the oly one in this case, i assumed that the following construction is still a homomorphism

[tex]\theta(a^i)=b^j[/tex] where i is different from j. Without loss of generality, i supposed that i>j=> there exists an integer k such that i=j+k, so the above mapping would look sth like this:

[tex]\theta(a^i)=b^{i+k}[/tex] THen i showed that this is not a homomorphism.

Well thnx for this.
 
  • #4
Well, for (ii) Here it is again what i think, for <= part.

SInce q|p=> p=kq for some integer k.

Now, if we construct a mapping [tex]\theta(a^i)=b^i[/tex] similar to what we did before,( whic i am not sure we an do here too), then we would have:

[tex]\theta(e)=\theta(a^p)=(b')^p=(b')^kq=(e')^k=e'[/tex] so this means that by this kind of mapping the identity is preserved.

So, now let x,y be in G. with [tex]x=a^m,y=a^n[/tex] so

[tex]\theta(xy)=\theta(a^ma^n)=\theta(a^{m+n})=(b')^{m+n}=(b')^m(b')^n=\theta(a^m)\theta(a^n)[/tex] so would this prove it?

I am not sure this is correct, since i think i didn't use the fact that q|p anywhere in this last part? MOreove, i don't even see how to use it...
 
  • #5
,,,,,,,,,,,,,,,,,,,,,,,,,,,...?
 
Last edited:
  • #6
sutupidmath said:
,,,,,,,,,,,,,,,,,,,,,,,,,,,...?

Yeah, i think you are right...But, wait until someone else confirms it..:biggrin:
 

1. What is a homomorphism?

A homomorphism is a mathematical concept that describes a structure-preserving map between two objects. In simpler terms, it is a function that preserves the mathematical structure of its input.

2. How is a homomorphism different from an isomorphism?

A homomorphism preserves the structure of its input, while an isomorphism also preserves the bijective properties of the input. This means that an isomorphism is a one-to-one and onto function, while a homomorphism does not necessarily have these properties.

3. What are some common examples of homomorphisms?

Some common examples of homomorphisms include group homomorphisms, ring homomorphisms, and vector space homomorphisms. These can be used to describe the relationship between mathematical structures such as groups, rings, and vector spaces.

4. How do homomorphisms relate to the concept of symmetry?

Homomorphisms can be thought of as a type of symmetry, as they preserve the structure of the input object. This means that if the input object has any symmetries, the homomorphism will also preserve those symmetries.

5. Are there any real-world applications of homomorphisms?

Yes, homomorphisms have many real-world applications in fields such as cryptography, computer science, and physics. For example, in cryptography, homomorphic encryption allows for computations to be performed on encrypted data, preserving its privacy.

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