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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 queston 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 recieved 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????