# Homomorphisms!Is this right?

1. Nov 23, 2008

### sutupidmath

1. The problem statement, all variables and given/known data 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 $$\theta(a)=b'$$
(ii)Show that there is a homomorphism $$\theta$$ from G to G' with $$\theta(a)=b'$$ 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.

2. Relevant equations

3. The attempt at a solution
(i) $$\theta:G->G'$$ $$\theta(a)=b'$$ 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 $$x=a^m,y=a^m$$ we would have

$$\theta(a^ma^n)=\theta(a^{m+n})=e'=e'e'=\theta(a^m)\theta(a^n)$$ 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 $$\theta:G->G'$$ $$\theta(a)=b'$$ be a homomorphism. Let o(a)=p and o(b')=q.WE want to show that q|p??

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

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

2. Nov 23, 2008

### Vid

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. Nov 23, 2008

### sutupidmath

AHA! So, you are saying to construct my isomorphism for part (i) something like this:

$$\theta(a^i)=b^i$$ 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

$$\theta(a^i)=b^j$$ 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:

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

Well thnx for this.

4. Nov 23, 2008

### sutupidmath

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 $$\theta(a^i)=b^i$$ similar to what we did before,( whic i am not sure we an do here too), then we would have:

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

So, now let x,y be in G. with $$x=a^m,y=a^n$$ so

$$\theta(xy)=\theta(a^ma^n)=\theta(a^{m+n})=(b')^{m+n}=(b')^m(b')^n=\theta(a^m)\theta(a^n)$$ 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. Nov 23, 2008

### sutupidmath

,,,,,,,,,,,,,,,,,,,,,,,,,,,......???

Last edited: Nov 24, 2008
6. Nov 24, 2008

### sutupidmath

Yeah, i think you are right...But, wait untill someone else confirms it..