1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homomorphisms!Is this right?

  1. Nov 23, 2008 #1
    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 [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.



    2. Relevant equations



    3. 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????
     
  2. jcsd
  3. Nov 23, 2008 #2

    Vid

    User Avatar

    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.
     
  4. Nov 23, 2008 #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.
     
  5. Nov 23, 2008 #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...
     
  6. Nov 23, 2008 #5
    ,,,,,,,,,,,,,,,,,,,,,,,,,,,......???
     
    Last edited: Nov 24, 2008
  7. Nov 24, 2008 #6
    Yeah, i think you are right...But, wait untill someone else confirms it..:biggrin:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Homomorphisms!Is this right?
  1. Homomorphism of groups (Replies: 2)

  2. Group Homomorphism? (Replies: 3)

Loading...