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!

Homomorphism Problem - Need work checked

  1. Jun 30, 2011 #1
    1. The problem statement, all variables and given/known data
    Let [itex]f : G \rightarrow H[/itex] be a surjective homomorphism. Prove that if G is cyclic, then H is cyclic, and if G is abelian, then H is abelian.


    2. Relevant equations
    Notation: Let it be understood that [itex]f^n(x) = [f(x)]^n[/itex].


    3. The attempt at a solution

    Just would like to see if I am right.

    Proof. (of G cyclic [itex]\Rightarrow[/itex] H cyclic)Suppose that [itex]f[/itex] is a surjective homomorphism, and suppose that [itex]G[/itex] is cyclic. Let [itex]G[/itex] be generated by [itex]x[/itex]. I want to show that there exists [itex]y[/itex] such that [itex]y[/itex] generates [itex]H[/itex].

    First I will prove that [itex]\forall x^n \in G[/itex] [itex]\Rightarrow[/itex] [itex]f(x^n) = f^n(x)[/itex]. As our base case for [itex]n=1[/itex] we cite that [itex]f(x e_G) = f(x) f(e_G) = f(x)[/itex]. Next suppose that [itex]f(x^n) = f^n(x)[/itex] (Inductive hypothesis). Then [itex]f(x^{n+1}) = f(x^n x) = f(x^n) f(x) = f^n(x) f(x) =f^{n+1}(x)[/itex], which from our inductive hypothesis.

    The proof that [itex]\forall x^{-n} \in G[/itex] [itex]\Rightarrow[/itex] [itex]f(x^{-n}) = f^{-n}(x)[/itex] is similar.

    Now since [itex]f[/itex] is surjective [itex]\Rightarrow[/itex] [itex]\mathrm{im}(f) = H[/itex] and therefore [itex]H[/itex] is generated by [itex]f(x)[/itex] and is therefore cyclic.

    Proof. (of G abelian [itex]\Rightarrow[/itex] H abelian) Suppose now that [itex]G[/itex] is abelian and let [itex]x, y \in G[/itex]. I want to show that [itex]\forall f(x) , f(y)[/itex] [itex]\Rightarrow[/itex] [itex]f(x)f(y) = f(y)f(x)[/itex]. Then [itex]f(xy) = f(x)f(y)[/itex] and [itex]f(yx) = f(y)f(x)[/itex], but [itex]f(xy) = f(yx)[/itex] [itex]\Rightarrow[/itex] [itex]f(x)f(y) = f(y)f(x)[/itex]. Since [itex]f[/itex] is surjective [itex]\mathrm{im}(f) = H[/itex] [itex]\Rightarrow[/itex] [itex]\forall f(x), f(y) \in H[/itex] [itex]f(x)f(y) = f(y)f(x)[/itex] [itex]\Rightarrow[/itex] [itex]H[/itex] is abelian.

    My intuition says I'm right, but it seems my intuition is wrong more than right this quarter. :biggrin:
     
  2. jcsd
  3. Jun 30, 2011 #2

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    Hi Samuelb88! :smile:

    Could you explain in more detail why H is generated by f(x). I'm sure you know it, but it's a crucial detail that should be mentioned.

    That proof is correct! But it's perhaps a bit chaotic, here's a proposal to clean it:

    Take h and h' in H. We want to show that hh'=h'h. Because of surjectivity, we know that there are g and g' in G such that f(g)=h and f(g')=h'. Since G is abelian, we know that gg'=g'g. But then it follows that hh'=f(g)f(g')=f(gg')=f(g'g)=f(g')f(g)=h'h.

    I'm not saying your proof is wrong, though.
     
  4. Jun 30, 2011 #3
    Since [itex]f[/itex] is surjective [itex]\Rightarrow[/itex] [itex]\mathrm{im}(f)=\{ ..., f^{-1}(x), e_H , f(x) , f^2(x) , ... \} = H[/itex]. Doesn't this imply that [itex]H[/itex] is generated by [itex]f[/itex]?
     
  5. Jun 30, 2011 #4

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    Yes, of course, I just wanted some more explanation on that part :smile:
    Well, it seems it is all correct!
     
  6. Jun 30, 2011 #5
    Great to hear! Hehe. Thanks much, micromass!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Homomorphism Problem - Need work checked
  1. A homomorphism problem (Replies: 2)

Loading...