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Homework Help: 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
    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
    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!
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