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Describe all homomorphisms

  1. Jul 5, 2011 #1
    1. The problem statement, all variables and given/known data
    Find all homomorphisms [itex]f: \mathbb{Z},+ \rightarrow \mathbb{Z},+[/itex]. Determine which are injective, which are surjective, and which are isomorphisms.

    Note. I must prove everything.


    2. Relevant equations
    Notation. [itex]\mathbb{Z}n = \{ p : p = kn, \, \, \, \mathrm{k} \in \mathbb{Z} \}[/itex]


    3. The attempt at a solution
    I'm not sure if I have found all homomorphisms from [itex]\mathbb{Z},+[/itex] to itself, but thus far I found a family of homomorphisms from [itex]\mathbb{Z},+[/itex] to itself.

    Define [itex]f: \mathbb{Z},+ \rightarrow \mathbb{Z},+[/itex] such that [itex]f(x) = nx[/itex], where [itex]n \in \mathbb{Z}[/itex]. This is the homomorphism from [itex]\mathbb{Z},+[/itex] to it's subgroup [itex]\mathbb{Z}n[/itex].

    Proof that f is a homomorphism. Let both [itex]x, y \in \mathbb{Z}[/itex]. Then [itex]f(x+y) = n(x+y) = nx + ny = f(x) + f(y)[/itex]. Done.

    I've determined that every homomorphism of form [itex]f[/itex], as defined above, is injective. I'll claim that [itex]\mathrm{ker}(f) = \{ 0 \}[/itex]. To check this, I'll suppose that there exists another element, say [itex]k[/itex], such that [itex]f(k) = 0[/itex]. Using the definition of [itex]f[/itex], it follows that [itex]nk = 0[/itex] [itex]\Rightarrow[/itex] [itex] k = 0[/itex]. Thus [itex]\mathrm{ker}(f) = \{ 0 \}[/itex].

    I've also determined that the only [itex]f[/itex] that is surjective is [itex]f(x) = x[/itex]. To prove this, I'll suppose that there exists another such [itex]f[/itex] that is surjective, say [itex]f(x) = kx[/itex], with [itex]k \neq 1[/itex]. Note that [itex]k \in \mathbb{N}[/itex]. Set [itex]f(x) = k+1[/itex] [itex]\Rightarrow[/itex] [itex]kx = k+1[/itex] [itex]\Rightarrow[/itex] [itex]x = 1 + \frac{1}{k}[/itex] which is never an integer if [itex]k \neq 1[/itex], a contradiction since [itex]x \in \mathbb{Z}[/itex].

    Thus from the information above, it follows that the only isomorphism is the homomorphism [itex]f[/itex] as defined above, such that [itex]f(x) = x[/itex].

    My question is have I done my work correctly and are there any other homomorphisms from [itex]\mathbb{Z},+[/itex] to itself?
     
  2. jcsd
  3. Jul 6, 2011 #2

    Dick

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    Science Advisor
    Homework Helper

    No, there are no other homomorphisms, but you seem to be making it more complicated than it needs to be. If you know what f(1) is, and you assume f is a homomorphism, then it's pretty easy to say what f(n) is for any n, right?
     
  4. Jul 6, 2011 #3
    Well if I know there f(1) goes, then f(n) would go to nf(1). Thanks for the clarification!
     
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