Describe all homomorphisms from \mathbb{Z},+ to \mathbb{Z},+

[Samuelb88]

Homework Statement

Find all homomorphisms f: \mathbb{Z},+ \rightarrow \mathbb{Z},+. Determine which are injective, which are surjective, and which are isomorphisms.

Note. I must prove everything.

Homework Equations

Notation. \mathbb{Z}n = \{ p : p = kn, \, \, \, \mathrm{k} \in \mathbb{Z} \}

The Attempt at a Solution

I'm not sure if I have found all homomorphisms from \mathbb{Z},+ to itself, but thus far I found a family of homomorphisms from \mathbb{Z},+ to itself.

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

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

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

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

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

My question is have I done my work correctly and are there any other homomorphisms from \mathbb{Z},+ to itself?

 

[Dick]

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?

 

[Samuelb88]

Well if I know there f(1) goes, then f(n) would go to nf(1). Thanks for the clarification!