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!