How many homomorphisms are there of
These two questions are from exercise 13, from book by John B. Fraleigh.
Answer of 1. is "two homomorphisms" and 2. is "infinite number of homomorphisms". I want to know difference between these, how to approach these problems.
Note that a homomorphism from $f:\mathbb{Z}\rightarrow G$ to another group $G$ is completely determined by where it maps $1$. Indeed, take $n\in\mathbb{Z}$, then $f(n) = f(1+\cdots+1) = f(1)+\cdots+f(1) = nf(1)$. Now suppose we want a surjective homomorphism $f:\mathbb{Z}\rightarrow\mathbb{Z}$. Then some element must be mapped to $1$. Suppose $|f(1)|>1$. Then $|f(n)| = n|f(1)|>n$, so if we want surjectiveness we must have $|f(1)|\le 1$. The only options left are $-1,0,1$. Check that only two of these work. Can you do the injective part yourself now?
If $f\colon\mathbb{Z}\to\mathbb{Z}$ is a group homomorphism, then, setting $n=f(1)$, you have $$ f(x)=nx $$ for every $x\in\mathbb{Z}$ (prove it). So the image of $f$ is $n\mathbb{Z}$ and it should be clear for what $n\in \mathbb{Z}$ we have $n\mathbb{Z}=\mathbb{Z}$.
For $f$ being injective you need $\ker f=\{0\}$; but $$ \ker f=\{x\in\mathbb{Z}:nx=0\} $$ Can you finish?
For the final counting, note that, when you fix $n\in\mathbb{Z}$, the map $x\mapsto nx$ is a homomorphism of $\mathbb{Z}$ to $\mathbb{Z}$.
Just rewriting other people's answers. Let's say that $f(1) = m$ (it's gotta be something and I am just calling that something $m$, so I haven't made any assumptions yet).
I claim we know $f$ on the nose just from this data of $m$. Indeed, $f(2) = f(1 + 1) = 2m$, $f(3) = f(1 + 1 + 1) = 3m$, etc. It's clear that $f(n) = nm$ for $n$ positive (you can prove this by induction if you are feeling excessively formal). I'll leave it to you to prove this for $n$ negative (which is true).
So we have a formula for $f(n)$ automatically, just in terms of $f(1)$. (And conversely, this formula determines a homomorphism, so anything can be $f(1)$). This answers the second question. To see which ones are "onto," consider when $1$ can be in the image. That says that $1 = mn$ for some $n$. Since $m$ and $n$ are integers, the only possibilities are $m = n = -1$ and $m = n = 1$. That answers the first question (if $1$ is in the image, anything is in the image -- do you see why?).
For the onto case since $1$ is a generator of the cyclic group $(Z,+)$, $f(1)$ is also a generator of the image group $Z$. Now there are only two generators of $Z$ and they are $1$ and $-1$. So either $f(1)=1$ or $f(1)=-1$. If $f(1)=1$, then $f(n)=n$ for all $n$. If $f(1)=-1$ then $f(n)=-n$ for all $n$. Thus there are only two homomorphisms
An onto homomorphism $f$ has the additional requirement that the image of $f$ is $\mathbb{Z}$ itself.
