Does There Exist of a Unique Group Homomorphism between the Additive Groups $(\mathbb{Q},+)$ and $(\mathbb{Z},+)$: $\mathbb{Q}\rightarrow\mathbb{Z}$? Note (!) I have no confidence in the solution I'm putting forth here.
I had this question on an algebra exam and I would like to ask:

*

*Is my conclusion is correct?

*Is there a better way to go about this?

*Am I allowed to multiply $\varphi(x)$ by integers because it is equivalent to addition (i.e., a shorthand)? Does this mean that I cannot divide by integers because there is no equivalence to the group operation defined? When I was writing this down, my immediate thought was that the multiplication notation was dubious and it bothered me.

For clarity, I will divide this into parts.
Let $\mathbb{Q}\xrightarrow{\varphi}\mathbb{Z}$ be any group homomorphism from $(\mathbb{Q},+)$ to $(\mathbb{Z},+)$. I claim there is only the trivial homomorphism.
Part (a)
First, I claim that if $\mathbb{Q}\ni A>0$ and $\varphi(A)=n$, then $\varphi(-A)=-n$ for some $n\in\mathbb{Z}$. We observe that if $\varphi(A)=n$, then $\varphi(A)+\varphi(-A)=n+\varphi(-A)$, so
$$\varphi(A)+\varphi(-A)=\varphi(A+(-A))=\varphi(0)=n+\varphi(-A)$$
If $\varphi(0)=0$, we are done. Suppose $\varphi(0)=x$. Then $\varphi(y+0)=\varphi(y)+\varphi(0)=\varphi(y)+x$, hence $\varphi(y)-\varphi(y)=x=0$, so $\varphi(-A)=-n$.
Part (b)
I claim, for any $A\in\mathbb{Q}$, that $\varphi(A)=n\implies\left|n\right|\varphi(\frac{A}{n})=n$ for some $n\in\mathbb{Z}$ satisfying $\varphi(A)=n$.
Inducting upon $n>0$, let $a\in\mathbb{Q}$, and suppose that $n=2$ (the case of $n=1$ is trivial but I'll do this as well), then
$$2\varphi(\frac{a}{2})=\varphi(\frac{a}{2})+\varphi(\frac{a}{2})=\varphi(a)=2$$
As claimed since a homomorphism here satisfies $\varphi(x+y)=\varphi(x)+\varphi(y)$.
Now suppose this works for all $m\le N-1$ and consider $N$. Let $b,c,d\in\mathbb{Q}$ such that $b+c=d$ and $\frac{b}{N-1}+\frac{c}{N-1}=\frac{a}{N}$. Then
$$N\varphi(\frac{a}{N})=(N-1)\varphi(\frac{b+c}{N-1})+\varphi(\frac{b+c}{N-1})$$
and by inductive assumption, for any element in $\mathbb{Q}$, $N\varphi{(\frac{a}{N})}=N$, and $(b+c)\in\mathbb{Q}$ as both $b,c\in\mathbb{Q}$, hence, $(N-1)\varphi(\frac{d}{N-1})=(N-1)$. Moreover, by inductive assumption, if $(N-1)\varphi(\frac{d}{N-1})=N-1$, then $\varphi(\frac{d}{N-1})=1$. Hence, we have
$$(N-1)\varphi(\frac{b+c}{N-1})+\varphi(\frac{b+c}{N-1})=N-1+1=N$$
If $n<0$, then we can pull out the negative sign and cancel it because of part (a) and so we arrive at the same result.
Part (c)
Then for every $A\in\mathbb{Q}$ and any $n\in\mathbb{Z}$, we have that $\varphi(\frac{A}{n})=1$ (it is unecessary to consider $n<0$, as we have shown the case for any $A$, so $A<0$ takes care of this). Then every rational $b\in\mathbb{Q}$ satisfies $\varphi(b)=1$ as, if $A\in\mathbb{Q}$, then certainly $A/n\in\mathbb{Q}$ as well for any $n\in\mathbb{N}$. However, this means that both every $b\in\mathbb{Q}$ satisfies $\varphi(b)=1$ and some rationals satisfy $\varphi(A)=n$, a contradiction.
So the only homomorphism that will satisfy this condition is the homomorphism $\mathbb{Q}\xrightarrow{\phi}\mathbb{Z}$ satisfying $\ker(\phi)=\mathbb{Q}$, that is, $n=0$.
I apologize in advance for the highly verbose (and likely wrong!) work I've done here.
 A: It looks pretty good, although it's quite wordy.  How about the following?
1. The trivial map $\varphi: \Bbb{Q} \to \Bbb{Z}$, $\varphi(r) = 0$ for all $r$ is a homomorphism.  (So we have one.)
2. Are there other homomorphisms?  Say that $\varphi(1) = n \ne 0 \in \Bbb{Z}$.  What is $\varphi(\frac{1}{2n})$ then?  Calculate:
$$
\begin{align}
2n \, \varphi \bigl( \tfrac{1}{2n} \bigr) &= \underset{2n}{\underbrace{\varphi \bigl( \tfrac{1}{2n} \bigr) + \cdots + \varphi \bigl( \tfrac{1}{2n} \bigr)}} \\
&= \varphi \bigl( \underset{2n}{\underbrace{\tfrac{1}{2n} + \cdots + \tfrac{1}{2n}}} \bigr) \\
&= \varphi(1) = n.
\end{align}
$$
So,
$$
\varphi \bigl( \tfrac{1}{2n} \bigr) = \tfrac{1}{2} \notin \Bbb{Z}.
$$
A: An easy way for me (check if this suits):
Let $f:\mathbb Q\rightarrow  \mathbb Z$ be a  homomorphism.
Then for any $\frac{p}{q}$ we have $p.f(1)=f({p})=f(q.\frac{p}{q})=q.f(\frac{p}{q})$
So  $f(\frac{p}{q})=\frac{p}{q}f(1) \rightarrow (1)$
Now $f(1)\in \mathbb Z$But R.H.S of $(1)$does not belong to $\mathbb Z$ for any value other than $f(1)=0$.Hence zero morphism only morphism
A: Yes, the general statement seems correct, but your argument is a huge mess. For example you're proving that $\phi(0)=0$ in part a), whereas it's a general fact that group homomorphisms map the identity element to the identity, with the proof being essentially the same as what you gave except in total generality. These sorts of redundancies make your arguments unclear.
The argument in part b) is more or less correct but very messy and painful to read. The argument in part c) appears incorrect. You prove in part b) that if $\phi(A)=n$ then $n\phi(\frac{A}{n})=n$. Then in part $c)$ you're claiming for any $n$ and any $A$, $\phi(\frac{A}{n})=1$. This doesn't follow from what you proved in part $b)$, since there your claim had the assumption $\phi(A)=n$. This makes the rest of your argument incorrect, though there is some semblance of the right idea there.
One way to think about this, and you've already started thinking along these lines in part (c), is using the concept of a divisible group. An abelian group $(G,+ )$ is called divisible, if for every element $g\in G$ and every positive integer $n$, there exists an element $h\in G$ such that $n\cdot h = g$ (Yes, it's okay to multiply by $n$. It's shorthand for applying the operation $+h$ to $h$, $n-1$ times). 
The group $\mathbb{Q}$ is divisible, since given any rational number $r$, and positive integer $n$, we have $s=\frac{r}{n} \in \mathbb{Q}$ and $ns = r$. The group $\mathbb{Z}$ is not divisible, and neither is any non-trivial subgroup of $\mathbb{Z}$. Any such subgroup of $\mathbb{Z}$ is of the form $n\mathbb{Z}$ for some integer $n\neq 0$, and is isomorphic to $\mathbb{Z}$ itself via the map "multiplication by n". The only divisible subgroup of $\mathbb{Z}$ is then the trivial subgroup.
Now for exercise, try to prove this statement: If $\phi: G \rightarrow H$ is a homomorphism of abelian groups, and $G$ is divisible, then the subgroup $\phi(G)$ of $H$ is also divisible. 
This is close in spirit to what you're proving in b), but there's no need for induction. 
It follows immediately that $\phi: \mathbb{Q} \rightarrow \mathbb{Z}$ must be trivial. Indeed, $\mathbb{Q}$ is divisible, hence so is $\phi(\mathbb{Q})$, which must be the trivial subgroup of $\mathbb{Z}$. Then $\phi(\mathbb{Q})=\{0\}$ and $\phi$ is the trivial homomorphism.
A: You should be able to shortcut a lot of this. For example, it is true for any group homomorphism that it takes the neutral to the neutral element and inverses to invresews. These two points ought to have been covered in your course, definitely!
The same holds for integer powers of elements (or in additive groups as here: for integer multiples of elements).
So in short (but with the very same idea): Assume there is $a\in\mathbb Q$ with $\phi(a)\ne 0$. Then wlog $\phi(a)>0$. With  $n=\phi(a)$ let $b=\frac {a}{n+1}$. Then $n=\phi(a)=\phi((n+1)b)=(n+1)\phi(b)$, which is absurd (as either $(n+1)\phi(b)\le (n+1)\cdot 0=0<n$ or $(n+1)\phi(b)\ge (n+1)\cdot 1\ge n+1>n$)
