Every isomorphism of subgroups of $\mathbb{Q}$ is of the form $\varphi(x)=qx$ for some $q \in \mathbb{Q}$. Let $A$ and $B$ be subgroups of the additive group $\mathbb{Q}$. If $A$ is isomorphic to $B$ and $\varphi : A \rightarrow B$ an isomorphism, then show there is $q \in \mathbb{Q}$ such that $\varphi(x)=qx$ for all $x \in A$.
Okay so this is what I did:
If $A=\{0\}$ then this is vacuously true so assume $A \neq \{0\}$. If $\frac{a}{b} \in A$ then $\frac{a}{b}+\frac{a}{b}+...+\frac{a}{b}=a$ where there were $b$ summands in the previous sum. Therefore $A$ contains all integers which appear as numerators of fractions in $A$. Let $d$ be the GCD of all such integers. Note that $d$ is defined since $A \neq \{0\}$. Since $d$ is a $\mathbb{Z}$-linear combination of these integers we have $d \in A$. Therefore every $\frac{a}{b} \in A$ can be written as $\frac{nd}{b}$ for some some $n \in \mathbb{Z}$ and in fact $\frac{nd}{b} \in A$ for all $n \in \mathbb{Z}$. Therefore the mapping $\frac{nd}{b} \mapsto \frac{n}{b}$ is surjective and is easily shown to be a well defined isomorphism of $A$ with some subgroup $A'$ of $\mathbb{Q}$ containing $\mathbb{Z}$. Therefore we may assume WLOG that $1 \in A$ or, equivalently $d = 1$. Then for any $a \in A$ we have $\varphi(a)=\varphi(a\cdot 1)=a\varphi(1)$. 
We now claim that $q=\varphi(1)$ is the desired $q$. Indeed if $\frac{a}{b}$ is any element of $A$ then $\varphi\big(\frac{a}{b}\big)=a\varphi \big(\frac{1}{b}\big) \implies \varphi \big(\frac{a}{b}\big)=1\cdot\varphi \big(\frac{a}{b}\big)=\frac{b}{b}\varphi \big(\frac{a}{b}\big)=\frac{ab}{b}\varphi \big(\frac{1}{b}\big)=\frac{a}{b}\varphi \big(\frac{b}{b}\big)=\frac{a}{b}\varphi(1)$.

I am fairly certain that my proof is correct, but I am curious if there is a better way of doing this without the part about the isomorphic copy of $A$ containing $1$. I was playing around with it for awhile and it seems that the $\varphi(1)$ should be replaced with some fraction involving $\varphi(d)$ in such a proof. Also if someone has another shorter or cleaner proof I am very interested to see it. 

 A: I'm not sure of how much this helps, but I'll just show why you can take $q=\varphi(d)$ without needing to use the isomorphism with a subgroup of $\Bbb Q$ containing $1$. 
Suppose $A\neq \{0\}$ and let $d$ be as in your proof, or, as lhf commented, the integer such that $A \cap \Bbb Z = d\Bbb Z$. 
Then for an arbitrary element $\frac {nd} b \in A$, we have 
$$
\varphi\left(nd\right)=\varphi\left(\frac {nd}b+\cdots +\frac {nd}b\right)=\varphi\left(\frac {nd}b\right)+\cdots+\varphi\left(\frac {nd}b\right)= b\varphi\left(\frac {nd}b\right) \\\implies \varphi\left(\frac {nd}b\right)=\frac 1b \varphi({nd}).
$$
Also,
$$
\varphi(nd)=\varphi(d+\cdots+d)=\varphi(d)+\cdots+\varphi(d)=n\varphi(d)
$$
Putting these together, we get that for an arbitrary element $\frac {nd}b \in A$
$$ 
\varphi\left(\frac {nd}b\right)=\frac nb\varphi(d) = \frac {nd}{b}\left(\frac{\varphi(d)}{d}\right).
$$
This means that we can take $q=\frac {\varphi(d)}{d}$.
A: Having blundered at first, I fear another error, but here goes...
pick any $x \in A$ (other than x = $0$) and define $q \in \mathbb Q$ by $q = \frac{\phi(x)}{x}$ Then at least we have $$\phi(nx) = n\phi(x)=nqx\;\;\;n\in \mathbb Z$$.
Now take any $y \in A$.  Certainly $y = \frac mn x$ for some integers m and n.  But then $ny = mx$ so we have $$n\phi(y)=\phi(ny) = \phi(mx) = m\phi(x)\;\;\;\Rightarrow\;\;\;\phi(y)=\frac mn\phi(x)= \frac mnqx=qy$$
