Here is a question:
Show that the additive subgroups of $\mathbb Z [\frac{1}{2}]$ and $\mathbb Z [\frac{1}{3}]$ are not isomorphic. Where $\mathbb Z [\frac{1}{n}] = \{ \frac{a}{n^k} | a, k \in \mathbb Z\}$
This is a trial, consider the isomorphism $\phi(\frac{a}{2^k}) = \frac{a}{3^k},$ then upon taking $a = 2 $ and $k = 1$ we get $\phi(1) = \frac{2}{3}$ which is a contradiction to $\phi$ being a ring isomorphism.
Is this trial correct? I am worried because the statement of the question said additive subgroups but I used a ring homomorphism instead.
Am I correct?
I don’t think your argument is correct, exactly for the reason you state. It seems like the problem wants you to work in the category of groups, in which case $1$ need not go to $1$. Instead notice that $1$ is infinitely $2-$divisible in $\mathbb Z\left[\frac12\right]$, but $\mathbb Z\left[\frac13\right]$ does not contain any infinitely $2-$divisible element.