Are there examples of subgroups of $\Bbb Z[\frac1n]/q\Bbb Z$ not totally ordered by inclusion? I have the theorem that the Prufer P-groups (of which $\Bbb Z[\frac12]/\Bbb Z$ is one example) are the only infinite groups whose subgroups are ordered by inclusion.  That this property holds for $\Bbb Z[\frac1p]/\Bbb Z$ (with addition modulo 1) is easy to verify.
I suspect this theorem is trivially incorrect and should be stated proper subgroups otherwise every group with no nontrivial subgroups would be a counterexample - the trivial group nested within its group forming a totally ordered pair - or am I wrong about that?  Anyway, no matter, that's not the point of this question.
I'm trying to see how other examples of $\Bbb Z[\frac1n]/q\Bbb Z$ fail.  Clearly an early prerequisite for success is that $G$ surjects over itself with addition modulo $q$.

Are all $\Bbb Z[\frac1n]/q\Bbb Z$ which satisfy this surjection property groups, and if so, where do their subgroups fail to be totally ordered?

For example, what about $\Bbb Z[\frac16]/\frac43\Bbb Z$.  Is this a group and where does the sequence of subgroups (if there is one) fail to be totally ordered by inclusion?
 A: To answer the pentultimate question first, yes, they are groups. If $q\in\mathbb{Z}[\frac{1}{n}]$, then $\langle q\rangle = q\mathbb{Z}$, as this is the additive subgroup; the subgroup is normal, so you can certainly take the quotient.
Now on to the other questions.

I suspect this theorem is trivially incorrect and should be stated proper subgroups otherwise every group with no nontrivial subgroups would be a counterexample - the trivial group nested within its group forming a totally ordered pair - or am I wrong about that?

There are no infinite groups with no proper nontrivial subgroups. Let $G$ be an infinite group, and let $x\in G$, $x\neq e$. If $\langle x\rangle=G$, then $G$ is cyclic, isomorphic to $\mathbb{Z}$, and has plenty of proper nontrivial subgroups. If $\langle x\rangle \neq G$, then $\langle x\rangle$ is a proper nontrivial subgroup. If you allow finite groups, then the groups with this property are exactly the Prüfer $p$-group and the cyclic groups of prime power order (which includes the cyclic groups of prime order which have a subgroup structure as you describe).
As for the group $\mathbb{Z}[\frac{1}{n}]/q\mathbb{Z}$...
First, let's deal with $q=1$. If $n=p^a$ is a prime power, then since $\frac{1}{p^a}\in\mathbb{Z}[\frac{1}{p}]$ and $\frac{1}{p} = \frac{p^{a-1}}{p^a}\in\mathbb{Z}[\frac{1}{p}]$, you just get the Prüfer $p$-group back. And if $n$ is not a prime power, then it can be written as $n=ab$ with $1\lt a,b\lt n$, $\gcd(a,b)=1$. Then the subgroups of $\mathbb{Z}[\frac{1}{n}]/\mathbb{Z}$ generated by $\frac{1}{a}=\frac{b}{n}$ and $\frac{1}{b}=\frac{a}{n}$ are nontrivial and have relatively prime orders, and so they are not comparable. Hence this subgroup does not have the desired property.
Now let's consider $q$ a positive integer.
From the above considerations, it follows that if $n$ is not a prime power, then $\mathbb{Z}[\frac{1}{n}]/q\mathbb{Z}$ also cannot have the desired property; for this group has $\mathbb{Z}[\frac{1}{n}]/\mathbb{Z}$ as a quotient (moding out by $\mathbb{Z}/q\mathbb{Z}$); if any two subgroups of $\mathbb{Z}[\frac{1}{n}]/q\mathbb{Z}$ were comparable, then taking the inverse image of $\langle \frac{1}{a}\rangle$ and $\langle \frac{1}{b}\rangle$ we would obtain two comparable subgroups, and then the Fourth Isomorphism Theorem would tell you that $\langle \frac{1}{a}\rangle $ and $\langle \frac{1}{b}\rangle$ are comparable, which is a contradiction.
So this leaves the case in which $n$ is a prime power, which is the same as the case when $n$ is just a prime, as above. So what happens if we take $\mathbb{Z}[\frac{1}{p}]/q\mathbb{Z}$?
If $q$ is not a power of $p$, you run into trouble: if the prime $r\neq p$ divides $q$, then the element $\frac{q}{r}\in\mathbb{Z}$ has order $r$ in this group; while  $\frac{q}{p}$ has order $p$; this gives you two nontrivial subgroups of relatively prime order, showing the group does not have the property that its subgroups are totally ordered by inclusion.
Thus, we may assume that $q$ is a power of $p$. But in that case, you just get a group isomorphic to the Prüfer $p$-group again. To see this, say $q=p^a$. Consider the map $\mathbb{Z}[\frac{1}{p}]$ to itself given by $\frac{a}{p^n}\longmapsto \frac{a}{p^{n+a}}$. This is an additive group morphism that is surjective; the preimage of $\mathbb{Z}$ is $p^a\mathbb{Z}=q\mathbb{Z}$.
Composing the map with the quotient map $\mathbb{Z}[\frac{1}{p}]\to\mathbb{Z}[\frac{1}{p}]/\mathbb{Z}$ yields an isomorphism, by the First Isomorphism Theorem,
$$\frac{\mathbb{Z}[\frac{1}{p}]}{q\mathbb{Z}} \cong \frac{\mathbb{Z}[\frac{1}{p}]}{\mathbb{Z}}.$$
Thus, for $q$ an integer, the groups you are constructing either don't have the property that their subgroups are totally ordered by inclusion, or else they do... but they are just isomorphic to the Prüfer $p$-group.
Finally, say $q$ is a rational in $\mathbb{Z}[\frac{1}{n}]$. Then $q=\frac{a}{b}$, $\gcd(a,b)=1$, and $b|n^k$ for some $k\geq 1$, say $bx=n^k$. The group $\mathbb{Z}[\frac{1}{n}]/q\mathbb{Z}$ has $\mathbb{Z}[\frac{1}{n}]/a\mathbb{Z}$ as a quotient (taking the quotient modulo $\frac{1}{b}\mathbb{Z}=\frac{x}{n^k}\mathbb{Z}$), so again we are reduced to the case in which $n=p$ is prime and $b$ is a power of $p$.
Then we are looking at $\mathbb{Z}[\frac{1}{p}]/\frac{a}{p^k}\mathbb{Z}$. But $\mathbb{Z}[\frac{1}{p}]/\frac{1}{p^k}\mathbb{Z}$ is a quotient of the Prüfer $p$-group, which is isomorphic to the Prüfer $p$-group. So this is isomorphic to the quotient $\mathbb{Z}[\frac{1}{p}]/a\mathbb{Z}$, and we already dealt with that case as well. So you are still back to either your group does not have the property that the subgroups are totally ordered by inclusion, or else you just got a group isomorphic to the Prüfer $p$-group.
For your specific example of $\mathbb{Z}[\frac{1}{6}]/\frac{4}{3}\mathbb{Z}$, note that $\frac{1}{3}=\frac{2}{6}$ is nontrivial and has order $4$ in the quotient; and $\frac{4}{9} = \frac{16}{6^2}$ is in $\mathbb{Z}[\frac{1}{6}]$, and has order $3$. So this gives you two subgroups of relatively prime order, hence incomparable.
