About non existence of maximal subgroup in additive group $\mathbb{Q}$

It's known that the rationals $\mathbb{Q}$ as an additive abelian group have no maximal subgroup. I think I understand the proof provided here, which is elementary and IMO elegant. I'll re-write the answer again here as both for convenience and if there's anything wrong someone could point out. (Can skip the following paragraph if one's familiar with the fact)

Suppose $N$ is maximal subgroup of $\mathbb{Q}$, by abelian of $\mathbb{Q}$, $N$ is normal, and hence quotient $\mathbb{Q}/N$ is well-defined. By correspondence theorem, since $N$ maximal, the quotient have no proper subgroup, hence by Lagrange's theorem the quotient must be isomorphic to some cyclic group of prime order, i.e. $\mathbb{Q}/N \simeq \mathbb{Z}/p\mathbb{Z}$ where $p$ is prime. By maximal-ity of $N$, there exists some $q\in\mathbb{Q}\setminus N$, then consider coset $\frac{q}{p}+N$ in the quotient $\mathbb{Q}/N$. Since it's isomorphic to a cyclic group of $p$ order, $\sum\limits_{i=1}^{p}{\left( \frac{q}{p}+N \right)} = N$, and by how operations on cosets is defined, $p\cdot \frac{q}{p}$ is hence forced to be in $N$, i.e. $q\in N$, a contradiction. Hence the assumption that $N$ is maximal is false.

My question is about the following group. Consider the set $P = \{\frac{1}{p} ~\vert~ p\text{ prime}\}$. An observation is that $P$ generates $\mathbb{Q}$. Next we consider the group generated by $P_3 = \{\frac{1}{p} ~\vert~ p\text{ prime} \land p >= 3\}$; let the group generated by $P_3$ be $H$. My question is, what are examples of proper subgroups of $\mathbb{Q}$ such that it contains $H$ as a proper subgroup?

I can't find such a group. Since for $p$ and $q$ primes, $\frac{n}{p}+\frac{m}{q} = \frac{nq+pm}{pq}$, i.e. the denominator of the sum have same set of prime factors as that of the denominator of the addends, we should have $\frac{1}{2} \not\in H$. By similar reasoning, $H$ should not contain any $q\in\mathbb{Q}$ of which simplest form have factor $2$ in its denominator; in fact $H$ should be exactly the set of rationals of which simplest form do not have factor $2$ in their denominators. Since $H$ is not maximal, there exists $G$ s.t. $H<G<\mathbb{Q}$, where $<$ denotes proper subgroup relation. There exists some $g\in G\setminus H$, WLOG assume $g>0$, let simplest form of $g$ be $\frac{\prod\limits_{i=1}^{m}q_i}{\prod\limits_{i=1}^{n}p_i}$. By above deduction, WLOG we can write $g$ as $\frac{\prod\limits_{i=1}^{m}q_i}{2\cdot \prod\limits_{i=2}^{n}p_i}$ and it's simplest form.

Consider $\sum\limits_{i=1}^{\left(\prod\limits_{i=2}^{n}{p_i}\right)}{g} = \frac{\prod\limits_{i=1}^{m}q_i}{2}$. The value cannot be a integer, since if it's integer then $2 ~\vert~ q_i$ for some $1\leq i \leq m$, violating the assumption we're writing $g$ in simplest form. Hence it's $n+\frac{1}{2}$ for some $n\in\mathbb{N}$. But since obviously all integers are in $H$ and hence in $G$, now we have $\frac{1}{2}$ should be in $G$. By $H = \langle P_3 \rangle$ and $\frac{1}{2}\in G$, $G$ should contain $P$ as subset, which means $G$ should be $\mathbb{Q}$. A contradiction.

Which part of the above derivation had gone wrong?

Solution 1:

The error lies at the very top: you assert that $P$ generates $\mathbb{Q}$, but it doesn't. Note that if $\frac{a}{b}$ and $\frac{s}{t}$ are in least form, then the denominator of the reduced form of $\frac{a}{b}+\frac{s}{t}$ must divide $\mathrm{lcm}(b,t)$, since you can certainly write the sum as a (possibly non-reduced) fraction with denominator $\mathrm{lcm}(b,t)$.

In particular, any sum obtained from the fractions $\frac{1}{p}$ with $p$ a prime must have denominator which is squarefree, so you cannot generate $\mathbb{Q}$ using only $P$: you only obtain fractions whose denominators are squarefree. Thus, for example, $\frac{1}{4}\notin \langle \frac{1}{p}\mid p\text{ prime}\rangle$.

You would need to extend your generating set to include all fractions of the form $\frac{1}{p^n}$. Then you could take all the rationals with odd denominator when written in least terms (which I guess is what you wanted to obtain). Then there are lots of proper subgroups that contain it. For example, for each natural number $n$, the collection of all rationals whose nominator is not divisible by $2^{n+1}$ would be a proper subgroup that contains the subgroup of rationals with odd denominator.