Clarification regarding the choice $x/a$ for an element that's not in $H + \langle x\rangle$ (Proof that $\Bbb Q$ has no maximal subgroup)

Question

This answer gives a proof of the fact that $(\Bbb Q, +)$ has no maximal subgroup.

Suppose $H$ is any nonzero proper subgroup of $\mathbb Q$ and let $x \in \mathbb Q \setminus H$ and $y \in H, y \neq 0$

Write $\dfrac {y}{x} = \dfrac {a}{b}$ with integers $a,b$. Then $a \neq 0$ and $\dfrac {x}{a} \notin H + \langle x \rangle$ : Suppose $\dfrac {x}{a} = h + nx$ for some $n \in \mathbb Z$ and $h \in H$. Then $x = ah+anx = ah+nby \in H$, which contradicts the hypothesis on $x$. Thus $H$ is not maximal.

Questions:

I understand that the idea is that for any non-zero proper subgroup $H$ of $G$, we can show that there exists another non-zero proper subgroup of $G$ called $H' = H + \langle x \rangle$ that strictly contains $H$.

However, how does the author guess out of the blue that $\frac{x}{a}$ is a good candidate for a number that may not be contained in $H + \langle x \rangle$? The choice of $\frac{x}{a} = \frac{yb}{a^2}$ appears rather arbitrary and is not well explained.

Answer 1

We want to find a group that is larger than $H$ but still isn't all of $\Bbb Q$. That is, we want a group $H'$ that contains $H$ as well as some $x\notin H$ and we also want to show that there exists some $z\in\Bbb Q$ such that $z\notin H'$; in particular $z\notin H$. If $H\le \Bbb Q$ and we know $x\notin H$, then the only numbers $z$ of which we can immediately tell that they are also not in $H$ are those which would immediately force $x\in H$. The latter would be the case if $x$ is an integer multiple of $z$, i.e., $x=az$ for some integer $a$. This motivates the search for $z$ of the form $\frac xa$ with $a\in\Bbb Z$.

Unfortunately, it might still happen that $z\in H':=H+\langle x\rangle$, namely precisely if there exists $h\in H$ and an integer $n$ with $z=h+nx$. We want to make this impossible. How? We want to make sure that $ \frac xa=h+nx$ leads to a contradiction to what we started with, which is $x\notin H$. So we want to find some non-zero integer $a$ such that for all $h\in H$ and all integers $n$, $\frac xa=h+nx$ implies $x\in H$.

Now, $\frac xa=h+nx$ implies $x=ah+nax$, where certainly $ah\in H$. so in order to conclude $x\in H$, it would be nice to have $nax\in H$ for all $n$, in particular, $ax\in H$. So we want $ax=y$ for some $y\in H$. This makes $a=\frac yx$. In general, however, if we pick some $0\ne y\in H$, then $\frac yx$ will only be a rational, not an integer. Fortunately, if $\frac yx=\frac ab$ in shortest terms, then also $by\in H$ and so $\frac{by}x=a$ gives us the desired result.