$(\mathbb{Q},+)$ has no maximal subgroups

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.

Fixed it

Assume by contradiction that $H$ is a maximal subgroup of $\mathbb Q$.

As for $r \neq 0$ the function $f(x)=rx$ is a group automorphism of $\mathbb Q$, by replacing $H$ by $f(H)$ we can assume without loss of generality that $1 \in H$.

Now, if $\frac{1}n \in H$ for each $n > 1$ it is easy to prove that $H =\mathbb Q$. Pick $n$ to be the smallest $n$ such that $\frac{1}{n} \notin H$.

Then $H + < \frac{1}{n} > =\mathbb Q$.

Now, for each positive integer $l$ if $l=qn+r$ we have $\frac{l}{n}=q+\frac{r}{m}$ and $q \in H$.

It follows from the above that each rational number can be written in the form $$r=h+\frac{k}{n} \, \mbox{ with } h \in H, 0 \leq k < n$$

Therefore, $$\frac{1}{n^2}= h+\frac{k}{n} \, \mbox{ with } h \in H, 0 \leq k < n$$

Multiplying both sides by $n$ we get $$\frac{1}{n}= nh+k \in H$$ as $nh \in H$ and $k \in h$.

This is a contradiction.

There is another simple proof, maybe an explanation of this answer:

Assume that $N$ is a maximal subgroup of $\mathbb{Q}$, then the quotient group $\mathbb{Q}/N$ must be simple abelian group, i.e. cyclic group with prime order. Let's say $\mathbb{Q}/N \cong \mathbb{Z}_p$ for some prime $p$. Since $N \neq \mathbb{Q}$, there exists $a \notin N$. Since $\mathbb{Q}/N \cong \mathbb{Z}_p$, we have $N = p(\frac{a}{p} + N) = a + N$ which implies $a \in N$, a contradiction.