TITLE: $(\mathbb{Q},+)$ has no maximal subgroups
QUESTION [27 upvotes]: I have a problem that I don't have any idea.

Show that group $(\mathbb{Q},+)$ has no maximal subgroups.

REPLY [13 votes]: 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.