Show that $M$ is not maximal in $\mathbb Q$ $\mathbb Q$ is a $\mathbb Z$ module. Let $M$ be a proper submodule of $\mathbb Q$.

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*Show that there is some $0 \neq b \in \mathbb Z$, such that $1/b \notin M$
If for all $b \in \mathbb Z$, $1/b \in M$, then $\mathbb Q = M$. contradiction. this is easy.


*Show that $M \cap \mathbb Z \neq 0$. This is also easy.


*Show that $M$ is not maximal in $\mathbb Q$. I can't figure this out.
i said that since $1/b \notin M$, we have $\mathbb Q = M + (1/b) \mathbb Z$.
i'm trying to find a contradiction from here. any help?
i also thought that if $M$ is maximal then $\mathbb Q/M$ is simple, then it is also cyclic. but this doesn't look useful.  thanks for help.
 A: $\;\Bbb Q\;$ is a divisible abelian group (=divisible $\;\Bbb Z-$module), and this means that for any $\;n\in\Bbb N,\,\,q\in\Bbb Q\;$ there exists $\;p\in\Bbb Q\;$ s.t. $\;np=q\;$ . 
Observe that any homomorphic image of a divisible group is also divisible, but no finite abelian group (in fact, no abelian group with finite exponent) can be divisible, so if $\;M\le\Bbb Q\;$ is maximal then $\;\Bbb Q/M\;$ is simple and thus finite (and of prime order, but we don't care about this), and thus it can't be divisible, contradiction.
A: A hugely elementary proof, since I don't know any algebra:
First, wlog $M\ne0$. Hence $M$ contains a non-zero integer.
Lemma Wlog $1\in M$.
Proof: Say $n\in M\cap\Bbb Z$, $n\ne 0$. Let $M'=\frac1n M$. Then
$M$ is maximal if and only if $M'$ is maximal. QED.
Lemma If $a,b\in\Bbb Z$, $a/b\in M$, $(a,b)=1$, then $1/b\in M$.
Proof: There exist $n,m\in\Bbb Z$ with $na+mb=1$. Hence $1/b=na/b+m\in M$, since $1\in M$. QED$
So $M\ne\Bbb Q$ implies that there exist $p$ and $n$ such that $1/p^n\in M$ but $1/p^{n+1}\notin M$ (otherwise the previous lemma shows that the reciprocal of any product of powers of primes is in $M$, so $M=\Bbb Q)$).
So $M$ is a proper submodule of $\frac1p M\ne\Bbb Q$.
