$(\Bbb Q,+)$ is not isomorphic to any of its proper subgroups. 
Show that 
   $(\Bbb Q,+)$ is not isomorphic to any of its proper subgroups. 

Let $H$ be  a proper subgroup of $\Bbb Q$ such that $H\cong \Bbb Q$ such that $\exists h\in \Bbb Q$ but $h\notin H$.
Let $f:H\to \Bbb Q$ be the isomorphism then $f(a)=h$ for some $a\in H$.
But how to proceed from here?
 A: Suppose $\phi$ is an isomorphism from $\mathbb{Q}$ to a subgroup $H$, and let $\phi(1)=\frac{a}{b}\neq 0$(this is the only place we need to use $\phi$ is an isomorphism). Let $\frac{c}{d}\in \mathbb{Q}$; then 
$$
\frac{c}{d}=\frac{1}{d}+\dots+\frac{1}{d},
$$
and thus 
$$
\phi\left(\frac{c}{d}\right)=c\phi\left(\frac{1}{d}\right).
$$
Furthermore, we have 
$$
\underbrace{\frac{1}{d}+\dots+\frac{1}{d}}_{d\text{ times.}}=1,
$$
and thus 
$$
d\phi\left(\frac{1}{d}\right)=\phi(1),
$$
so all in all we have $$\phi\left(\frac{c}{d}\right)=\frac{c}{d}\phi(1).$$
Now $\phi(1)=\frac{a}{b}$; let $\frac{c}{d}\in \mathbb{Q}$ be arbitrary. We have 
$$
\phi\left(\frac{cb}{ad}\right)=\frac{cb}{da}\phi(1)=\frac{c}{d},
$$
and thus $\phi$ must map surjectively to $\mathbb{Q}$.
A: $\Bbb Q$ has the property that it is divisible: if $a\in \Bbb Q$
and $n\in\Bbb N$ then there is $b\in\Bbb Q$ such that $nb=a$.
Here by $nb$ I mean $b+b+\cdots+b$, the $n$-fold sum of $b$ with itself.
This is a group-theoretic property.
But the only divisible subgroups of $\Bbb Q$ are $\{0\}$ and $\Bbb Q$
itself.
