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

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 $\mathbb{Q}$.

$\Bbb Q$ 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.