Isomorphism and $(\mathbb Q,+)$ 
Prove that $(\mathbb Q,+)$ is not isomorphic to $(H,+) \neq (\mathbb Q,+)$, a proper subgroup of $(\mathbb Q,+)$.

$\mathbb Q$ is the rationals.
I thought about taking the contradiction direction. If we do that then we have $f:\mathbb Q\to H$ such that $f$ is an isomorphism. That means that that it is surjective and injective, but $(H,+)\neq (\mathbb Q,+)$ so they have different sizes, therefore it can not be surjective and injective, therefore it is not invertible, therefore it is not isomorphic. What's wrong with this solution? It seems too easy to the point that it makes me think that I do not even understand the definitions.
 A: Re: your proposed solution, note that there can be a bijection between $A$ and $B$, even if $A$ is a proper subset of $B$; for instance, $f(x)=2x$ is a bijection between the set of integers and the set of even integers.
Indeed, there are proper subgroups of $(\mathbb{Q}, +)$ which are in bijection (not structure-preserving bijection, though) with $(\mathbb{Q}, +)$ - for instance, $(\mathbb{Z}, +)$. You'll need to use something special about the group $\mathbb{Q}$. 
HINT: Suppose I told you $H$ is a proper subgroup of $\mathbb{Q}$ which is isomorphic to $\mathbb{Q}$, and $7\in H$. Does ${7\over 2}$ have to be in $H$? Why? Do you see how to generalize this?
A: In the case of a finite group $G$, it's obvious that a proper subgroup of $G$ cannot be isomorphic to it, because it has strictly smaller cardinality. However, this is not necessarily true for infinite groups: for instance $2\mathbb{Z}$ is a proper subgroup of $\mathbb{Z}$, but it is isomorphic to $\mathbb{Z}$; the map $x\mapsto 2x$ is indeed an isomorphism from $\mathbb{Z}$ onto $2\mathbb{Z}$.
For $\mathbb{Q}$ the situation is different, because $\mathbb{Q}$ is divisible: for every $x\in\mathbb{Q}$ and every integer $n\ne0$, there exists $y\in\mathbb{Q}$ such that $ny=x$.
Such a property of course transfers to any group that's isomorphic to $\mathbb{Q}$. So you can prove your assertion by showing that no proper subgroup of $\mathbb{Q}$ is divisible (except for $\{0\}$, but this is clearly not isomorphic to $\mathbb{Q}$).
So, let $H\ne\{0\}$ be a divisible subgroup of $\mathbb{Q}$; then $H$ contains a nonzero element, say $a/b$. Now take any $p/q\in\mathbb{Q}$; we want to find $m,n$ such that
$$
m\cdot\frac{a}{nb}=\frac{p}{q}
$$
that is,
$$
maq=nbp
$$
and this is clearly possible by taking $m=bp$ and $n=aq$.
Since $H$ is divisible, then
$$
\frac{a}{nb}\in H
$$
because it's the only element in $\mathbb{Q}$ that, multiplied by $n$, yields $a/b$ (and $n=aq\ne0$). Therefore
$$
\frac{p}{q}=m\cdot\frac{a}{nb}\in H
$$
and so $H$ contains every rational number.

Perhaps the proof can be understood better by observing that a torsionfree divisible abelian group is in a natural way a vector space over $\mathbb{Q}$: torsion freeness ensures the division is unique. Once we know this, we can observe that $\mathbb{Q}$ has dimension $1$ over itself, so no nontrivial subspaces either.
Note that the argument above essentially proves that any nonzero element of $\mathbb{Q}$ spans it as a $\mathbb{Q}$-vector space.
