Prove that $\mathbb{Z}$ and $\mathbb{Q}$ are not isomorphic as groups I'm working on the following problem:

Prove that $\mathbb{Z}$ and $\mathbb{Q}$ are not isomorphic as groups.

Here is my attempt at a solution:
If $\mathbb{Z} \cong \mathbb{Q}$, then there must exist a bijective homomorphism $\varphi: \mathbb{Z} \to \mathbb{Q}$. Consider $\varphi(x) = x$. Clearly, this is a homomorphism as it is the identity map. Moreover, it is also clearly injective. However, it fails to be surjective since there exist elements of $\mathbb{Q}$ that are not mapped to by $\varphi$ (like $3/2$ or $1/4$). Hence, since $\varphi$ is an injective homomorphism, but not surjective, we see that there cannot exist a bijection between $\mathbb{Z}$ and $\mathbb{Q}$. Thus, $\mathbb{Z} \not \cong \mathbb{Q}$.
Could anyone critique this solution? Am I on the right path?
 A: Your solution proves that there is an injective, but not surjective homomorphism from $\mathbb Z\rightarrow \mathbb Q$. This is not the statement you wished to prove - it certainly doesn't prove that $\mathbb Z$ and $\mathbb Q$ are not isomorphic, because I can say the same about $f(x)=x+x$ mapping $\mathbb Z\rightarrow\mathbb Z$ - but that certainly doesn't mean that $\mathbb Z$ and $\mathbb Z$ are not isomorphic (because they obviously are).
A nice simple solution would be to notice there is no element $x$ in $\mathbb Z$ such that $x+x=1$, but for every $y$ in $\mathbb Q$, there is a $x$ such that $x+x=y$. The groups are therefore not isomorphic.
A: Another thought: Let $\phi:\Bbb Z\to\Bbb Q$ be a homomorphism, and let $\phi(1)=a$. Because, $\Bbb Z$ is cyclic, $\phi$ is completely determined. 
$$\phi(n)=na$$
It's easy now to show that $\phi$ is not surjective.
A: Your solution is incorrect. You have shown that one particular homomorphism is not an isomorphism. 
To show that they are not isomorphic you need to show that there is no possible isomorphism whatsoever. 
HINT: One of these groups is cyclic. 
A: $\mathbb Q$ has a subgroup $N:=\mathbb{Z}$ such that $\mathbb{Q}/N$ is infinite
