$2\mathbb{Z}\cong 3\mathbb{Z}$ as $\mathbb{Z}$-modules 
Why is $2\mathbb{Z}\cong 3\mathbb{Z}$ as $\mathbb{Z}$-modules?

I gave the following homomorphism:
$f:2\mathbb{Z}\to 3\mathbb{Z}$, $n\mapsto \frac32n$
Since $n\in 2\mathbb{Z}$ this function is well-defined, because $n=2m$ and $\frac{3}{2}n\in 3\mathbb{Z}$
f is a homomorphism of abelian groups, since
$f(a+b)=\frac32(a+b)=\frac32a+\frac32b=f(a)+f(b)$ and for every $z\in\mathbb{Z}$ it is 
$f(za)=\frac32za=z\cdot\frac32a=zf(a)$,
hence f is a $\mathbb{Z}$-homomorphism.
Also it is a bijection with the obvious inverse function $f^{-1}: 3\mathbb{Z}\to 2\mathbb{Z}$, $m\mapsto \frac23m$

Question: Can we define $f$ like this? Since it uses a operation which is not necessarly well-defined on $\mathbb{Z}$, since $\frac32\in\mathbb{Q}$, or does this simply not matter?

Thanks in advance.
 A: This is perfectly good, since it is well defined as a map from $2 \mathbb Z \to \mathbb 3 \mathbb Z$.
If it makes you more comfortable, you could remark that as a $\mathbb Z$ module, it is enough to specify maps of generators and take $2 \mathbb Z \to 3\mathbb Z$ by $2  \mapsto 3$, which extends uniquely to a $\mathbb Z$-linear homomorphism since both modules are free on one generator, and moreover it is exactly your map.

In light of Ennar's comment, I will include following observation for posterity:
the homomorphism is uniquely determined since $f(2n)=nf(2)=3 \cdot n$ by $\mathbb Z$ linearity.
A: Another way to think of the map $n\mapsto \frac 32 n$ is as composition $2\mathbb Z\to \mathbb Z\to 3\mathbb Z$ where the first map is the inverse of the map $n\mapsto 2n\colon\mathbb Z\to 2\mathbb Z$. 
You shouldn't be worried about fractional notation since it is perfectly reasonable to denote the inverse of $n\mapsto 2n$ by $n\mapsto \frac 12 n$. The notation fails if and only if $n\mapsto 2n$ isn't invertible.
