Find every homomorphism between groups $(\mathbb{Z}\times \mathbb{Z}, +,-,0)$ and $(\mathbb{Q}, +,-,0)$

Question

The problem is to find every group homomorphism $$\varphi:(\mathbb{Z}\times \mathbb{Z}, +,-,0)\longrightarrow(\mathbb{Q}, +,-,0)$$ and the other way around $$\varrho: (\mathbb{Q}, +,-,0) \longrightarrow(\mathbb{Z}\times \mathbb{Z}, +,-,0) $$

I started with the latter one, but I found only one:

1. $\forall x\in\mathbb{Q}: \varrho(x)=(0,0)$

I tried this one: 2. $\varrho(\frac{a}{b})=(a,b), \frac{a}{b}\in\mathbb{Q}$

but it doesn't work, because $$\varrho(\frac{a}{b})+\varrho(\frac{c}{d})=(a+c,b+d)=\varrho(\frac{a+c}{b+d})\not=\varrho(\frac{a}{b}+\frac{c}{d})$$

I couldn't think of any other.

When I tried to find a homomorphism $\varphi$, again, by trying, I came only with the null one:

$$\varphi((a,b))=0$$

How can I find all homomorphisms and be sure that I really found them all?

Answer 1

• $\mathbb{Q} \rightarrow \mathbb{Z}\times\mathbb{Z}$

Any $f: \mathbb{Q} \rightarrow \mathbb{Z}\times\mathbb{Z}$ is trivial since division by arbitrary integers is possible in $\mathbb{Q}$ but not in $\mathbb{Z} \times \mathbb{Z}$.

In details, for a group homomorphism $f: \mathbb{Q} \rightarrow \mathbb{Z}\times\mathbb{Z}$, let $f(1) = (a,b)$, where $a,b \neq 0$. Then, for example, $f\left(\frac{1}{2ab}\right) = (c,d)$ should be such that $2ab \cdot (c,d) = (a,b)$, but this can happen only if $c = \frac{1}{2b}, d = \frac{1}{2a}$, but $c$ and $d$ are integers, a contradiction.

If one of $a, b$ equals zero, a similar argument shows that there is no homomorphism other then the trivial one.

• $\mathbb{Z}\times\mathbb{Z} \rightarrow \mathbb{Q}$

$\mathbb{Z} \times \mathbb{Z}$ is the free abelian group with two generators, which completely describes the homomorphisms from it.

A homomorphism $f: \mathbb{Z}\times\mathbb{Z} \rightarrow \mathbb{Q}$ is fully determined by $f(1,0)$ and $f(0,1)$. Given any rational numbers $r_1, r_2$ you can set $f(1,0) = r_1$, $f(0,1) = r_2$ and extend it by linearity: $f(a,b) = a\cdot r_1 + b \cdot r_2$. So, for any such pair $r_1,r_2$ there is a homomorphism.

Answer 2

Let’s start with $\varrho$.

We need $\varrho(0) = (0, 0)$, and $\varrho(a + b) = \varrho(a) + \varrho(b)$.

Now, suppose we pick some element of $\mathbb Z \times \mathbb Z$ for $\varrho(1)$. Say, $(p, q)$.

Then $(p, q) = \varrho(1) = \varrho(\frac 12) + \varrho(\frac 12)$, so $p$ and $q$ must both be even...

And $(p, q) = \varrho(1) = \varrho(\frac 13) + \varrho(\frac 13) + \varrho(\frac 13)$, so $p$ and $q$ must both be multiples of three...

Et cetera, for all primes. So clearly, $p = q = 0$, the only number that’s a multiple of everything!

In general, $\varrho(\frac ab)$ must satisfy that

$$\underbrace{\varrho(\frac ab) + \varrho(\frac ab) + \dots + \varrho(\frac ab)}_{b \text{ times}} = \varrho(a) = a \cdot \varrho(1) = 0.$$

Now try $\varphi$.