Homomorphism $\phi : (\mathbb{Z} \oplus \mathbb{Z}) \rightarrow (G, +)$ via $(3,2) \mapsto x$ and $(2,1) \mapsto y$. Find $\phi((4,4))$

Question

$\phi : (\mathbb{Z} \oplus \mathbb{Z}) \rightarrow (G, +)$ via $(3,2) \mapsto x$ and $(2,1) \mapsto y$. Find $\phi((4,4))$ where $\phi$ is a homomorphism

Since we know $\phi$ is a homomorphism:
$\phi((3,2),(2,1)) = \phi((3,2))\phi(2,1)$
$\phi(3+2, 2+1) = \phi(3,2)\phi(2,1)$

And here's where I'm stuck. We have found what $\phi(5,3) = x+y$, but I don't see any steps I can take to find what $\phi(4,4)$ equals in terms of $x$ and $y$

Answer 1

Observe that $4(3, 2) - 4(2, 1) = (4, 4)$.

Answer 2

Note that a map $\varphi : \mathbb{Z} \oplus \mathbb{Z} \to G$ is determined by $\varphi(1,0)$ and $\varphi(0,1)$, as

$$ \varphi(a,b) = \varphi(a,0) + \varphi(0,b) = \overbrace{\varphi(1,0) + \dots \varphi(1,0)}^{\text{$a$ times}} + \overbrace{\varphi(0,1) + \dots \varphi(0,1)}^{\text{$b$ times}}. $$

Now, notice that

$$ (1,0) = 2(2,1) - (3,2) \text{ and } (0,1) = 2(3,2) - 3(2,1) $$

and apply $\varphi$ to both sides.

This gives $\varphi(1,0) = 2y-x$ and $\varphi(0,1) = 2x-3y$ where we note $nx$ summing (by that I mean using the $+$ we have in $G$) $n$ times $x \in G$. Hence, $\varphi(a,b) = a(2y-x) + b(2x-3y) = x(2b-a) + y(2a-3b)$ and in particular, $\varphi(4,4) = 4x -4y = 4(x-y)$.