# Show that $\ker(f)$ is isomorphic to $\mathbb Z^{2}$

I have a function $$f: \mathbb Z^4 \to \mathbb Z^2$$ defined by: $$f(x,y,u,v):= (x+2y+u,y-3v)$$ I have to show that it's a homomorphism and that $$\ker(f)$$ is isomorphic to $$\mathbb Z^2$$.

I showed the first part just by looking at $$f((x+x'),(y+y'),(u+u'),(v+v'))$$ and got that it was equal to $$f(x,y,u,v)+f(x',y',u',v')$$ which by definition mean, that it's homomorphic.

But I have trouble showing that $$\ker(f)$$ is isomorphic to $$\mathbb Z^2$$, I want to define a homomorphism from the kernel to $$\mathbb Z^2$$ and show that it's a isomorphism, but don't know how to proceed , any suggestions?

Lastly, What possibilities are there generally for the structure of the kernel for a homomorphism $$g: \mathbb Z^4 \to \mathbb Z^4$$? Could we use the structure theorem for finitely generated abelian groups maybe?

Well, $$(x, y, u, v)\in\ker f \iff f(x, y, u, v)=(0, 0)\iff (x+2y+u, y-3v)=(0, 0)$$, which is equivalent to the system of equations: $$\begin{cases} x+2y+u=0 \\ y-3v=0 \end{cases}$$ From the second equation we get $$y=3v$$ so $$x+6v+u=0$$ or $$x=-6v-u$$ and $$y=3v$$. This implies $$\ker f=\{(-6v-u, 3v, u, v):v, u\in\mathbb{Z}^2\}$$ and from here it is clear $$\ker f\cong\mathbb{Z}^2$$ by the isomorphism $$\varphi:\mathbb{Z}^2\to\ker f$$ given by $$\varphi(u, v)=(-6v-u, 3v, u, v)$$ (do check that this is indeed an appropriate isomorphism).
Hint: $$f$$ is surjective because $$f(1,0,0,0)=(1,0)$$ and $$f(-2,1,0,0)=(0,1)$$.