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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?

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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).

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  • $\begingroup$ Thanks! Very useful. Could u maybe help with the last part as well about the structure of the kernel? $\endgroup$ – Lil coNe Apr 2 at 21:26
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Hint: $f$ is surjective because $f(1,0,0,0)=(1,0)$ and $f(-2,1,0,0)=(0,1)$.

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