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Let $\phi(x,y,u,v):=(x+2y+u, y-3v)$ from $\mathbb{Z}^4 \rightarrow \mathbb{Z}^2$

How do I show that? $\ker(\phi) \cong \mathbb{Z}^2$

I have tried to set $y=3v$ and then $x+2 \cdot 3v+u = 0 \Leftrightarrow u=-x-6v$ but cannot get any further.

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2 Answers 2

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So you have proved that $\ker(\phi)=\{(x,3v,-x-6v,v)\mid x,v\in \mathbb{Z}\}$. Now consider $f:\ker (\phi)\to \mathbb{Z}^2$ defined by $f(x,3v,-x-6v,v)=(x,v)$, clearly $f$ is a bijection, and it is also easy to prove that is an homomorphism (w.r.t addition, I suppose).

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You can choose any $x\in \mathbb{Z}$ and $v\in \mathbb{Z}$, and then set $u=x+6v$ and $y=3v.$ Then $\phi(x,y,u,v)=0.$ Thus the kernel is $\mathbb{Z}^2.$

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  • $\begingroup$ I think you mean $u=-x-6v$, right? I have also updated the description because of typo $\endgroup$
    – Daniel
    Dec 30, 2020 at 9:59

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