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The structure $(\mathbb{Z}^2,+)$, where the addition on $\mathbb{Z}^2$ is defined by $(a,b)+(c,d) = (a + c,b+ d)$, forms an additive abelian group.

The map $\phi:\mathbb{Z}^2\rightarrow \mathbb{Z}$ is defined by $\phi (a,b)=b-a$.

I have shown that $\phi$ is an homomorphism from $(\mathbb{Z}^2, +)$ to $(\mathbb{Z}, +)$.

I have calculated the kernel of the map $\ker \phi =\{(a,a)\mid a\in \mathbb{Z}\}$.

Since the kernel doesn't contain only the zero vector, $\phi$ is not injective and so $\phi$ is not an isomorphism.

It is given that $\ker\phi$ is a normal subgroup $(\mathbb{Z}^2, +)$. Why does this holds? Is this always true for the kernel?

The coset of $\ker\phi$ with representative $(3,7)$ is \begin{align*}(3,7)+\ker\phi&=(3,7)+\{(a,a)\mid a\in \mathbb{Z}\}\\ & =\{(3,7)+(a,a)\mid a\in \mathbb{Z}\}\\ & =\{(x,y)\mid y-x=7\}\end{align*}

The coset of $\ker\phi$ with representative $(\alpha, \beta)$ is \begin{align*}(\alpha, \beta)+\ker\phi&=(\alpha, \beta)+\{(a,a)\mid a\in \mathbb{Z}\}\\ & =\{(\alpha, \beta)+(a,a)\mid a\in \mathbb{Z}\}\\ & =\{(x,y)\mid y-x=\beta-\alpha\} \\ & = \{(x,y)\mid y-x=\phi (\alpha, \beta)\}\end{align*}

Do we use that to get an isomorphism between $\mathbb{Z}^2/\ker\phi$ and $\text{im}(\phi)$ ? But how?

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  • $\begingroup$ If you know that $\varphi$ is a homomorphism, then $\mathbb Z^2/\mathrm {Ker}(\varphi) \cong \mathrm {Im} (\varphi)$ by fundamental homomorphism theorem. Such isomorphism could be $(0,a) + \{(x,y) \colon x= y\} \mapsto a$. $\endgroup$ – xbh Aug 25 '18 at 14:48
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    $\begingroup$ Yes. It is always true that the kernel of any group homomorphism is a normal subgroup. In your case though it is even simpler: any subgroup of any abelian group is normal $\endgroup$ – DonAntonio Aug 25 '18 at 14:48
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It is given that $\kerϕ$ is a normal subgroup $(\mathbb{Z}^2,+). Why does this holds? Is this always true for the kernel?

Yes, the kernel of a group homomorphism is always normal. Additionally we are dealing with abelian groups here, so every subgroup is normal.

Do we use that to get an isomorphism between $\mathbb{Z}^2/\kerϕ$ and $\text{im}(ϕ)$ ? But how?

From your last point we can conclude $$ (\alpha, \beta) + \ker\phi = \{(x,y) \mid y = \phi(\alpha,\beta) + x \} = \{(x,x + \phi (\alpha,\beta)\mid x \in \mathbb{Z}\}.$$

The induced map $\bar{f}:\mathbb{Z}^2 / \ker\phi \rightarrow \mathbb{Z}$ is defined by $\bar{f}((\alpha,\beta) + \ker\phi) = \phi(\alpha,\beta)$.

To show that it is an ismorphism onto $\text{im}\phi$ we only have to show injectivity. Point here is, that two cosets $(\alpha,\beta) + \ker\phi$ and $(\gamma, \delta) + \ker\phi$ have the same image if and only if $(\alpha,\beta) - (\gamma,\delta) \in \ker\phi$. This happens if and only if $(\alpha, \beta) + \ker \phi = (\gamma, \delta) + \ker\phi$.

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  • $\begingroup$ Let $f:\mathbb{Z}^2/\ker \phi\rightarrow \text{im}\phi$. The elements in $\mathbb{Z}^2/\ker \phi$ are in th eform $(\alpha,\beta)+\ker\phi$, or not? Do we have to define the map $f$ so that we get the isomorphism? I got stuck right now. $\endgroup$ – Mary Star Aug 25 '18 at 15:21
  • $\begingroup$ Sorry, I confused myself and my writing was not clear. I tried to explain the isomorphism, is this better? $\endgroup$ – red_trumpet Aug 25 '18 at 15:38
  • $\begingroup$ Why do we have to show only the injectivity? $\endgroup$ – Mary Star Aug 25 '18 at 16:57
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For $f:A\to B$ homomorphism of groups $\ker f \lhd A$ since

$\ker f=\{x\in A:f(x)= 1\}\Rightarrow f(gxg^{-1})=f(g)f(x)f(g^{-1})=f(g)f(g^{-1})=f(g)f(g)^{-1}=1\quad \forall x\in\ker(f),g\in A \Rightarrow gxg^{-1}\in\ker(f)\quad \forall x\in\ker(f),g\in A$

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