# Isomorphism between $\mathbb{Z}^2/\ker\phi$ and $\text{im}(\phi)$

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?

• 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 $(0,a) + \{(x,y) \colon x= y\} \mapsto a$. – xbh Aug 25 '18 at 14:48
• 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 – DonAntonio Aug 25 '18 at 14:48

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$(\alpha, \beta) + \ker \phi = (\gamma, \delta) + \ker\phi$. • 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. – Mary Star Aug 25 '18 at 15:21 • Sorry, I confused myself and my writing was not clear. I tried to explain the isomorphism, is this better? – red_trumpet Aug 25 '18 at 15:38 • Why do we have to show only the injectivity? – Mary Star Aug 25 '18 at 16:57 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\$