Are $\mathbb Z\times \mathbb Z/(4\mathbb Z\times 5 \mathbb Z)\simeq \mathbb Z$ as groups? Is $(\mathbb Z\times \mathbb Z)/(4\mathbb Z\times 5 \mathbb Z)\simeq \mathbb Z$,
where $4\mathbb Z\times 5 \mathbb Z$ is given as a subgroup of $\mathbb Z\times \mathbb Z$?
My feelings says that I need to find well defined $f:\mathbb Z\times \mathbb Z\to \mathbb Z$ which has a kernel $4\mathbb Z\times 5\mathbb Z$
So using first isomorphism theorem I can show the isomorphism.
I could not find the map and how can I show it in different way?
 A: By definition of the quotients:
\begin{alignat}{1}
(\mathbb Z\times \mathbb Z)/(4\mathbb Z\times 5 \mathbb Z) &= \{(m,n)+4\Bbb Z\times5\Bbb Z\mid m,n\in\Bbb Z\} \\
&= \{(m+4\Bbb Z,n+5\Bbb Z)\mid m,n\in\Bbb Z\} \\
&= \{(k,l)\mid k\in\Bbb Z/4\Bbb Z,l\in\Bbb Z/5\Bbb Z\} \\
&= \Bbb Z/4\Bbb Z \times\Bbb Z/5\Bbb Z \\
\end{alignat}
So, your quotient is precisely (not just isomorphic to) $\Bbb Z/4 \Bbb Z\times\Bbb Z/5\Bbb Z $, which is not isomorphic to $\Bbb Z$, if only for reasons of cardinality.
A: No.
Since $$G:=\Bbb Z\times \Bbb Z\cong\langle a,b\mid ab=ba\rangle,$$
we can write the subgroup $H:=4\Bbb Z\times 5\Bbb Z$ as the normal subgroup $$\langle \langle a^4,b^5\rangle \rangle, $$ so that
$$\begin{align}
G/H &\cong \langle a,b\mid a^4, b^5, ab=ba\rangle\\
&\cong \langle a,b,c\mid a^4, b^5, ab=ba, a=cb^{-1}\rangle \\
&\cong \langle b,c\mid (cb^{-1})^4, b^5, c=b(cb^{-1})\rangle\\
&\cong\langle b,c\mid c^4=b^4, b^5, cb=bc\rangle\\
&\cong\langle b,c\mid c^4=b^{-1}, b^5, cb=bc\rangle \\
&\cong \langle b,c\mid b=c^{-4}, (c^{-4})^5, c^{-3}=c^{-3}\rangle \\
&\cong\langle c,d\mid d=c^{-1}, d^{20}\rangle \\
&\cong\langle d\mid d^{20}\rangle\\
&\cong\Bbb Z/ 20\Bbb Z\\
&\not\cong\Bbb Z.
\end{align}$$
A: What is the kernel of the group homomorphism $\mathbb Z \times \mathbb Z \rightarrow \mathbb Z/4\mathbb Z \times \mathbb Z/5\mathbb Z$ given by
$$(x,y) \mapsto (x+4\mathbb Z, y +5\mathbb Z)?$$
A: Consider the map
$$\mathbb{Z}\times \mathbb{Z}\to \mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z}: (x,y) \mapsto (x+ 4 \mathbb{Z}, y + 5\mathbb{Z})$$
This map is surjective and has  kernel  $\mathbb{4}\mathbb{Z}\times 5\mathbb{Z}$, so by the first isomorphism theorem,
$$(\mathbb{Z}\times \mathbb{Z})/(4\mathbb{Z}\times 5 \mathbb{Z})\cong \mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z}\cong \mathbb{Z}/20\mathbb{Z}$$
The latter group contains $20$ elements and is finite, so the group you are interested in is finite. Hence, it cannot be isomorphic to $\mathbb{Z}$.
A: There shouldn't be such a map. You may want to check for elements with finite order in both groups.
A: There are a number of answers explaining why your isomorphism isn't true, but I am going to try to answer to the question maybe you wanted yo ask: let $A = \mathbb{Z}\times \mathbb{Z}$, and $B = \{(4x,5x)\in A\mid x\in \mathbb{Z}\}$. Note that $B$ is not the subgroup $4\mathbb{Z}\times 5\mathbb{Z}$, and is not even isomorphic to it since it is instead isomorphic to $\mathbb{Z}$.
Then indeed $A/B\simeq \mathbb{Z}$, because $4$ and $5$ are coprime. Explicitly, the surjective morphism $A\to \mathbb{Z}$ given by $(a,b)\mapsto 5a-4b$ has kernel $B$.
