Use first isomorphism theorem to show that $(\mathbb{Z}\oplus\mathbb{Z})/\langle(1,3)\rangle$ is isomorphic to $\mathbb{Z}$.
Explain why $(\mathbb{Z}\oplus\mathbb{Z})/\langle(3,6)\rangle$ is not isomorphic to $\mathbb{Z}$.
For the first part I think I have to take the mapping $\mathbb{Z}\oplus\mathbb{Z}\rightarrow\mathbb{Z}$ defined by $(a,b)\mapsto (3a - b)$ then kernel will be $\langle (1,3)\rangle$.
For the second part I have no idea how do I show that it is not a isomorphism. What I think is since here $\gcd(3,6)= 3$ but in the first case $1$ and $3$ where coprime (this is somewhere related with this kind of isomorphism). May be i am wrong. Any hint will be enough. Thank you.
Hint: Unlike $\Bbb Z$, the group $\Bbb Z \oplus \Bbb Z/ \langle (3,6) \rangle$ contains a non-identity element of finite order.
In other words, its torsion subgroup is non-trivial.