Let $f: \Bbb{Z} \times \Bbb{Z} \to G $ be an epimorphism, $\ker f$ be generated by $\langle (3,0),(0,5) \rangle$. What type of abelian group is $G$?

Question

Let $f: \Bbb{Z} \times \Bbb{Z} \to G $ be an epimorphism, $\ker f$ be generated by $\langle (3,0),(0,5) \rangle$. What type of abelian group is $G$?

I don't even know the context of that question since I found it on the back of my abstract algebra book (without a connection to any particular topic). Any hint would be much appreciated.

EDIT (based on comment below): I know that since f is surjective then $G$ is isomorphic to $\Bbb Z\times \Bbb Z/\ker(f)$ so it is isomorphic to $\Bbb Z\times \Bbb Z/\langle(3,0),(0,5)\rangle$. I see that $3$ and $5$ are prime numbers, but I don't know what to say more.

Answer 1

Since $f$ is an epimorphism, we can apply the First Isomorphism Theorem to get $(\Bbb Z\times\Bbb Z)/\ker f\cong G$.

Let $x=(1,0), y=(0,1)$. Then the presentation

$$\Bbb Z\times\Bbb Z\cong \langle x,y\mid x+y=y+x\rangle$$

is standard.

Note that $(3,0)=3x$ and $(0,5)=5y$. Then

$$\begin{align} \Bbb Z\times\Bbb Z/\ker f &\cong \langle x,y\mid 3x, 5y, x+y=y+x\rangle\\ &\cong\Bbb Z_3\times \Bbb Z_5\\ &\cong \Bbb Z_{15}, \end{align}$$

where the final isomorphism holds by the Chinese remainder theorem and the fact that $\gcd(3,5)=1$.