Let $G/H = \mathbb{Z}_{4} \times \mathbb{Z}_{10} / \langle (2, 4) \rangle$. I know that $|G/H|$ = 4, so $G/H \simeq \mathbb{Z}_{2} \times \mathbb{Z}_{2}$ or $\mathbb{Z}_{4}$. Since $G/H$ has an element of order 4, namely $(0, 1) + \langle (2, 4) \rangle$, $G/H \simeq \mathbb{Z}_{4}$. Is my reasoning correct?
Also why is $\mathbb{Z} \times \mathbb{Z}_{6}/ \langle (1, 2) \rangle \simeq \mathbb{Z}_{6}$ and not $\mathbb{Z}$?
Edit
So for the first question, $\mathbb{Z}_{4} \times \mathbb{Z}_{10} / \langle (2, 4) \rangle$ becomes $\mathbb{Z}_{4} \times \mathbb{Z}_{10} / \langle (2, 0), (0, 2) \rangle \simeq \mathbb{Z}_{10}/ \langle 2 \rangle \times \mathbb{Z}_{4} / \langle 2 \rangle \simeq \mathbb{Z}_{2} \times \mathbb{Z}_{2}$.
Could you elaborate on how you approached my second question? I still do not follow.
2nd Edit
I see how $\mathbb{Z} \times \mathbb{Z}_{6} / \langle(3,0)\rangle \simeq \mathbb{Z}_{3} \times \mathbb{Z}_{6}$. And we want to do that because it is simple and $(3,0) \in \langle (1, 2) \rangle$. Then I think your next step is to compute $\mathbb{Z}_{3} \times \mathbb{Z}_{6} / \langle (1,2 )\rangle$. But then $|\mathbb{Z}_{3} \times \mathbb{Z}_{6} / \langle (1,2 )\rangle|$ = 6. So the resulting quotient must be $\mathbb{Z}_{6}$.