Computing Quotient Groups $\mathbb{Z}_4 \times \mathbb{Z}_{10} / \langle (2, 4) \rangle$, $\mathbb{Z} \times \mathbb{Z}_{6}/ \langle (1, 2) \rangle$ 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 $\mathbb{Z}_{6}$.
 A: Actually, $(0,1)+\langle(2,4)\rangle$ is not of order $4$. Note that 
$$\langle (2,4)\rangle = \{ (2,4), (0,8), (2,2), (0,6), (2,0), (0,4), (2,8), (0,2), (2,6), (0,0)\}$$
so
$$\bigl( (0,1)+\langle (2,4)\rangle\bigr) + \bigl( (0,1)+\langle (2,4)\rangle\bigr) = (0,2)+\langle(2,4)\rangle = (0,0)+\langle(2,4)\rangle.$$
In fact, note that $\langle(2,4)\rangle = \langle (2,0), (0,2)\rangle$, which should make the isomorphism type of the quotient very clear.
For the second, notice that $(3,0)\in\langle(1,2)\rangle$. So you can first mod out by $(3,0)$ as a first approximation; we have $\mathbb{Z}\times\mathbb{Z}_6/\langle(3,0)\rangle \cong \mathbb{Z}_3\times\mathbb{Z}_6$. Now you want to quotient out this by the subgroup generated by (the image of) $(1,2)$. 
$$\langle(1,2)\rangle = \{ (1,2), (2,4), (0,0)\}.$$
So $(0,1)+\langle (1,2)\rangle$ is of order $6$, which shows that the quotient is cyclic of order $6$.
Intuitively, taking the quotient modulo $\langle(1,2)\rangle$ "identifies" the $1$ in $\mathbb{Z}$ with the $2$ in $\mathbb{Z}_6$; that means that the $1$ in $\mathbb{Z}$ is of order $3$ in the quotient (as we saw), and that twice $(0,1)$ is the same as $(1,0)$. So from $(0,1)$ has order $6$ in the quotient; since $(1,0)$ and $(0,1)$ generate $\mathbb{Z}\times\mathbb{Z}_6$, knowing their images tells you exactly what happens to the whole group.
