Computing Factor Group step-by-step manually I am reading Fraleigh's Abstract Algebra $\S$15 on factor group, Example #15.11: 

Compute factor group $(\mathbb Z_4 \times \mathbb Z_6) / \langle (2, 3)\rangle.$

The text gives a solution that is only a shortcut and I think it's applicable only to this problem. I would like to do it in manually, dumbly, in step-by-step fashion such that I can internalize the concept better; admittedly factor group is a hard concept for beginners.
I did the following tediously:
$$\begin{align}
\mathbb Z_4 \times \mathbb Z_6 = \{ &(0, 0), (0,1), (0,2), (0, 3),(0, 4), (0,5),\\
&(1, 0), (1,1), (1,2), (1, 3),(1, 4), (1,5),\\
&(2, 0), (2,1), (2,2), (2, 3),(2, 4), (2,5),\\
&(3, 0), (3,1), (3,2), (3, 3), (3, 4), (3,5)\}
\end{align}$$
Questions: (1) Is it possible to manually solve the problem like this? 
(2) If it is, how do I mode out each of them with $\langle (2, 3) \rangle$? Just give me a few examples.
The text says the factor group is isomorphic to either $\mathbb Z_{12}$ or $\mathbb Z_4 \times \mathbb Z_3$. Thank you for yoru time and effort. 

Post Script: I must admit that there is a possible duplicate here which I found out belated only after I posted mine. But keep in mind that my question is about how to solve it manually to enforce my understanding. Thank you.
 A: I'll show you how to compute it manually. 
The elements of the factor (or quotient) group are the cosets of the subgroup $H=\langle (2,3)\rangle$, which is normal in $\Bbb Z_4 \times \Bbb Z_6$ because $\Bbb Z_4 \times \Bbb Z_6$ is abelian. 
To find all the cosets, we take each element $(h,k) \in \Bbb Z_4 \times \Bbb Z_6$ and  compute $$(h,k)+H =\{ (h,k)+(0,0), (h,k)+(2,3)\} = \{(h,k), (h+2,k+3)\}.$$ Doing this, we get the following:
$$
H = \{(0,0), (2,3)\}\\
(0,1)+H=\{(0,1),(2,4)\}\\
(0,2)+H=\{(0,2), (2,5)\}\\
(0,3)+H=\{(0,3), (2,0)\}\\
(0,4)+H = \{(0,4), (2,1)\}\\
(0,5)+H = \{(0,5), (2,2)\}\\
(1,0)+H = \{(1,0), (3,3)\}\\
(1,1)+H = \{(1,1), (3,4)\}\\
(1,2)+H = \{(1,2), (3,5)\}\\
(1,3)+H = \{(1,3), (3,0)\}\\
(1,4)+H = \{(1,4), (3,1)\}\\
(1,5)+H = \{(1,5), (3,2)\}
$$
We know that we're done once we have a partition of $\Bbb Z_4 \times \Bbb Z_6$. To check this, just check that every element of $\Bbb Z_4 \times \Bbb Z_6$ is in one of the cosets. 
(This idea of the cosets being the equivalence classes of the equivalence relation $a \sim b \iff a-b \in H$ is used in the proof of Langrange's theorem, and it also explains why two cosets are either disjoint or equal; it's a property of equivalence classes.)
We could also argue that we've found all the cosets by using Lagrange's theorem, which tells us that the  factor (or quotient) group should have 12 elements: $$[\Bbb Z_4 \times \Bbb Z_6 : \langle (2,3)\rangle] = \frac{|\Bbb Z_4 \times \Bbb Z_6|}{|\langle (2,3)\rangle|} = \frac {24}{2}=12.$$

Now, to prove this group is isomorphic to $Z_{12}$, you could, as A.P. did, use the fundamental theorem of finitely generated abelian groups, or you could show directly that the group is cyclic (because all cyclic groups of a given order are isomorphic):
To show $G= \Bbb Z_4 \times Z_6 / \langle(2,3)\rangle$ is cyclic, we will show that the element $(1,1)+H$ has order 12:
We know its order must divide $|G|=12$, so it can be $1,2,3,4, 6,$ or $12$.
Its order is not 1 because $(1,1)+H \neq H$, the identity of $G$.
Its order is not $2,3,$ or $6$, because $6((1,1)+H)= (6,6)+H=(2,0)+H \neq H$
Its order is not $4$ because $4((1,1)+H)=(4,4)+H = (0,4)+H \neq H$.
Thus, its order is 12, meaning  $\Bbb Z_4 \times Z_6 / \langle(2,3)\rangle = \langle (1,1)+H \rangle$ is cyclic, so $$\Bbb Z_4 \times Z_6 / \langle(2,3)\rangle \cong \Bbb Z_{12} \cong \Bbb Z_3 \times \Bbb Z_4$$.
A: We can definitely compute quotients of finite groups by inspecting each element, but one almost never does, because it is a tedious and error prone process.
In this case, note that
$$
(2,3) + (2,3) = (4,6) \equiv (0,0)
$$
so $(2,3)$ has order $2$, i.e. that $\langle (2,3) \rangle = \{(0,0),(2,3)\}$. This means that the quotient group has cardinality $24/2 = 12$, so by the fundamental theorem of finitely generated abelian groups the quotient is either $\Bbb{Z}_{12} \simeq \Bbb{Z}_4 \times \Bbb{Z}_3$ (the isomorphism is due to the Chinese Remainder Theorem) or $\Bbb{Z}_2 \times \Bbb{Z}_2 \times \Bbb{Z}_3$.
We can easily distinguish between those two cases, because the latter doesn't have any elements of order $4$. Indeed, observe that, say
$$
(1,0),(2,0),(3,0) \notin \langle (2,3) \rangle
$$
which means that the class of $(1,0)$ has order $4$ in the quotient. Thus $(\Bbb{Z}_4 \times \Bbb{Z}_6) / \langle (2,3) \rangle \simeq \Bbb{Z}_{12}$.
