$(Z_4 \oplus Z_{12})/\langle(2,2) \rangle $ is isomorphic to which group out of $Z_8, Z_4\oplus Z_2,Z_2\oplus Z_2\oplus Z_2$? 
The quotient group $(Z_4 \oplus Z_{12})/\langle(2,2) \rangle $ is isomorphic to which group out of $Z_8, Z_4\oplus Z_2,Z_2\oplus Z_2\oplus Z_2$?

I proceeded as follows: 
Let $G=(Z_4 \oplus Z_{12})$.  We have $|G|=12\times 4=48$ and let $$H=\langle (2,2)\rangle=\{(2,2),(0,4),(2,6),(0,8),(2,10),(0,0)\}\,.$$ Hence, $|G/H|=48/6=8$.
Since $H$ is normal subgroup of $G$, factor group $G/H$ can be defined. Let's write out all elements of $G/H$. Clearly, $H\in G/H$. Note that:
$$G/H=\{H, (1,0)+H, (0,1)+H, (0,2)+H, (0,3)+H, (3,0)+H, (1,1)+H, (1,3)+H\} \tag{1}$$
There are $48$ elements in $G$ out of which I considered around $40$ elements to ensure that $G/H$ in $(1)$ has only $8$ nos. of elements by using contradiction. If $a+H=b+H$ then $a-b\in H$. For example:  I was not sure if $(2,0)$ is a distinct element of $G/H$ or not so I assumed on the contrary that let $(2,0)\in H$ in addition to already existing $8$ elements of $G/H$ then I noticed that $(0,2)+H=(2,0)+H$, which is obvious by adding $(2,0)$ on both sides. I repeated this procedure several times to discard rest of $42$ elements. It's a very lengthy process. 
I was wondering if there is any alternative (much less tedious)to write elements of $G/H$. 
Once I am done with writing elements of $G/H$ then procedure is simple. I'll check if there's any element of order $8$ in $G/H$, if yes then it's cyclic and hence isomorphic to $8$. 
If there's any element of order $4$ in $G/H$, this will knock out $Z_2\oplus Z_2 \oplus Z_2$ and so $G/H$ is isomorphic to $Z_4\oplus Z_2$ etc.  
In view of the above, I request your help in the following
A) Big trouble that I have typed in bold above. 
B) Noticing that $G/H$ absorbs any element which belongs to $H$, writing $G/H$ elements requires that each element is independent from one another (i.e., they must not be the same), which gives a feeling of "finding linearly independent vectors in linear algebra" and I suspect that there might be some way involving bases, dimension of vector space etc., to make my problem in $(A)$ solvable in much lesser time. 
Thanks for your time.
 A: Note that $G$ is generated by $a:=(1,0)$ and $b:=(0,1)$.  Let $\pi:G\to (G/H)$ be the canonical projection.  Then, $\pi(a)$ and $\pi(b)$ generate the factor group $G/H$.
Observe that $\alpha:=\pi(a)$ generates a subgroup $M\cong Z_4$ of $G/H$, while $\beta:=\pi(b)$ generates a subgroup $N\cong Z_4$ of $G/H$.  From this information, we know that $G/H$ contains a subgroup isomorphic to $Z_4$.  That means the possible choices are $Z_8$ and $Z_4\times Z_2$.  Now, since $\alpha$ and $\beta$ generate the abelian group $G/H$, with both having order $4$, we conclude that elements of $G/H$ have orders dividing $4$.  Thus, $Z_8$ is not possible.  This implies $G/H\cong Z_4\times Z_2$.
Indeed, $G/H$ is the abelian group with the presentation $$G/H=\langle \alpha,\beta\,|\,4\alpha=4\beta=2\alpha+2\beta=0\rangle\,.$$  If $L:=\langle \alpha+\beta\rangle$, then $G$ is given by the internal direct product $M\times L$, where $M=\langle \alpha\rangle$ as defined in the previous paragraph.
A: Here is another approach.
Let $G=(\mathbb{Z}_4\oplus\mathbb{Z}_{12})/H$ and $H=\langle(2,2)\rangle$.
Since $G$ is a quotient, we can try to find a surjective homorphism from $\mathbb{Z}_4\oplus\mathbb{Z}_{12}$ to one of the groups that has $H$ as kernel. If we look at $H$ we see that every element $(x,y)\in H$ satisfies $x-y\equiv 0 \text{ mod } 4$, so we can guess that our homomorphism $\varphi$ has to encode some information mod $4$. This rules out two of the three groups. Now we have to find a surjective homomorphism
$$\varphi:\mathbb{Z}_4\oplus\mathbb{Z}_{12}\rightarrow \mathbb{Z}_4\oplus\mathbb{Z}_2$$
This is a bit of trial and error. One way to define this is by
$$
\varphi(x,y)=(x-y,y)
$$
This is well defined and to see that it is surjective we take $(a,b)\in\mathbb{Z}_4\oplus\mathbb{Z}_2$. Then we can first choose $y=b$ (not really as these guys live in different groups). Then we can choose $x$ such that $x-y=a$.
Now for the kernel of $\varphi$. It is not hard to see that $H\subseteq\text{ker}(\varphi)$. Conversely let $(x,y)\in\text{ker}(\varphi)$. Then $y\equiv 0\text{ mod }2$ i.e. $y=2i$. We also have $x\equiv y\text{ mod } 4$, that is $x-y=4k$. Combining these two equations gives $x=4k+2i$. Thus we get $(x,y)=(4k+2i,2i)=(2i,2i)\in H$
A: Let $G=Z_4 \oplus Z_{12}$ and $M=\langle (2,2) \rangle.$ You can actually consider the matrix $[2\,\,\, 2]$ and reduce this matrix to smith normal form to get $[0 \,\,\, 2].$ So invariant factors are $0$ and $2$. Hence it follows that
$$G/M \cong Z_4/0Z \oplus Z_{12}/2Z \cong Z_4 \oplus Z_2$$
A: Let $G:=\mathbb{Z}_4\oplus\mathbb{Z}_{12}$ and $H\subset G$ the subgroup $\langle (2,2)\rangle$.
Because $G$ has no elements of order $8$, $G/H$ doesn't either, so it can't be $\mathbb{Z}_8$.
Note that $(0,3)$ has order $4$ in $G$ and $(0,3),(0,6)\not\in H$, so $(0,3)$ has order $4$ in $G/H$. Hence, $G/H$ has elements of order $4$ and cannot be isomorphic to $\mathbb{Z}_2\oplus\mathbb{Z}_2\oplus\mathbb{Z}_2$.
Therefore, since it must be one of the three, $G/H\cong \mathbb{Z}_2\oplus\mathbb{Z}_4$
A: One option is to use Tietze transformations.
Let's start with the presentation found in Batominovski's answer,
$$
A = \langle \alpha, \beta \mid 4\alpha, 4\beta, 2(\alpha + \beta) \rangle.
$$
Let $\gamma = \alpha + \beta$. Using this substitution, we get:
$$
A = \langle \alpha, \gamma \mid 4\alpha, 4(\gamma - \alpha), 2\gamma \rangle. 
$$
The element $4(\gamma - \alpha)$ is superfluous, it follows that
$$
A = \langle \alpha, \gamma \mid 4\alpha, 2\gamma \rangle. 
$$
We conclude that your original group is isomorphic to $\mathbb Z/(4) \times \mathbb Z/(2)$.
