Find all subgroups of $(\Bbb{Z}_2\times\Bbb{Z}_4,+)$ 
Find all subgroups of $(\Bbb{Z}_2\times\Bbb{Z}_4,\overline{+})$.

I could find the following subgroups:
$$\begin{array}{ll}
H_1=\langle(0,0)\rangle=\{(0,0)\}&\text{(Trivial subgroup)}\\
H_2=\langle(0,1)\rangle=\{(0,1),(0,2),(0,3),(0,0)\}=\langle(0,3)\rangle=H_4\\
H_3=\langle(0,2)\rangle=\{(0,2),(0,0)\}\\
H_5=\langle(1,0)\rangle=\{(1,0),(0,0)\}\\
H_6=\langle(1,1)\rangle=\{(1,1),(0,2),(1,3),(0,0)\}=\langle(1,3)\rangle=H_8\\
H_7=\langle(1,2)\rangle=\{(1,2),(0,0)\}\\
H_9=\Bbb{Z}_2\times\Bbb{Z}_4&\text{(Improper subgroup)}
\end{array}$$
That is, $7$ subgroups in total.
I found all the CYCLIC subgroups, but since the group IS NOT CYCLIC, then I need to find the NOT CYCLIC subgroups.
The final answer should be $8$ subgroups; the only subgroup that is NOT CYCLIC is: $$\text{Subgroup that is not cyclic}=\{(0,0),(0,2),(1,0),(1,2)\}.$$ So there are $8$ subgroups in total.
My question is:

How can we find $\{(0,0),(0,2),(1,0),(1,2)\}$? I mean, what should we put inside $\color{red}{\langle(\ldots)\rangle}$?

 A: We write $\langle A \rangle$ for a subset (not necessarily subgroup) $A \subseteq G$ to mean the smallest subgroup which contains $A$. So it would be correct in your case to write 
$$ \{(0,0),(0,2),(1,0),(1,2)\} = \langle \{(0,0),(0,2),(1,0),(1,2)\} \rangle $$
However, this is somewhat unsatisfying, since in the cyclic case we only needed one generator (that is, we only needed one element inside the $\langle - \rangle$).
If you are looking for a small set of elements to put inside, you can get away with
$$ \{(0,0),(0,2),(1,0),(1,2)\} = \langle (0,2), (1,0) \rangle.$$
Any subgroup contains $(0,0)$, and any subgroup containing $(1,0)$ and $(0,2)$ must also contain their sum $(1,2)$. Thus {(0,0),(0,2),(1,0),(1,2)} is the smallest subgroup containing $(0,2)$ and $(1,0)$.
Edit: 
In a comment you ask if we should operate on the elements inside the $\langle - \rangle$. The answer is "no". As you have noticed, if we reduce the elements inside, we can only ever get cyclic subgroups. We allow ourselves multiple elements inside so that we can express more complicated subgroups, such as the example above.
$\langle x + y \rangle$ is always a subgroup of $\langle x, y \rangle$. (Since $x+y \in \langle x, y \rangle$). It is very rarely the case that they are equal, however.

I hope this helps ^_^
A: If it's not cyclic there's going to be more than one element inside $\langle\rangle$; e.g.,   $\langle(0,2),(1,0)\rangle$.
