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 $\color{red}{\langle(\ldots)\rangle}$?