How to find a subgroup of the external direct product $\Bbb{Z}/4\Bbb{Z}\oplus\Bbb{Z}/2\Bbb{Z}$ not of the form $H\oplus K$, where $H$ and $K$ are subgroups of $\Bbb{Z}/4\Bbb{Z}$ and $\Bbb{Z}2/\Bbb{Z}$, respectively.
My progress: I find this question rather weird, the only thing that makes sense is to find a subgroup up to isomorphism, since I actually (easily) managed to prove that for each $H\oplus K$ being a subgroup of $G_1\oplus G_2$, the groups $H$ and $K$ are subgroups of $G_1$ and $G_2$. Wouldn't this make it impossible to give an example?
There seems to be confusion about subgroups direct products, or what direct sums are. You claim:
But here $H$ and $K$ are supposed to be subgroups of $G_1$ and $G_2$ to begin with; how else can $H\oplus K$ be considered a subgroup of $G_1\oplus G_2$? Besides, the point is to show that there are subgroups of $G_1\oplus G_2$ that are not of this form. So why prove something about subgroups that are of this form?
An example of such a subgroup is the subgroup $G$ generated by $(2,1)\in\Bbb{Z}/4\Bbb{Z}\oplus\Bbb{Z}/2\Bbb{Z}$. That is, $$G:=\langle(2,1)\rangle=\{(0,0),(2,1)\}.$$ If this subgroup is of the form $G=H\oplus K$ for some subgroups $H$ of $\Bbb{Z}/4\Bbb{Z}$ and $K$ of $\Bbb{Z}/2\Bbb{Z}$, then $$|H\oplus K|=|G|=2,$$ where $|H\oplus K|=|H|\times|K|$. Then either $H=\{0\}$ or $K=\{0\}$, but then $(2,1)\notin H\oplus K$, a contradiction. This shows that $G$ is not of the form $H\oplus K$.