Prove/Disprove: If $K\leq G \times H$ Then $K\in G'\times H'$ Where $G'\leq G$ And $H'\leq H$

Question

Let there $K\leq G \times H$ Then $K$ is of the form $G'\times H'$ where $G'\leq G$ And $H'\leq H$

Intuitively the definition of $G\times H$ a cartesian product of two groups, so there can not be an element in it, which is not a product of two subgroups.

By contradiction , Let assume that there is $(k_1,k_2)\in (A\times B)$ and $(k_3,k_4)\in (C\times D)$ Where $A,B,C,D$ are not a subgroup of $G$ or $H$ and not of each other, then $$(k_1,k_2)(k_3,k_4)=(k_1k_2,k_3k_4)\in K$$ but $k_1k_2\notin G$ and $k_3k_4\notin H$ contradiction

Is it correct?

Answer 1

Consider $G$ and $H$ be vector spaces of dimension $2$ (over a finite field with their structure of additive group) and $K$ a subspace of dimension $3$ of $G\times H$.

Answer 2

Hint(Counterexample):

Consider $G=H=\Bbb{Z}_2$ and $K=\{(0,0),(1,1)\}$.