Let $K=\{(g,g):g \in G\}$ where $G$ is a group. There are no subgroups $H_1,H_2$ of $G$ such that $K=H_1\times H_2$

Question

How to formally prove the following:

Let $K=\{(g,g):g \text{ is a member of G}\}$ where $G$ is a group

There are no sub-groups $H_1,H_2$ of $G$ such that $K=H_1 \times H_2$

If I think about it it's so easy to prove (even without using groups) as we can't construct ONLY members of the following shape $(g,g)$ as multiplication of 2 groups... but that's not formal.

Answer 1

Hint. Show if one of $H_1$ or $H_2$ is nontrivial, then $H_1\times H_2$ has an "off-diagonal" element.