The External Direct Product $G\oplus H$ is a Group of order $|G||H|$

Let $G$ and $H$ be groups. The binary operation $(g,h)\times(a,b) = (ga,hb)$ is associative as we can compute $$((g,h)(a,b))(m,n)=(ga,hb)(m,n) = ((ga)m,(hb)n)$$ and similarly $$(g,h)((a,b)(m,n))=(g,h)(am,bn) = (g(am),h(bn)).$$ These two quantities are equal because the operation in both $G$ and $H$ is associative. The identity in $G\times H$ is simply $(e_g,e_H)$, and the inverse of $(g,h)$ is $(g^{-1},h^{-1})$. I'll let you show that these satisfy the requirements to be the identity and inverse.

If $G$ has order $|G|$ and $H$ has order $|H|$, and both are finite, then there are clearly $|G|$ choices for the first component, and $|H|$ choices for the second component. Hence the total number of elements in $G\times H$ is precisely $|G\times H|=|G||H|$.