Order of semidirect product is product of orders

For external semidirect product
Suppose $$N$$ and $$H$$ are groups, $$\rho: H \to \operatorname{Aut}(N)$$ is a homomorphism of groups, and $$G = N \rtimes H$$ is the external semidirect product of $$N$$ by $$H$$ for the action $$\rho$$.

Then, the order of $$G$$ is the product (in the sense of multiplication) of the order of $$N$$ and the order of $$H$$.

When both $$N$$ and $$H$$ are finite groups, so is $$G$$ and the above statement is true in the sense of multiplication of finite numbers. When either $$N$$ or $$H$$ is an infinite group, so is $$G$$, and the above statement is true in the sense of multiplication of cardinals.

For internal semidirect product
Suppose $$G$$ is a group, $$N$$ is a complemented normal subgroup of $$G$$, and $$H$$ is a complement to $$N$$ in $$G$$. Thus, $$G$$ is the internal semidirect product of $$N$$ and $$H$$.

Then, the order of $$G$$ is the product of the order of $$N$$ and the order of $$H$$.