Why $G=HK$ and $H\cap K=\{e\} \implies G = H\times K$?

Question

Given that $G$ is finite abelian, in order to show that $G = H\times K$ one only needs to show that $G=HK$ and $H\cap K=\{e\}$. What motivates this and why is it only true for finite abelian groups?

Answer 1

Let define: $$\varphi:\left\{\begin{array}{ccc}H\times K & \rightarrow & G\\(h,k) & \mapsto & hk\end{array}\right.$$ $\varphi$ is a group homomorphism, since $G$ is abelian and your assumptions imply that $\varphi$ is bijective. Indeed, if $hk=e$, then $H\ni h=k^{-1}\in K$.

Going further:

Proposition. Let $G$ be a group and $H,K$ normal subgroups of $G$. If $H\cap K=\{e\}$, then: $$H\times K\cong\langle H\cup K\rangle.$$

One derives the:

Corollary. Let $G$ be a finite group. If all Sylow's subgroups of $G$ are normal, then $G$ is isomorphic to the direct product of its Sylow's subgroups.