Edit: As pointed out by Mark, this question is obviously false when things are abelian. Please assume things are non-abelian.
Let $G$ and $H$ be subgroups of a larger group, and define $GH = \{gh \,|\, g \in G, h \in H\}$. Then it is clear that for $gh = hg$ to be satisfied for all $g \in G, h \in H$, it is a sufficient condition that $G$ and $H$ are normal subgroups of $GH$ and that $G \cap H = \{e\}$. My question is whether this is a necessary condition. In other words, is it true that $$\forall g \in G, \forall h \in H\qquad gh = hg \implies G \lhd GH, H \lhd GH, G \cap H = \{e\}?$$
Here is what I was able to do. It is very easy to prove that $G \lhd GH, H \lhd GH$. Proving that $G \cap H = \{e\}$ is turning out to be challenging for me. I was trying to prove the contrapositive, i.e. that if $x \in G\cap H$ and $x \neq e, G \lhd GH, H \lhd GH$, then we can show that $$\exists g \in G, \exists h \in H\text{ such that }gh \neq hg.$$
I am trying to understand why internal direct and semidirect products are defined the way they are. It is clear that we need $GH = HG$, and the sufficient condition for that is what I said it is. I just want to know if it is a necessary condition.
No it's not a necessary condition. A central product of two groups $G$ and $H$ satisfies the condition, and $G$ and $H$ could both be nonabelian.
For example, you could take a central product of two nonabelian groups of order $8$, giving an extraspecial group of order $32$.