When is the decomposition of a group element as a product of elements of two subgroups unique?

Question

Let $G$ and $H$ be subgroups of $P$. Assume that every element of $P$ can be expressed as the product of an element of $G$ and an element of $H$.

Under what conditions is this decomposition unique?

This question came up while reading the Wikipedia entry on direct product of groups, but in the wikipedia entry there are two additional assumptions whereas I would rather understand assumption 2. in isolation

Let $G$ and $H$ be groups, let $P = G \times H$ and consider the following two subsets of $P$:

$$G' = \{ (g,1): g \in G \} \ \text{and}\ H' = \{ (1,h): h \in H \}. $$

[...]

1. The intersection $G ∩ H$ is trivial.
2. Every element of $P$ can be expressed as the product of an element of $G$ and an element of $H$.
3. Every element of $G$ commutes with every element of $H$.

Answer 1

If every element of $P$ can be written as a product of an element of $H$ and an element of $G$, then $P=GH$ as sets; but it is a well-known exercise in group theory that given two subgroups $G$ and $H$, the set of products $GH = \{gh\mid g\in G, h\in H\}$ is a (sub)group of $P$ if and only if $GH=HG$ as sets. In particular, if every element of $P$ can be written as a product of an element of $H$ and an element of $G$ in at least one way, then $GH=HG$. Note that this is not the same as your property 3: we just need that for every $g\in G$ and $h\in H$, there exist elements $h’\in H$ and $g’\in G$ such that $gh=h’g’$ and elements $h’’\in H$ and $g’’\in G$ such that $hg=g’’h’’$.

Now, suppose we have uniqueness of the expression. If $x\in G\cap H$, then $x^{-1}\in G\cap H$ (since $G\cap H$ is a subgroup). So $e = ee = xx^{-1}$ gives two ways of expressing $e$ as a product of elements of $G$ and elements of $H$, hence $x=e$. That is, $G\cap H\subseteq \{e\}$. Thus, if you have uniqueness, then the intersection is trivial.

Conversely, suppose the intersection is trivial, and $g,g’\in G$ and $h,h’\in H$ are such that $gh=g’h’$. Then $(g’)^{-1}g = h’h^{-1}\in G\cap H=\{e\}$, hence $(g’)^{-1}g=e$, so $g=g’$; and $h’h^{-1} = e$, so $h’=h$. That is, we get uniqueness of the representation.

In short: you get uniqueness of the representation if and only if the intersection is trivial. But this is weaker than the condition of $G$ being isomorphic to the direct product of $G$ and $H$. We say, in this case, that $G$ and $H$ are complementary, or that $G$ is a complement of $H$ and vice-versa.