 Solution

Assume (I). Existence of a decomposition is part of the hypothesis. If $h_1{h}_2=k_1{k}_2$ are two decompositions, multiply on the left by $k_1^{-1}$ and on the right by $h_2^{-1}$ to obtainThe left side belongs to $H_1$ and the right side to $H_2$, since they are subgroups. Their common value belongs to $H_1\cap H_2=\{e\}$, so $h_1=k_1$ and $h_2=k_2$. This proves uniqueness, hence (II).

Conversely, assume (II). Existence immediately gives $G=H_1{H}_2$. If $a\in H_1\cap H_2$, thenare two allowed factorizations, because the identity is in each subgroup. Uniqueness makes their first components equal, so $a=e$. Thus $H_1\cap H_2=\{e\}$ and (I) follows.

Therefore (I) and (II) are equivalent. This is an exact factorization of a group. For finite groups it gives $|G|=|H_1||H_2|$, but it does not by itself identify $G$ with a direct product of groups; multiplication between the two factors can be noncommutative.