Do you mean that since every element of H \times G_1 is unique and that every element of H \times G_2 can be defined as (c, f(a)) = (c, b), \forall c \in H, a \in G_1, b \in G_2 such that it is also unique, we have F: (c, a) \rightarrow (c, f(a)) = (c, b) And thus, there is a one-to-one...