Number of subgroups of direct product is bounded below by product of number of subgroups in each factor

Statement
Suppose $$G_1$$ and $$G_2$$ are groups. Then, the number of subgroups of the external direct product $$G_1 \times G_2$$ is at least equal to the product of the (number of subgroups of $$G_1$$) and the (number of subgroups of $$G_2$$):

(Number of subgroups of $$G_1 \times G_2$$) $$\ge$$ (Number of subgroups of $$G_1$$)(Number of subgroups of $$G_2$$)

Proof
We will construct an injective set map:

(Set of subgroups of $$G_1$$) $$\times$$ (Set of subgroups of $$G_2$$) $$\to$$ (Set of subgroups of $$G_1 \times G_2$$)

The map is as follows: for subgroups $$H_1 \le G_1$$ and $$H_2 \le G_2$$, we have:

$$(H_1,H_2) \mapsto H_1 \times H_2$$