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:

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