Subgroup Product is Internal Group Direct Product iff Surjective

Theorem
Let $G$ be a group.

Let $\sequence {H_n}$ be a sequence of subgroups of $G$.

Let $\ds \phi: \prod_{k \mathop = 1}^n H_k \to G$ be a mapping defined by:
 * $\ds \map \phi {h_1, h_2, \ldots, h_n} = \prod_{k \mathop = 1}^n h_k$

Then $\phi$ is surjective :
 * $\ds G = \prod_{k \mathop = 1}^n H_k$

That is, $G$ is the internal group direct product of $H_1, H_2, \ldots, H_n$.

Necessary Condition
Let $\phi$ be a surjection.

Then $\Img \phi$ consists of all the products $\ds \prod_{k \mathop = 1}^n h_k$ such that:
 * $\forall k \in \closedint 1 n: h_k \in H_k$

Thus, as $\phi$ is surjective, every element of $G$ must be representable in this form.

Using the notation:


 * $\ds \prod_{k \mathop = 1}^n H_k = \set {\prod_{k \mathop = 1}^n h_k: \forall k \in \closedint 1 n: h_k \in H_k}$

the result follows immediately.

Thus every element of $G$ is the product of an element of each of $H_k$.

Sufficient Condition
Let $\ds G = \prod_{k \mathop = 1}^n H_k$.

If this is the case, then $g$ can be written as:
 * $\ds g = \prod_{k \mathop = 1}^n h_k: \forall k \in \closedint 1 n: h_k \in H_k$

Thus:
 * $g = \map \phi {h_1, h_2, \ldots, h_n}$

and $\phi$ is surjective.