Subgroup Product is Internal Group Direct Product iff Surjective

Theorem
Let $$G$$ be a group.

Let $$\left \langle {H_n} \right \rangle$$ be a sequence of subgroups of $$G$$.

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

Then $$\phi$$ is surjective iff $$G = \prod_{k=1}^n H_k$$.

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

Proof

 * First we show that if $$\phi$$ is surjective then $$G = \prod_{k=1}^n H_k$$.

So, suppose $$\phi$$ is a surjection.

Then $$\operatorname{Im} \left({\phi}\right)$$ consists of all the products $$\prod_{k=1}^n h_k$$ such that $$\forall k \in \left[{1 \,. \, . \, n}\right]: h_k \in H_k$$.

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

If we use the notation:


 * $$\prod_{k=1}^n H_k = \left\{{\prod_{k=1}^n h_k: \forall k \in \left[{1 \, . \, . \, n}\right]: h_k \in H_k}\right\}$$

we see that the result follows immediately.

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


 * Now suppose that $$G = \prod_{k=1}^n H_k$$.

If this is the case, then $$g$$ can be written as $$g = \prod_{k=1}^n h_k: \forall k \in \left[{1 \,. \, . \, n}\right]: h_k \in H_k$$.

Thus, $$g = \phi \left({\left({h_1, h_2, \ldots, h_n}\right)}\right)$$ and $$\phi$$ is surjective.