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

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

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 $\displaystyle G = \prod_{k=1}^n H_k$.

So, suppose $\phi$ is a surjection.

Then $\operatorname{Im} \left({\phi}\right)$ consists of all the products $\displaystyle \prod_{k=1}^n h_k$ 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:


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

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 $\displaystyle G = \prod_{k=1}^n H_k$.

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

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