# 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 $\displaystyle \phi: \prod_{k \mathop = 1}^n H_k \to G$ be a mapping defined by:

- $\displaystyle \map \phi {h_1, h_2, \ldots, h_n} = \prod_{k \mathop = 1}^n h_k$

Then $\phi$ is surjective if and only if:

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

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

## Proof

### Necessary Condition

Let $\phi$ be a surjection.

Then $\Img \phi$ consists of all the products $\displaystyle \prod_{k \mathop = 1}^n h_k$ 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:

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

the result follows immediately.

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

$\Box$

### Sufficient Condition

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

If this is the case, then $g$ can be written as:

- $\displaystyle g = \prod_{k \mathop = 1}^n h_k: \forall k \in \closedint 1 n: h_k \in H_k$

Thus:

- $g = \map \phi {h_1, h_2, \ldots, h_n}$

and $\phi$ is surjective.

$\blacksquare$