Subgroup Product is Internal Group Direct Product iff Surjective
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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 if and only if:
- $\ds G = \prod_{k \mathop = 1}^n H_k$
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 $\ds \prod_{k \mathop = 1}^n h_k$ 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 $H_k$.
$\Box$
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.
$\blacksquare$