Internal Direct Product Theorem/Proof 1

Proof
From Conditions for Internal Group Direct Product it is sufficient to show that if:
 * $\quad G = H \circ K$

and:
 * $\quad H \cap K = \set e$

then:
 * $\struct {H, \circ {\restriction_H} }$ and $\struct {H_2, \circ {\restriction_K} }$ are both normal subgroups of $\struct {G, \circ}$


 * $\forall \tuple {h, k} \in H \times K: h \circ k = k \circ h$

Sufficient Condition
Let $\struct {H, \circ {\restriction_H} }$ and $\struct {H_2, \circ {\restriction_K} }$ both be normal subgroups of $\struct {G, \circ}$.

Let $x \in H$ and $y \in K$.

Then:

$x$ and $y$ are arbitrary, so:


 * $\forall \tuple {h, k} \in H \times K: h \circ k = k \circ h$

Necessary Condition
Let:
 * $\forall \tuple {h, k} \in H \times K: h \circ k = k \circ h$

Let $z = G$.

We have:

Then we have:

Similarly:

Thus, by definition, $H$ and $K$ are both $\struct {H, \circ {\restriction_H} }$ and $\struct {H_2, \circ {\restriction_K} }$ are both normal subgroups of $\struct {G, \circ}$.