Internal Direct Product Theorem

Theorem
Let $G$ be a group whose identity is $e$.

Let $H_1, H_2 \le G$.

Then $G$ is the internal group direct product of $H_1$ and $H_2$ iff:


 * $(1): \quad G = H_1 H_2$;
 * $(2): \quad H_1 \cap H_2 = \left\{{e}\right\}$;
 * $(3): \quad H_1, H_2 \triangleleft G$.

Generalized Theorem
Let $G$ be a group whose identity is $e$.

Let $\left \langle {H_k} \right \rangle_{1 \le k \le n}$ be a sequence of subgroups of $G$.

Then $G$ is the internal group direct product of $\left \langle {H_k} \right \rangle_{1 \le k \le n}$ iff:


 * $(1): \quad G = H_1 H_2 \cdots H_n$;
 * $(2): \quad \left \langle {H_k} \right \rangle_{1 \le k \le n}$ is a sequence of independent subgroups;
 * $(3): \quad$ For each $k \in \left[{1 . . n}\right]$, $H_k \triangleleft G$.

Alternative Formulation
Let $H_1, H_2, \ldots, H_n$ be normal subgroups of $G$, such that:


 * $(1): \quad \forall i \in \left\{{2, 3, \ldots, n}\right\}: \left({H_1 H_2 \ldots H_{i-1}}\right) \cap H_i = \left\{{e}\right\}$;
 * $(2): \quad G = H_1 H_2 \cdots H_n$

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

Proof
By definition, $G$ is the internal group direct product of $\left \langle {H_k} \right \rangle_{1 \le k \le n}$ iff the mapping:


 * $C: \prod_{k=1}^n H_k \to G: C \left({h_1, \ldots, h_n}\right) = \prod_{k=1}^n h_k$

is a group isomorphism from the cartesian product $\left({H_1, \circ \! \restriction_{H_1}}\right) \times \cdots \times \left({H_n, \circ \! \restriction_{H_n}}\right)$ onto $\left({G, \circ}\right)$.

Sufficient Condition
Let $G$ be the internal group direct product of $\left \langle {H_k} \right \rangle_{1 \le k \le n}$.


 * $(1): \quad$ From Internal Group Direct Product Surjective, $G = H_1 H_2 \cdots H_n$.
 * $(2): \quad$ From Internal Group Direct Product Injective, $\left \langle {H_k} \right \rangle_{1 \le k \le n}$ is a sequence of independent subgroups.
 * $(3): \quad$ From Internal Group Direct Product Isomorphism, $\forall k \in \left[{1 . . n}\right]: H_k \triangleleft G$.

Necessary Condition

 * Now suppose the three conditions hold.


 * $(1): \quad$ From Internal Group Direct Product Surjective, $C$ is surjective.
 * $(2): \quad$ From Internal Group Direct Product Injective, $C$ is injective.
 * $(3): \quad$ From Internal Group Direct Product of Normal Subgroups, $C$ is a group homomorphism.

Putting these together, we see that $C$ is a bijective homomorphism, and therefore an isomorphism.

So by definition, $G$ is the internal group direct product of $\left \langle {H_k} \right \rangle_{1 \le k \le n}$.

Proof of Alternative Formulation
The second condition ensures that every element can be written in the form:
 * $g = h_1 h_2 \ldots h_n: h_i \in H_i, i \in \N^*_n$

We need to show that this expression is unique.

Suppose:
 * $g = g_1 g_2 \ldots g_n = h_1 h_2 \ldots h_n$

where $g_i, h_i \in H_i, i \in \N^*_n$ and $g_j \ne h_j$ for some $j$.

Let $j$ be the largest integer such that $g_j \ne h_j$, so $g_i = h_i$ for all $i > j$.

Cancelling $g_i$ for all $i > j$, we have $g_1 g_2 \ldots g_j = h_1 h_2 \ldots h_j$.

Thus:
 * $g_j h_j^{-1} = \left({g_1 g_2 \ldots g_{j-1}}\right)^{-1} \left({h_1 h_2 \ldots h_{j-1}}\right) \in \left({H_1 H_2 \ldots H_{j-1}}\right) \cap H_j$

But:
 * $\left({H_1 H_2 \ldots H_{j-1}}\right) \cap H_j = \left\{{e}\right\}$

by hypothesis.

So $h_j = g_j$ contrary to the definition of $j$.

Thus the decomposition is unique.