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

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) $$G = H_1 H_2 \cdots H_n$$;
 * 2) $$\left \langle {H_k} \right \rangle_{1 \le k \le n}$$ is a sequence of independent subgroups;
 * 3) 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) $$\forall i \in \left\{{2, 3, \ldots, n}\right\}: \left({H_1 H_2 \ldots H_{i-1}}\right) \cap H_i = \left\{{e}\right\}$$;
 * 2) $$G = H_1 H_2 \cdots H_n$$

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) From Internal Group Direct Product Surjective, $$G = H_1 H_2 \cdots H_n$$.
 * 2) From Internal Group Direct Product Injective, $$\left \langle {H_k} \right \rangle_{1 \le k \le n}$$ is a sequence of independent subgroups.
 * 3) 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) From Internal Group Direct Product Surjective, $$C$$ is surjective.
 * 2) From Internal Group Direct Product Injective, $$C$$ is injective.
 * 3) 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.