Internal Direct Product Theorem

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

Let $$H_1, H_2 \le G$$.

Then the mapping $$\phi: H_1 \times H_2 \to G$$ given by:


 * $$\forall \left({h_1, h_2}\right) \in H_1 \times H_2: \phi \left({\left({h_1, h_2}\right)}\right) = h_1 h_2$$

is an isomorphism iff:


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

If this is the case, then $$G \cong H_1 \times H_2$$ and $$G$$ is the internal group direct product of $$H_1$$ and $$H_2$$.

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 $$G \cong H_1 \times H_2$$ and $$G$$ is the internal group direct product of $$\left \langle {H_k} \right \rangle_{1 \le k \le n}$$ iff:


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

Alternative Formulation
Let $$G_1, G_2, \ldots, G_n$$ be normal subgroups of $$G$$, such that:


 * 1) $$\forall i \in \left\{{2, 3, \ldots, n}\right\}: \left({G_1 G_2 \ldots G_{i-1}}\right) \cap G_i = \left\{{e}\right\}$$;
 * 2) $$G = G_1 G_2 \ldots G_n$$

Then $$G$$ is the internal group direct product of $$G_1, G_2, \ldots, G_n$$.

Proof

 * Let $$\phi$$ be an isomorphism.

Then:
 * 1) From Internal Group Direct Product Surjective, $$G = H_1 H_2$$.
 * 2) From Internal Group Direct Product Injective, $$H_1 \cap H_2 = \left\{{e}\right\}$$.
 * 3) From Internal Group Direct Product Isomorphism, $$H_1, H_2 \triangleleft G$$.


 * Now suppose the three conditions hold.


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

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

Generalized Proof
By Internal Direct Product Generated by Subgroups, it is sufficient to prove that if 2 and 3 hold, then 1 holds iff every element of $$H_i$$ commutes with every element of $$H_j$$ whenever $$1 \le i < j \le n$$.


 * Necessity:

Let $$x_i \in H_i$$ and $$x_j \in H_j$$.

Let $$z = x_i x_j x_i^{-1} x_j^{-1}$$.

As $$z = \left({x_i x_j x_i^{-1}}\right) x_j^{-1} = x_i \left({x_j x_i^{-1} x_j^{-1}}\right)$$ it is clear from 1 and Normal Subgroup Equivalent Definitions: 5 that $$z \in H_j \cap H_i$$.

Let $$z_k = \begin{cases} z & : k = i \\ z^{-1} & : k = j \\ e & : k \ne i, k \ne j \end{cases} $$

Then $$z_1 z_2 \cdots z_n = z z^{-1} = e$$, so by 3, $$z = z_i = e$$.

Therefore $$x_j x_i = z x_j x_i = x_i x_j$$.


 * Sufficiency:

By Normal Subgroup Equivalent Definitions: 5 it is enough to show that:

$$z = G, g \in H_k \implies z g z^{-1} \in H_k$$

By 2, there exist $$h_1, \ldots, h_n$$ in $$H_1, \ldots, H_n$$ respectively such that $$z = h_1 h_2 \cdots h_n$$.

Let $$g' = h_k g h_k^{-1}$$.

Then $$g' \in H_k$$.

By Associativity and Commutativity Properties and the hypothesis, $$g$$ commutes with $$h_{k+1} \cdots h_n$$ and $$g'$$ commutes with $$h_1 \cdots h_{k-1}$$. Therefore:

$$ $$ $$ $$

Thus $$z g z^{-1} = g' \in H_k$$.

Proof of Alternative Formulation
The second condition ensures that every element can be written in the form:


 * $$g = g_1 g_2 \ldots g_n: g_i \in G_i, i \in \N^*_n$$

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 $$g_i, h_i \in G_i, i \in \N^*_n$$ 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:
 * $$g_j h_j^{-1} = \left({g_1 g_2 \ldots g_{j-1}}\right)^{-1} \left({h_1 h_2 \ldots h_{j-1}}\right) \in \left({G_1 G_2 \ldots G_{j-1}}\right) \cap G_j$$

But:
 * $$\left({G_1 G_2 \ldots G_{j-1}}\right) \cap G_j = \left\{{e}\right\}$$

by hypothesis.

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

Thus the decomposition is unique.