Internal Direct Product Generated by Subgroups

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

Let $$\left \langle {H_n} \right \rangle$$ be a sequence of subgroups of $$G$$.

Then the subgroup generated by $$\bigcup_{k=1}^n H_k$$ is the internal direct product of $$\left \langle {H_n} \right \rangle$$ iff $$\left \langle {H_n} \right \rangle$$ is an independent sequence of subgroups such that every element of $$H_i$$ commutes with every element of $$H_j$$ whenever $$1 \le i < j \le n$$.

Proof
For each $$k \in \left[{1 \,. \, . \, n}\right]$$, let $$L_k = \prod_{j=1}^k H_j$$ be the cartesian product of the subgroups $$H_1, H_2, \ldots, H_k$$ of $$G$$.

Let $$C_k: L_k \to G: C_k \left({x_1, x_2, \ldots, x_k}\right) = \prod_{j=1}^k x_j$$.


 * Necessity:

The kernel of $$C_n$$ is $$\left\{{\left({e, e, \ldots, e}\right)}\right\}$$.

Therefore $$\left \langle {H_n} \right \rangle$$ is an independent sequence.

Let $$x \in H_i$$ and $$y \in H_j$$ where $$1 \le i < j \le n$$.

For each $$k \in \left[{1 \,. \, . \, n}\right]$$, let $$x_k$$ and $$y_k$$ be defined as:


 * $$x_k =

\begin{cases} e & : k \ne i \\ x & : k = i \end{cases} \qquad y_k = \begin{cases} e & : k \ne j \\ y & : k = j \end{cases} $$

Then:

$$ $$ $$ $$ $$ $$


 * Sufficiency:

Let $$S$$ be the set of all $$k \in \left[{1 \,. \, . \, n}\right]$$ such that $$C_k: L_k \to G$$ is a (group) homomorphism.

Clearly $$1 \in S$$.

Now let $$k \in S$$ such that $$k < n$$.

Let $$\left({x_1, \ldots, x_k, x_{k+1}}\right), \left({y_1, \ldots, y_k, y_{k+1}}\right) \in L_k$$.

By the General Associativity Theorem and Associativity and Commutativity Properties:

$$ $$ $$ $$ $$ $$ $$ $$

Thus $$C_{k+1}: L_{k+1} \to G$$ is a homomorphism, so $$k+1 \in S$$.

So by induction, $$S = \left[{1 \,. \, . \, n}\right]$$.

In particular, $$C_n: L_n \to G$$ is a homomorphism.

By Morphism Property Preserves Closure and Homomorphism Preserves Subsemigroups, the range $$\prod_{k=1}^n H_k$$ of $$C_n$$ is therefore a subgroup of $$G$$ containing $$\bigcup_{k=1}^n {H_k}$$.

However, any subgroup containing $$\bigcup_{k=1}^n H_k$$ must clearly contain $$\prod_{k=1}^n H_k$$.

Therefore, $$\prod_{k=1}^n H_k$$ is the subgroup of $$G$$ generated by $$\bigcup_{k=1}^n H_k$$.

As $$\left \langle {H_n} \right \rangle$$ is an independent sequence of subgroups, the kernel of $$C_n$$ is $$\left\{{\left({e, \ldots, e}\right)}\right\}$$.

Hence by the Quotient Theorem for Group Epimorphisms, $$C_n$$ from $$L_n$$ to the subgroup of $$G$$ generated by $$\bigcup_{k=1}^n {H_k}$$ is an isomorphism.