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 $\displaystyle \bigcup_{k=1}^n H_k$ is the internal group 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
In the following, the notation $\left[{m \,.\,.\, n}\right]$ is to be understood to mean a (closed) integer interval:
 * $\left[{m \,.\,.\, n}\right] := \left\{{x \in \Z: m \le x \le n}\right\}$

for $m, n \in \Z$.

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

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

Necessary Condition
Suppose that the subgroup generated by $\displaystyle \bigcup_{k=1}^n H_k$ is the internal group direct product of $\left \langle {H_n} \right \rangle$.

We have that Internal Group Direct Product Injective.

Hence by definition $C_k$ is a monomorphism.

It follows from Kernel of Monomorphism is Trivial that 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:

Sufficient Condition
Suppose that $\left \langle {H_n} \right \rangle$ is an independent sequence of subgroups such that every element of $H_i$ commutes with every element of $H_j$ whenever $1 \le i < j \le n$.

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 codomain $\displaystyle \prod_{k=1}^n H_k$ of $C_n$ is therefore a subgroup of $G$ containing $\displaystyle \bigcup_{k=1}^n {H_k}$.

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

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

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 $\displaystyle \bigcup_{k=1}^n {H_k}$ is an isomorphism.