# Internal Direct Product Generated by Subgroups

## Theorem

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

Let $\sequence {H_n}$ be a sequence of subgroups of $G$.

Then:

the subgroup generated by $\displaystyle \bigcup_{k \mathop = 1}^n H_k$ is the internal group direct product of $\sequence {H_n}$
$\sequence {H_n}$ 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 $\closedint m n$ is to be understood to mean a (closed) integer interval:

$\closedint m n := \set {x \in \Z: m \le x \le n}$

for $m, n \in \Z$.

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

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

### Necessary Condition

Suppose that the subgroup generated by $\displaystyle \bigcup_{k \mathop = 1}^n H_k$ is the internal group direct product of $\sequence {H_n}$.

We have that Internal Group Direct Product is Injective.

Hence by definition $C_k$ is a monomorphism.

It follows from Kernel is Trivial iff Monomorphism that the kernel of $C_n$ is $\set {\tuple {e, e, \ldots, e} }$.

Therefore $\sequence {H_n}$ is an independent sequence.

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

For each $k \in \closedint 1 n$, 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:

 $\displaystyle x y$ $=$ $\displaystyle \paren {y_i x_i} \paren {y_j x_j}$ $\displaystyle$ $=$ $\displaystyle \prod_{k \mathop = 1}^n y_k x_k$ $\displaystyle$ $=$ $\displaystyle \map {C_n} {y_1 x_1, \ldots, y_n x_n}$ $\displaystyle$ $=$ $\displaystyle \map {C_n} {\tuple {y_1, \ldots, y_n} \tuple {x_1, \ldots, x_n} }$ $\displaystyle$ $=$ $\displaystyle \map {C_n} {y_1, \ldots, y_n} \map {C_n} {x_1, \ldots, x_n}$ $\displaystyle$ $=$ $\displaystyle y x$

$\Box$

### Sufficient Condition

Suppose that $\sequence {H_n}$ 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 $k \in \closedint 1 n$ 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 $\tuple {x_1, \ldots, x_k, x_{k + 1} }, \tuple {y_1, \ldots, y_k, y_{k + 1} } \in L_k$.

 $\displaystyle \map {C_{k + 1} } {\tuple {x_1, \ldots, x_k, x_{k + 1} } \tuple {y_1, \ldots, y_k, y_{k + 1} } }$ $=$ $\displaystyle \map {C_{k + 1} } {x_1 y_1, \ldots, x_k y_k, x_{k + 1} y_{k + 1} }$ $\displaystyle$ $=$ $\displaystyle \paren {x_1 y_1} \cdots \paren {x_k y_k} \paren {x_{k + 1} y_{k + 1} }$ $\displaystyle$ $=$ $\displaystyle \map {C_k} {\paren {x_1 y_1} \cdots \paren {x_k y_k} } \paren {x_{k + 1} y_{k + 1} }$ $\displaystyle$ $=$ $\displaystyle \map {C_k} {\tuple {x_1, \ldots, x_k} \tuple {y_1, \ldots, y_k} } \paren {x_{k + 1} y_{k + 1} }$ $\displaystyle$ $=$ $\displaystyle \map {C_k} {x_1, \ldots, x_k} \map {C_k} {y_1, \ldots, y_k} \paren {x_{k + 1} y_{k + 1} }$ $\displaystyle$ $=$ $\displaystyle \paren {\prod_{k \mathop = 1}^n{x_j} } \paren {\prod_{k \mathop = 1}^n{y_j} } x_{k + 1} y_{k + 1}$ $\displaystyle$ $=$ $\displaystyle \paren {\paren {\prod_{k \mathop = 1}^n x_j} x_{k + 1} } \paren {\paren {\prod_{k \mathop = 1}^n y_j} y_{k + 1} }$ $\displaystyle$ $=$ $\displaystyle \map {C_{k + 1} } {x_1, \ldots, x_k, x_{k + 1} } \map {C_{k + 1} } {y_1, \ldots, y_k, y_{k + 1} }$

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

So by induction:

$S = \closedint 1 n$

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

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

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

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

As $\sequence {H_n}$ is an independent sequence of subgroups, the kernel of $C_n$ is $\set {\tuple {e, \ldots, e} }$.

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

$\blacksquare$