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 $\ds \bigcup_{k \mathop = 1}^n H_k$ 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 $\ds L_k = \prod_{j \mathop = 1}^k H_j$ be the cartesian product of the subgroups $H_1, H_2, \ldots, H_k$ of $G$.
Let $\ds C_k: L_k \to G: \map {C_k} {x_1, x_2, \ldots, x_k} = \prod_{j \mathop = 1}^k x_j$.
Necessary Condition
Suppose that the subgroup generated by $\ds \bigcup_{k \mathop = 1}^n H_k$ 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:
\(\ds x y\) | \(=\) | \(\ds \paren {y_i x_i} \paren {y_j x_j}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \prod_{k \mathop = 1}^n y_k x_k\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map {C_n} {y_1 x_1, \ldots, y_n x_n}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map {C_n} {\tuple {y_1, \ldots, y_n} \tuple {x_1, \ldots, x_n} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map {C_n} {y_1, \ldots, y_n} \map {C_n} {x_1, \ldots, x_n}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds y x\) |
$\Box$
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 $\tuple {x_1, \ldots, x_k, x_{k + 1} }, \tuple {y_1, \ldots, y_k, y_{k + 1} } \in L_k$.
By the General Associativity Theorem and Element Commutes with Product of Commuting Elements:
\(\ds \map {C_{k + 1} } {\tuple {x_1, \ldots, x_k, x_{k + 1} } \tuple {y_1, \ldots, y_k, y_{k + 1} } }\) | \(=\) | \(\ds \map {C_{k + 1} } {x_1 y_1, \ldots, x_k y_k, x_{k + 1} y_{k + 1} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x_1 y_1} \cdots \paren {x_k y_k} \paren {x_{k + 1} y_{k + 1} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map {C_k} {\paren {x_1 y_1} \cdots \paren {x_k y_k} } \paren {x_{k + 1} y_{k + 1} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map {C_k} {\tuple {x_1, \ldots, x_k} \tuple {y_1, \ldots, y_k} } \paren {x_{k + 1} y_{k + 1} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map {C_k} {x_1, \ldots, x_k} \map {C_k} {y_1, \ldots, y_k} \paren {x_{k + 1} y_{k + 1} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\prod_{k \mathop = 1}^n{x_j} } \paren {\prod_{k \mathop = 1}^n{y_j} } x_{k + 1} y_{k + 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\paren {\prod_{k \mathop = 1}^n x_j} x_{k + 1} } \paren {\paren {\prod_{k \mathop = 1}^n y_j} y_{k + 1} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map {C_{k + 1} } {x_1, \ldots, x_k, x_{k + 1} } \map {C_{k + 1} } {y_1, \ldots, y_k, y_{k + 1} }\) |
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 $\ds \prod_{k \mathop = 1}^n H_k$ of $C_n$ is therefore a subgroup of $G$ containing $\ds \bigcup_{k \mathop = 1}^n H_k$.
However, any subgroup containing $\ds \bigcup_{k \mathop = 1}^n H_k$ must clearly contain $\ds \prod_{k \mathop = 1}^n H_k$.
Therefore, $\ds \prod_{k \mathop = 1}^n H_k$ is the subgroup of $G$ generated by $\ds \bigcup_{k \mathop = 1}^n H_k$.
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 $\ds \bigcup_{k \mathop = 1}^n H_k$ is an isomorphism.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 18$: Induced $N$-ary Operations: Theorem $18.13$