Internal Direct Product Theorem/General Result/Proof 1
Theorem
Let $G$ be a group whose identity is $e$.
Let $\sequence {H_k}_{1 \mathop \le k \mathop \le n}$ be a sequence of subgroups of $G$.
Then $G$ is the internal group direct product of $\sequence {H_k}_{1 \mathop \le k \mathop \le n}$ if and only if:
- $(1): \quad G = H_1 H_2 \cdots H_n$
- $(2): \quad \sequence {H_k}_{1 \mathop \le k \mathop \le n}$ is a sequence of independent subgroups
- $(3): \quad \forall k \in \set {1, 2, \ldots, n}: H_k \lhd G$
where $H_k \lhd G$ denotes that $H_k$ is a normal subgroup of $G$.
Proof
The validity of the material on this page is questionable. In particular: This uses the erroneous result Subgroup Product is Internal Group Direct Product iff Surjective which is up for deletion. Instead we will follow Warner closely and base the result on Internal Direct Product Generated by Subgroups. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by resolving the issues. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Questionable}} from the code.If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page. |
By definition, $G$ is the internal group direct product of $\sequence {H_k}_{1 \mathop \le k \mathop \le n}$ if and only if the mapping:
- $\ds C: \prod_{k \mathop = 1}^n H_k \to G: \map C {h_1, \ldots, h_n} = \prod_{k \mathop = 1}^n h_k$
is a group isomorphism from the cartesian product $\struct {H_1, \circ {\restriction_{H_1} } } \times \cdots \times \struct {H_n, \circ {\restriction_{H_n} } }$ onto $\struct {G, \circ}$.
Necessary Condition
Let $G$ be the internal group direct product of $\sequence {H_k}_{1 \mathop \le k \mathop \le n}$.
- $(1): \quad$ From Subgroup Product is Internal Group Direct Product iff Surjective, $G = H_1 H_2 \cdots H_n$.
- $(2): \quad$ From Internal Group Direct Product is Injective: General Result, $\sequence {H_k}_{1 \mathop \le k \mathop \le n}$ is a sequence of independent subgroups.
- $(3): \quad$ From Internal Group Direct Product Isomorphism:
- $\forall k \in \set {1, 2, \ldots, n}: H_k \lhd G$
$\Box$
Sufficient Condition
Now suppose the three conditions hold.
- $(1): \quad$ From Subgroup Product is Internal Group Direct Product iff Surjective, $C$ is surjective.
- $(2): \quad$ From Internal Group Direct Product is Injective: General Result, $C$ is injective.
- $(3): \quad$ From Internal Group Direct Product of Normal Subgroups, $C$ is a group homomorphism.
Putting these together, we see that $C$ is a bijective homomorphism, and therefore an isomorphism.
So by definition, $G$ is the internal group direct product of $\sequence {H_k} _{1 \mathop \le k \mathop \le n}$.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 18$: Induced $N$-ary Operations: Theorem $18.14$