description/proof of that group as direct sum of finite number of normal subgroups is group as direct sum of any reordered and combined normal subgroups
Topics
About: group
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description
- 2: Natural Language Description
- 3: Proof
Starting Context
- The reader knows a definition of group as direct sum of finite number of normal subgroups.
- The reader admits the proposition that for any group, the product of any finite number of subgroups is associative.
- The reader admits the proposition that for any group, the product of any finite number of normal subgroups is commutative and is a normal subgroup.
Target Context
- The reader will have a description and a proof of the proposition that any group as direct sum of finite number of normal subgroups is the group as direct sum of any reordered and combined normal subgroups.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(G\): \(\in \{\text{ the groups }\}\)
\(\{G_1, ..., G_n\}\): \(\subseteq \{\text{ the normal subgroups of } G\}\)
\(\{G_{\sigma_1}, ..., G_{\sigma_n}\}\): \(\sigma \in \{\text{ the permutations of } (1, ..., n)\}\)
\(\{G_{\sigma_1} ... G_{\sigma_{l_1}}, ..., G_{\sigma_{l_1 + ... + l_{k - 1} + 1}} ... G_{\sigma_n}\}\): \(n = l_1 + ... + l_k\)
//
Statements:
\(G = \text{ the direct sum of } \{G_1, ..., G_n\}\)
\(\implies\)
\(G = \text{ the direct sum of } \{\{G_{\sigma_1} ... G_{\sigma_{l_1}}, ..., G_{\sigma_{l_1 + ... + l_{k - 1} + 1}} ... G_{\sigma_n}\}\).
//
2: Natural Language Description
For any group, \(G\), any normal subgroups, \(G_1, ..., G_n\), of \(G\) such that \(G\) is the direct sum of \(\{G_1, ..., G_n\}\), \(\{G_{\sigma_1}, ..., G_{\sigma_n}\}\), where \(\sigma\) is any permutation of \((1, ..., n)\), and any \(\{G_{\sigma_1} ... G_{\sigma_{l_1}}, ..., G_{\sigma_{l_1 + ... + l_{k - 1} + 1}} ... G_{\sigma_n}\}\), where \(n = l_1 + ... + l_k\), \(G\) is the direct sum of \(\{G_{\sigma_1} ... G_{\sigma_{l_1}}, ..., G_{\sigma_{l_1 + ... + l_{k - 1} + 1}} ... G_{\sigma_n}\}\).
3: Proof
Let us see that \(G\) is the direct sum of \(\{G_{\sigma_1}, ..., G_{\sigma_n}\}\).
For each \(G_{\sigma_j} \in \{G_{\sigma_1}, ... , G_{\sigma_n}\}\), \(\sigma_j = k \in \{1, ..., n\}\), and \(G_{\sigma_j} \cap (G_{\sigma_1} ... G_{\sigma_{j - 1}} \hat{G_{\sigma_j}} G_{\sigma_{j + 1}} ... G_{\sigma_n}) = G_k \cap (G_1 ... G_{k - 1} \hat{G_k} G_{k + 1} ... G_n) = \{1\}\), by the proposition that for any group, the product of any finite number of normal subgroups is commutative and is a normal subgroup.
\(G = G_1 ... G_n = G_{\sigma_1} ... G_{\sigma_n}\), by the proposition that for any group, the product of any finite number of normal subgroups is commutative and is a normal subgroup.
For the simplicity of notations, hereafter, let us think of \(\{G_1 ... G_{l_1}, ..., G_{l_1 + ... + l_{k - 1} + 1} ... G_n\}\) instead of \(\{G_{\sigma_1} ... G_{\sigma_{l_1}}, ..., G_{\sigma_{l_1 + ... + l_{k - 1} + 1}} ... G_{\sigma_n}\}\), because we already know that the permutation does not cause any problem.
Let us see that \(G\) is the direct sum of \(\{G_1 ... G_{l_1}, ..., G_{l_1 + ... + l_{k - 1} + 1} ... G_n\}\).
Let us see that for each \(G_{l_1 + ... + l_{j - 1} + 1} ... G_{l_1 + ... + l_{j - 1} + l_j}\), \(S := G_{l_1 + ... + l_{j - 1} + 1} ... G_{l_1 + ... + l_{j - 1} + l_j} \cap ((G_1 ... G_{l_1}) ... (G_{l_1 + ... + l_{j - 2} + 1} ... G_{l_1 + ... + l_{j - 2} + l_{j - 1}}) \widehat{(G_{l_1 + ... + l_{j - 1} + 1} ... G_{l_1 + ... + l_{j - 1} + l_j})} (G_{l_1 + ... + l_j + 1} ... G_{l_1 + ... + l_j + l_{j + 1}}) ... (G_{l_1 + ... + l_{k - 1} + 1} ... G_n)) = \{1\}\).
For each \(p \in S\), \(p = p_{l_1 + ... + l_{j - 1} + 1} ... p_{l_1 + ... + l_{j - 1} + l_j} = (p_1 ... p_{l_1}) ... (p_{l_1 + ... + l_{j - 2} + 1} ... p_{l_1 + ... + l_{j - 2} + l_{j - 1}}) \widehat{(p_{l_1 + ... + l_{j - 1} + 1} ... p_{l_1 + ... + l_{j - 1} + l_j})} (p_{l_1 + ... + l_j + 1} ... p_{l_1 + ... + l_j + l_{j + 1}}) ... (p_{l_1 + ... + l_{k - 1} + 1} ... p_n)\), where \(p_j \in G_j\).
\(p_{l_1 + ... + l_{j - 1} + 1} = (p_1 ... p_{l_1}) ... (p_{l_1 + ... + l_{j - 2} + 1} ... p_{l_1 + ... + l_{j - 2} + l_{j - 1}}) \widehat{(p_{l_1 + ... + l_{j - 1} + 1} ... p_{l_1 + ... + l_{j - 1} + l_j})} (p_{l_1 + ... + l_j + 1} ... p_{l_1 + ... + l_j + l_{j + 1}}) ... (p_{l_1 + ... + l_{k - 1} + 1} ... p_n) ({p_{l_1 + ... + l_{j - 1} + l_j}}^{-1} ... {p_{l_1 + ... + l_{j - 1} + 2}}^{-1}) \in G_{l_1 + ... + l_{j - 1} + 1} \cap ((G_1 ... G_{l_1}) ... (G_{l_1 + ... + l_{j - 2} + 1} ... G_{l_1 + ... + l_{j - 2} + l_{j - 1}}) \widehat{(G_{l_1 + ... + l_{j - 1} + 1} ... G_{l_1 + ... + l_{j - 1} + l_j})} (G_{l_1 + ... + l_j + 1} ... G_{l_1 + ... + l_j + l_{j + 1}}) ... (G_{l_1 + ... + l_{k - 1} + 1} ... G_n) (G_{l_1 + ... + l_{j - 1} + l_j} ... G_{l_1 + ... + l_{j - 1} + 2}) \subseteq G_{l_1 + ... + l_{j - 1} + 1} \cap G_1 ... G_{l_1 + ... + l_{j - 1}} \widehat{G_{l_1 + ... + l_{j - 1} + 1}} G_{l_1 + ... + l_{j - 1} + 2} ... G_n) = \{1\}\), which implies that \(p_{l_1 + ... + l_{j - 1} + 1} = (p_1 ... p_{l_1}) ... (p_{l_1 + ... + l_{j - 2} + 1} ... p_{l_1 + ... + l_{j - 2} + l_{j - 1}}) \widehat{(p_{l_1 + ... + l_{j - 1} + 1} ... p_{l_1 + ... + l_{j - 1} + l_j})} (p_{l_1 + ... + l_j + 1} ... p_{l_1 + ... + l_j + l_{j + 1}}) ... (p_{l_1 + ... + l_{k - 1} + 1} ... p_n) ({p_{l_1 + ... + l_{j - 1} + l_j}}^{-1} ... {p_{l_1 + ... + l_{j - 1} + 2}}^{-1}) = 1\).
\(p_{l_1 + ... + l_{j - 1} + 2} = (p_1 ... p_{l_1}) ... (p_{l_1 + ... + l_{j - 2} + 1} ... p_{l_1 + ... + l_{j - 2} + l_{j - 1}}) \widehat{(p_{l_1 + ... + l_{j - 1} + 1} ... p_{l_1 + ... + l_{j - 1} + l_j})} (p_{l_1 + ... + l_j + 1} ... p_{l_1 + ... + l_j + l_{j + 1}}) ... (p_{l_1 + ... + l_{k - 1} + 1} ... p_n) ({p_{l_1 + ... + l_{j - 1} + l_j}}^{-1} ... {p_{l_1 + ... + l_{j - 1} + 3}}^{-1}) \in G_{l_1 + ... + l_{j - 1} + 2} \cap ((G_1 ... G_{l_1}) ... (G_{l_1 + ... + l_{j - 2} + 1} ... G_{l_1 + ... + l_{j - 2} + l_{j - 1}}) \widehat{(G_{l_1 + ... + l_{j - 1} + 1} ... G_{l_1 + ... + l_{j - 1} + l_j})} (G_{l_1 + ... + l_j + 1} ... G_{l_1 + ... + l_j + l_{j + 1}}) ... (G_{l_1 + ... + l_{k - 1} + 1} ... G_n) (G_{l_1 + ... + l_{j - 1} + l_j} ... G_{l_1 + ... + l_{j - 1} + 3})) \subseteq G_{l_1 + ... + l_{j - 1} + 2} \cap (G_1 ... G_{l_1 + ... + l_{j - 1} + 1} \widehat{G_{l_1 + ... + l_{j - 1} + 2}} G_{l_1 + ... + l_{j - 1} + 3} ... G_n) = \{1\}\), which implies that \(p_{l_1 + ... + l_{j - 1} + 2} = (p_1 ... p_{l_1}) ... (p_{l_1 + ... + l_{j - 2} + 1} ... p_{l_1 + ... + l_{j - 2} + l_{j - 1}}) \widehat{(p_{l_1 + ... + l_{j - 1} + 1} ... p_{l_1 + ... + l_{j - 1} + l_j})} (p_{l_1 + ... + l_j + 1} ... p_{l_1 + ... + l_j + l_{j + 1}}) ... (p_{l_1 + ... + l_{k - 1} + 1} ... p_n) ({p_{l_1 + ... + l_{j - 1} + l_j}}^{-1} ... {p_{l_1 + ... + l_{j - 1} + 3}}^{-1}) = 1\).
And so on. After all, \(p_{l_1 + ... + l_{j - 1} + 1} = ... = p_{l_1 + ... + l_{j - 1} + l_j} = 1\), so, \(p = p_{l_1 + ... + l_{j - 1} + 1} ... p_{l_1 + ... + l_{j - 1} + l_j} = 1\). So, \(S = \{1\}\).
\(G = G_1 ... G_n = (G_1 ... G_{l_1}) ... (G_{l_1 + ... + l_{k - 1} + 1} ... G_n)\), by the proposition that for any group, the product of any finite number of subgroups is associative.
