description/proof of that for group as direct sum of finite number of normal subgroups, product of subset of normal subgroups is group as direct sum of subset
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
Target Context
- The reader will have a description and a proof of the proposition that for any group as direct sum of finite number of normal subgroups, the product of any subset of the normal subgroups is the group as direct sum of the subset.
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:
\(G\): \(\in \{\text{ the groups }\}\)
\(\{G_1, ..., G_n\}\): \(\subseteq \{\text{ the normal subgroups of } G\}\)
\(\{G_{s_1}, ..., G_{s_m}\}\): \(\subseteq \{G_1, ..., G_n\}\)
//
Statements:
\(G = \text{ the direct sum of } \{G_1, ..., G_n\}\)
\(\implies\)
\(G_{s_1} ... G_{s_m} = \text{ the direct sum of } \{G_{s_1}, ..., G_{s_m}\}\).
//
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 \(\{G_1, ..., G_n\}\), and any subset, \(\{G_{s_1}, ..., G_{s_m}\} \subseteq \{G_1, ..., G_n\}\), \(G_{s_1} ... G_{s_m}\) is the direct sum of \(\{G_{s_1}, ..., G_{s_m}\}\).
3: Proof
\(G_{s_1} ... G_{s_m}\) is a normal subgroup of \(G\), by the proposition that for any group, the product of any finite number of normal subgroups is commutative and is a normal subgroup, so, is a group.
Let us see that for each \(G_{s_j} \in \{G_{s_1}, ..., G_{s_m}\}\), \(G_{s_j} \cap (G_{s_1} ... G_{s_{j - 1}} \hat{G_{s_j}} G_{s_{j + 1}} ... G_{s_m}) = \{1\}\).
\(G_{s_j} = G_k\) for a \(k \in \{1, ..., n\}\), and \(G_{s_j} \cap (G_{s_1} ... G_{s_{j - 1}} \hat{G_{s_j}} G_{s_{j + 1}} ... G_{s_m}) \subseteq G_k \cap (G_1 ... G_{k - 1} \widehat{G_k} G_{k + 1} ... G_n) = \{1\}\). As obviously, \(1 \in G_{s_j} \cap (G_{s_1} ... G_{s_{j - 1}} \hat{G_{s_j}} G_{s_{j + 1}} ... G_{s_m})\), \(G_{s_j} \cap (G_{s_1} ... G_{s_{j - 1}} \hat{G_{s_j}} G_{s_{j + 1}} ... G_{s_m}) = \{1\}\).
\(G_{s_1} ... G_{s_m} = G_{s_1} ... G_{s_m}\) is immediate.
