description/proof of that for group as direct sum of finite number of normal subgroups, element is uniquely decomposed and decomposition is commutative
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 any group as direct sum of finite number of normal subgroups is the group as direct sum of any reordered and combined normal subgroups.
- The reader admits 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.
Target Context
- The reader will have a description and a proof of the proposition that for any group as the direct sum of any finite number of normal subgroups, each element is uniquely decomposed and the decomposition is commutative.
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:
\(G\): \(\in \{\text{ the groups }\}\)
\(\{G_1, ..., G_n\}\): \(\subseteq \{\text{ the normal subgroups of } G\}\)
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Statements:
\(G = \text{ the direct sum of } \{G_1, ..., G_n\}\)
\(\implies\)
\(\forall p_1 ... p_n = p'_1 ... p'_n \in G \text{ such that } p_j, p'_j \in G_j (p_j = p'_j \land \forall \sigma \in \{\text{ the permutations of } \{1, ..., n\}\} (p_1 ... p_n = p_{\sigma_1} ... p_{\sigma_n}))\).
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2: Natural Language Description
For any group, \(G\), and any normal subgroups, \(G_1, ..., G_n\), of \(G\), such that \(G\) is the direct sum of \(\{G_1, ..., G_n\}\), for each \(p_1 ... p_n \in G\) such that \(p_j \in G_j\), \(p_1 ... p_n\) is the unique decomposition of the element, and for each permutation of \(\{1, ..., n\}\), \(\sigma\), \(p_1 ... p_n = p_{\sigma_1} ... p_{\sigma_n}\).
3: Proof
Let us prove it inductively with respect to \(n\).
Let \(n = 2\).
As \(p_1 p_2 = p'_1 p'_2\), \({p_1}^{-1} p_1 p_2 {p'_2}^{-1} = {p_1}^{-1} p'_1 p'_2 {p'_2}^{-1}\), so, \(p_2 {p'_2}^{-1} = {p_1}^{-1} p'_1\), but the left hand side is in \(G_2\) and the right hand side is in \(G_1\), so, the both side is in \(G_1 \cap G_2 = \{1\}\), so, \(p_2 {p'_2}^{-1} = {p_1}^{-1} p'_1 = 1\), which implies that \(p'_1 = p_1\) and \(p'_2 = p_2\).
For each \(p_2 p_1 \in G\), \(p_2 p_1 = p'_1 p'_2\) for a \(p'_1 \in G_1\) and a \(p'_2 \in G_2\), because \(G = G_1 G_2\). \({p_2}^{-1} p_2 p_1 = {p_2}^{-1} p'_1 p'_2\), so, \(p_1 = {p_2}^{-1} p'_1 p'_2 = {p_2}^{-1} p'_1 p_2 {p_2}^{-1} p'_2 = p''_1 {p_2}^{-1} p'_2\) for a \(p''_1 \in G_1\), because \(G_1\) is a normal subgroup, but as the decomposition is unique, \(p_1 1 = p''_1 {p_2}^{-1} p'_2\) implies that \({p_2}^{-1} p'_2 = 1\), so, \(p'_2 = p_2\). Likewise, \(p_2 p_1 {p_1}^{-1} = p'_1 p'_2 {p_1}^{-1}\), so, \(p_2 = p'_1 p'_2 {p_1}^{-1} = p'_1 {p_1}^{-1} p_1 p'_2 {p_1}^{-1} = p'_1 {p_1}^{-1} p''_2\) for a \(p''_2 \in G_2\), but as the decomposition is unique, \(1 p_2 = p'_1 {p_1}^{-1} p''_2\) implies that \(p'_1 {p_1}^{-1} = 1\), so, \(p'_1 = p_1\). So, \(p_2 p_1 = p_1 p_2\).
Let us suppose that the proposition holds through \(n = n'\).
Let us suppose that \(G\) is the direct sum of \(\{G_1, ..., G_{n' + 1}\}\).
\(G\) is the direct sum of \(\{G_1, G_2 ... G_{n' + 1}\}\), by 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.
For each \(p = p_1 ... p_{n' + 1} = p'_1 ... p'_{n' + 1}\), \(p = p_1 (p_2 ... p_{n' + 1}) = p'_1 (p'_2 ... p'_{n' + 1})\), where \(p_2 ... p_{n' + 1}, p'_2 ... p'_{n' + 1} \in G_2 ... G_{n' + 1}\), so, \(p_1 = p'_1\) and \(p_2 ... p_{n' + 1} = p'_2 ... p'_{n' + 1}\), by applying this proposition for \(n = 2\).
\(G_2 ... G_{n' + 1}\) is the direct sum of \(\{G_2, ... , G_{n' + 1}\}\), by 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. So, applying this proposition for \(n = n'\), \(p_j = p'_j\) for each \(2 \le j \le n' + 1\).
So, \(p_j = p'_j\) for each \(1 \le j \le n' + 1\).
For each \(\sigma\), when \(\sigma_1 = 1\), \(p_1 ... p_{n' + 1} = p_{\sigma_1} ... p_{\sigma_{n' + 1}}\) by applying this proposition for \(n = n'\); when \(\sigma_1 \neq 1\), \(\sigma_1 = k\) for a \(2 \le k \le n' + 1\), and \(p_1 ... p_{n' + 1} = p_1 (p_2 ... p_{n' + 1}) = (p_2 ... p_{n' + 1}) p_1 = p_2 (p_3 ... p_{n' + 1} p_1) = (p_3 ... p_{n' + 1} p_1) p_2 = ... = p_k p_{k + 1} ... p_{n' + 1} p_1 ... p_{k - 1} = p_{\sigma_1} p_{k + 1} ... p_{n' + 1} p_1 ... p_{k - 1} = p_{\sigma_1} ... p_{\sigma_{n' + 1}}\), where the 'until (not including)'-the-last equals are by applying this proposition for \(n = 2\) and the last equal is by applying this proposition for \(n = n'\).
So, this proposition holds for each \(2 \le n\).
