description/proof of that finite product of normal subgroups is commutative and is normal subgroup
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 normal subgroup of group.
- 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, any subgroup of the group multiplied by any normal subgroup of the group is a subgroup of the group.
- The reader admits the proposition that for any group and its any subgroup, the subgroup is a normal subgroup if its conjugate subgroup by each element of the group is contained in it.
Target Context
- The reader will have a description and a proof of the proposition that for any group, the product of any finite number of normal subgroups is commutative and is a normal subgroup.
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:
\(G\): \(\in \{\text{ the groups }\}\)
\(\{G_1, ..., G_n\}\): \(\subseteq \{\text{ the normal subgroups of } G\}\)
\(G_1 ... G_n\):
\(G_{\sigma_1} ... G_{\sigma_n}\): \(\sigma\) is any permutation of \((1, ..., n)\)
//
Statements:
\(G_1 ... G_n = G_{\sigma_1} ... G_{\sigma_n}\)
\(\land\)
\(G_1 ... G_n \in \{\text{ the normal subgroups of } G\}\)
//
2: Natural Language Description
For any group, \(G\), and any normal subgroups, \(G_1, ..., G_n\), of \(G\), \(G_1 ... G_n = G_{\sigma_1} ... G_{\sigma_n}\) where \(\sigma\) is any permutation of \((1, ..., n)\), and \(G_1 ... G_n\) is a normal subgroup of \(G\).
3: Proof
Let us prove the commutativity inductively.
Let \(n = 2\). \(G_1 G_2 = G_2 G_1\)? For each \(g_1 g_2 \in G_1 G_2\), \(g_1 g_2 = g_1 g_2 {g_1}^{-1} g_1\), but \(g'_2 := g_1 g_2 {g_1}^{-1} \in G_2\), because \(G_2\) is a normal subgroup; so, \(g_1 g_2 = g'_2 g_1 \in G_2 G_1\). So, \(G_1 G_2 \subseteq G_2 G_1\). Likewise, \(G_2 G_1 \subseteq G_1 G_2\). So, \(G_1 G_2 = G_2 G_1\).
Let us suppose that the commutativity holds through \(n = n'\). \(G_1 ... G_{n' + 1} = G_{\sigma_1} ... G_{\sigma_{n' + 1}}\)? \(G_{\sigma_{n' + 1}} = G_k\). \(G_1 ... \hat{G_k} ... G_{n' + 1} = G_{\sigma_1} ... G_{\sigma_{n'}}\) by the supposition. \(G_1 ... \hat{G_k} ... G_{n' + 1} G_k = G_{\sigma_1} ... G_{\sigma_{n'}} G_{\sigma_{n' + 1}}\). As the multiplications are associative, by the proposition that for any group, the product of any finite number of subgroups is associative, \(G_1 ... \hat{G_k} ... G_{n'} G_{n' + 1} G_k = G_1 ... \hat{G_k} ... G_{n'} (G_{n' + 1} G_k) = G_1 ... \hat{G_k} ... G_{n'} (G_k G_{n' + 1}) = G_1 ... \hat{G_k} ... (G_{n'} G_k) G_{n' + 1} = G_1 ... \hat{G_k} ... (G_k G_{n'}) G_{n' + 1} = ... = G_1 ... G_k ... G_{n' + 1}\). So, \(G_1 ... G_{n' + 1} = G_{\sigma_1} ... G_{\sigma_{n' + 1}}\).
So, \(G_1 ... G_n = G_{\sigma_1} ... G_{\sigma_n}\) for each \(2 \le n\).
Let us prove that \(G_1 ... G_n\) is a normal subgroup inductively.
Let \(n = 2\). Is \(G_1 G_2\) a normal subgroup? \(G_1 G_2\) is a subgroup of \(G\), by the proposition that for any group, any subgroup of the group multiplied by any normal subgroup of the group is a subgroup of the group. For each \(g \in G\), \(g G_1 G_2 g^{-1} = G_1 G_2\)? For each \(g g_1 g_2 g^{-1} \in g G_1 G_2 g^{-1}\), \(g g_1 g^{-1} g g_2 g^{-1}\), but \(g g_1 g^{-1} \in G_1\) and \(g g_2 g^{-1} \in G_2\), so, \(g g_1 g_2 g^{-1} \in G_1 G_2\). So, \(g G_1 G_2 g^{-1} \subseteq G_1 G_2\). So, \(g G_1 G_2 g^{-1} = G_1 G_2\), by the proposition that for any group and its any subgroup, the subgroup is a normal subgroup if its conjugate subgroup by each element of the group is contained in it.
Let us suppose that the claim holds through \(n = n'\). \(G_1 ... G_{n' + 1} = (G_1 ... G_{n'}) G_{n' + 1}\), but \((G_1 ... G_{n'})\) is a normal subgroup by the supposition, and \((G_1 ... G_{n'}) G_{n' + 1}\) is a normal subgroup.
So, \(G_1 ... G_n\) is a normal subgroup for each \(2 \le n\).
