description/proof of that finite product of subgroups is associative
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: Note
- 4: Proof
Starting Context
- The reader knows a definition of group.
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 subgroups is associative.
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:
\(G\): \(\in \{\text{ the groups }\}\)
\(\{G_1, ..., G_n\}\): \(\subseteq \{\text{ the subgroups of } G\}\)
\(S\): \(= G_1 ... G_n\) with any association
\(S'\): \(= G_1 ... G_n\) with any association
//
Statements:
\(S = S'\)
//
2: Natural Language Description
For any group, \(G\), and any subgroups, \(G_1, ..., G_n\), of \(G\), \(S = G_1 ... G_n\) with any association is \(S' = G_1 ... G_n\) with any association.
3: Note
We are not claiming that \(S\) or \(S'\) is a group; \(S = S'\) is just set-wise.
4: Proof
For each \(p \in S\), \(p = g_1 ... g_n\) with the corresponding association. But as multiplications in \(G\) is associative, that multiplication can be done in the association that corresponds to that of \(S'\), and the result is on \(S'\). So, \(S \subseteq S'\). Likewise, \(S' \subseteq S\). So, \(S = S'\).
