description/proof of that finite product of subsets of group is associative
Topics
About: group
The table of contents of this article
Starting Context
- The reader knows a definition of finite product of subsets of group.
Target Context
- The reader will have a description and a proof of the proposition that for any group, any finite product of subsets of the group 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:
\(G\): \(\in \{\text{ the groups }\}\)
\(J\): \(\in \{\text{ the finite index sets }\}\), such that \(\vert J \vert = n\)
\(\{S_j \subseteq G \vert j \in J\}\):
\(S\): \(= S_{J_1} ... {J_n}\) with any association
\(S'\): \(= S_{J_1} ... {J_n}\) with any association
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Statements:
\(S = S'\)
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2: Proof
Whole Strategy: Step 1: see that \(S \subseteq S'\); Step 2: conclude the proposition.
Step 1:
Let \(s \in S\) be any.
\(s = s_{J_1} ... s_{J_n}\) where \(s_{J_1} \in S_{J_1}, ..., s_{J_n} \in S_{J_n}\), with the association that corresponds to that of \(S\).
But as multiplications in \(G\) are associative, \(s_{J_1} ... s_{J_n}\) is \(s_{J_1} ... s_{J_n}\) with the association that corresponds to that of \(S'\)
So, \(s = s_{J_1} ... s_{J_n} \in S'\).
So, \(S \subseteq S'\).
Step 2:
Symmetrically, \(S' \subseteq S\).
So, \(S = S'\).
