description/proof of that for group, product of element and intersection of subsets is intersection of products of element and subsets
Topics
About: group
The table of contents of this article
Starting Context
- The reader knows a definition of indexed set.
- 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 element, and any possibly uncountable number of subsets, the product of the element and the intersection of the subsets is the intersection of the products of the element and the subsets.
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:
\(G\): \(\in \{\text{ the groups }\}\)
\(g\): \(\in G\)
\(J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{S_j \subseteq G\}_{j \in J}\): \(\in \{\text{ the indexed sets }\}\)
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Statements:
\(g \cap_{j \in J} S_j = \cap_{j \in J} (g S_j)\)
\(\land\)
\((\cap_{j \in J} S_j) g = \cap_{j \in J} (S_j g)\)
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2: Note
Compare with the proposition that for any group, the product of any subset and the intersection of any 2 subsets is contained in but not necessarily equal to the intersection of the product of the 1st subset and the 2nd subset and the product of the 1st subset and the 3rd subset.
3: Proof
Whole Strategy: Step 1: see that for each \(g' \in g \cap_{j \in J} S_j\), \(g' \in \cap_{j \in J} (g S_j)\), and for each \(g' \in \cap_{j \in J} (g S_j)\), \(g' \in g \cap_{j \in J} S_j\); Step 2: see that for each \(g' \in (\cap_{j \in J} S_j) g\), \(g' \in \cap_{j \in J} (S_j g)\), and for each \(g' \in \cap_{j \in J} (S_j g)\), \(g' \in (\cap_{j \in J} S_j) g\).
Step 1:
Let \(g' \in g \cap_{j \in J} S_j\) be any.
There is a \(g'' \in \cap_{j \in J} S_j\) such that \(g' = g g''\).
For each \(j \in J\), \(g' \in g S_j\), because \(g'' \in S_j\).
So, \(g' \in \cap_{j \in J} (g S_j)\).
So, \(g \cap_{j \in J} S_j \subseteq \cap_{j \in J} (g S_j)\).
Let \(g' \in \cap_{j \in J} (g S_j)\) be any.
For each \(j \in J\), \(g' \in g S_j\), so, \(g' = g s_j\) for an \(s_j \in S_j\).
For each \(j \in J\), \(s_j = g^{-1} g'\).
So, \(g^{-1} g' \in \cap_{j \in J} S_j\).
\(g' = g g^{-1} g'\), where \(g^{-1} g' \in \cap_{j \in J} S_j\).
So, \(g' \in g \cap_{j \in J} S_j\).
So, \(\cap_{j \in J} (g S_j) \subseteq g \cap_{j \in J} S_j\).
So, \(g \cap_{j \in J} S_j = \cap_{j \in J} (g S_j)\).
Step 2:
Let \(g' \in (\cap_{j \in J} S_j) g\) be any.
There is a \(g'' \in \cap_{j \in J} S_j\) such that \(g' = g'' g\).
For each \(j \in J\), \(g' \in S_j g\), because \(g'' \in S_j\).
So, \(g' \in \cap_{j \in J} (S_j g)\).
So, \((\cap_{j \in J} S_j) g \subseteq \cap_{j \in J} (S_j g)\).
Let \(g' \in \cap_{j \in J} (S_j g)\) be any.
For each \(j \in J\), \(g' \in S_j g\), so, \(g' = s_j g\) for an \(s_j \in S_j\).
For each \(j \in J\), \(s_j = g' g^{-1}\).
So, \(g' g^{-1} \in \cap_{j \in J} S_j\).
\(g' = g' g^{-1} g\), where \(g' g^{-1} \in \cap_{j \in J} S_j\).
So, \(g' \in (\cap_{j \in J} S_j) g\).
So, \(\cap_{j \in J} (S_j g) \subseteq (\cap_{j \in J} S_j) g\).
So, \((\cap_{j \in J} S_j) g = \cap_{j \in J} (S_j g)\).
