description/proof of that for group, product of subset and intersection of 2 subsets is contained in but not necessarily equal to intersection of product of 1st subset and 2nd subset and product of 1st subset and 3rd subset
Topics
About: group
The table of contents of this article
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 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.
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:
\(G\): \(\in \{\text{ the groups }\}\)
\(S_1\): \(\subseteq G\)
\(S_2\): \(\subseteq G\)
\(S_3\): \(\subseteq G\)
//
Statements:
\(S_1 (S_2 \cap S_3) \subseteq (S_1 S_2) \cap (S_1 S_3)\)
//
2: Proof
Whole Strategy: Step 1: let \(g \in S_1 (S_2 \cap S_3)\) be any and see that \(g \in (S_1 S_2) \cap (S_1 S_3)\); Step 2: see an example that \(S_1 (S_2 \cap S_3) \subset (S_1 S_2) \cap (S_1 S_3)\).
Step 1:
Let \(g \in S_1 (S_2 \cap S_3)\) be any.
\(g = s_1 s\), where \(s_1 \in S_1\) and \(s \in S_2 \cap S_3\).
\(g = s_1 s \in S_1 S_2\), because \(s \in S_2\), and \(g = s_1 s \in S_1 S_3\), because \(s \in S_3\).
So, \(g \in (S_1 S_2) \cap (S_1 S_3)\).
Step 2:
For any \(g \in (S_1 S_2) \cap (S_1 S_3)\), \(g \in S_1 (S_2 \cap S_3)\) does not necessarily hold, because \(g = s_1 s_2 = s'_1 s_3\) where \(s_1, s'_1 \in S_2\) and \(s_2 \in S_2\) and \(s_3 \in S_3\), but as \(s_1 \neq s'_1\) in general, \(g\) cannot be expressed as \(s''_1 s\) in general.
For example, let \(G = \mathbb{Z}\) as the additive group, \(S_1 = \{0, 1\}\), \(S_2 = \{3, 4\}\), and \(S_3 = \{5, 6\}\).
Then, \(S_1 (S_2 \cap S_3) = \{0, 1\} (\{3, 4\} \cap \{5, 6\}) = \{0, 1\} \emptyset = \emptyset\) and \((S_1 S_2) \cap (S_1 S_3) = \{0, 1\} \{3, 4\} \cap \{0, 1\} \{5, 6\} = \{3, 4, 5\} \cap \{5, 6, 7\} = \{5\}\).
