description/proof of that for group, \(2\) subsets, and symmetric subset, if intersection of 1st subset and product of 2nd subset and symmetric subset is empty, intersection of product of 1st subset and symmetric subset and 2nd subset is empty
Topics
About: group
The table of contents of this article
Starting Context
- The reader knows a definition of symmetric subset of group.
- 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 \(2\) subsets, and any symmetric subset, if the intersection of the 1st subset and the product of the 2nd subset and the symmetric subset is empty, the intersection of the product of the 1st subset and the symmetric subset and the 2nd subset is empty.
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:
\(G\): \(\in \{\text{ the groups }\}\)
\(S\): \(\in \{\text{ the symmetric subsets of } G\}\)
\(S_1\): \(\subseteq G\)
\(S_2\): \(\subseteq G\)
//
Statements:
(
\(S_1 \cap (S_2 S) = \emptyset\)
\(\implies\)
\((S_1 S) \cap S_2 = \emptyset\)
)
\(\land\)
(
\(S_1 \cap (S S_2) = \emptyset\)
\(\implies\)
\((S S_1) \cap S_2 = \emptyset\)
)
//
2: Note
Also \((S_1 S) \cap S_2 = \emptyset\) implies \(S_1 \cap (S_2 S) = \emptyset\), because \(S_2 \cap (S_1 S) = \emptyset\), which implies that \((S_2 S) \cap S_1 = \emptyset\), which implies \(S_1 \cap (S_2 S) = \emptyset\).
Also \((S S_1) \cap S_2 = \emptyset\) implies \(S_1 \cap (S S_2) = \emptyset\), because \(S_2 \cap (S S_1) = \emptyset\), which implies \((S S_2) \cap S_1 = \emptyset\), which implies \(S_1 \cap (S S_2) = \emptyset\).
3: Proof
Whole Strategy: Step 1: suppose that \(S_1 \cap (S_2 S) = \emptyset\); Step 2: suppose that there was a \(g \in (S_1 S) \cap S_2\), and find a contradiction; Step 3: suppose that \(S_1 \cap (S S_2) = \emptyset\); Step 4: suppose that there was a \(g \in (S S_1) \cap S_2\), and find a contradiction.
Step 1:
Let us suppose that \(S_1 \cap (S_2 S) = \emptyset\).
Step 2:
Let us suppose that there was a \(g \in (S_1 S) \cap S_2\).
\(g \in S_1 S\), so, \(g = s_1 s\) where \(s_1 \in S_1\) and \(s \in S\).
\(g \in S_2\), so, \(g = s_2\) where \(s_2 \in S_2\).
\(s_1 s = g = s_2\), so, \(s_1 = s_2 s^{-1}\).
But as \(S\) was symmetric, \(s^{-1} \in S^{-1} = S\).
So, \(s_1 \in S_2 S\), so, \(s_1 \in S_1 \cap (S_2 S)\), a contradiction against \(S_1 \cap (S_2 S) = \emptyset\).
So, there is no \(g \in (S_1 S) \cap S_2\).
So, \((S_1 S) \cap S_2 = \emptyset\).
Step 3:
Let us suppose that \(S_1 \cap (S S_2) = \emptyset\).
Step 4:
Let us suppose that there was a \(g \in (S S_1) \cap S_2\).
\(g \in S S_1\), so, \(g = s s_1\) where \(s_1 \in S_1\) and \(s \in S\).
\(g \in S_2\), so, \(g = s_2\) where \(s_2 \in S_2\).
\(s s_1 = g = s_2\), so, \(s_1 = s^{-1} s_2\).
But as \(S\) was symmetric, \(s^{-1} \in S^{-1} = S\).
So, \(s_1 \in S S_2\), so, \(s_1 \in S_1 \cap (S S_2)\), a contradiction against \(S_1 \cap (S S_2) = \emptyset\).
So, there is no \(g \in (S S_1) \cap S_2\).
So, \((S S_1) \cap S_2 = \emptyset\).
