description/proof of that for group, subgroup, and \(2\) subsets, if intersection of 1st subset and product of 2nd subset and subgroup is empty, intersection of product of 1st subset and subgroup and product of 2nd subset and subgroup is empty
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 subgroup, and any \(2\) subsets, if the intersection of the 1st subset and the product of the 2nd subset and the subgroup is empty, the intersection of the product of the 1st subset and the subgroup and the product of the 2nd subset and the subgroup 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:
\(G'\): \(\in \{\text{ the groups }\}\)
\(G\): \(\in \{\text{ the subgroups of } G'\}\)
\(S_1\): \(\subseteq G'\)
\(S_2\): \(\subseteq G'\)
//
Statements:
(
\(S_1 \cap (S_2 G) = \emptyset\)
\(\implies\)
\((S_1 G) \cap (S_2 G) = \emptyset\)
)
\(\land\)
(
\(S_1 \cap (G S_2) = \emptyset\)
\(\implies\)
\((G S_1) \cap (G S_2) = \emptyset\)
)
//
2: Note
Also \((S_1 G) \cap S_2 = \emptyset\) implies \((S_1 G) \cap (S_2 G) = \emptyset\), because \(S_2 \cap (S_1 G) = \emptyset\), which implies that \((S_2 G) \cap (S_1 G) = \emptyset\), which implies \((S_1 G) \cap (S_2 G) = \emptyset\).
Also \((G S_1) \cap S_2 = \emptyset\) implies \((G S_1) \cap (G S_2) = \emptyset\), because \(S_2 \cap (G S_1) = \emptyset\), which implies that \((G S_2) \cap (G S_1) = \emptyset\), which implies \((G S_1) \cap (G S_2) = \emptyset\).
\(S_1 \cap (S_2 G) = \emptyset\) implies also \((S_1 G) \cap S_2 = \emptyset\), as a special case 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, because any subgroup is symmetric, as is mentioned in Note for the definition of symmetric subset of group.
\(S_1 \cap (G S_2) = \emptyset\) implies also \((G S_1) \cap S_2 = \emptyset\), likewise.
3: Proof
Whole Strategy: Step 1: suppose that \(S_1 \cap (S_2 G) = \emptyset\); Step 2: suppose that there was a \(g \in (S_1 G) \cap (S_2 G)\), and find a contradiction; Step 3: suppose that \(S_1 \cap (G S_2) = \emptyset\); Step 4: suppose that there was a \(g \in (G S_1) \cap (G S_2)\), and find a contradiction.
Step 1:
Let us suppose that \(S_1 \cap (S_2 G) = \emptyset\).
Step 2:
Let us suppose that there was a \(g \in (S_1 G) \cap (S_2 G)\).
\(g \in S_1 G\), so, \(g = s_1 g_1\) where \(s_1 \in S_1\) and \(g_1 \in G\).
\(g \in S_2 G\), so, \(g = s_2 g_2\) where \(s_2 \in S_2\) and \(g_2 \in G\).
\(s_1 g_1 = g = s_2 g_2\), so, \(s_1 = s_2 g_2 {g_1}^{-1}\).
But as \(G\) was a subgroup, \(g_2 {g_1}^{-1} \in G\).
So, \(s_1 \in S_1 \cap (S_2 G)\), a contradiction against \(S_1 \cap (S_2 G) = \emptyset\).
So, there is no \(g \in (S_1 G) \cap (S_2 G)\).
So, \((S_1 G) \cap (S_2 G) = \emptyset\).
Step 3:
Let us suppose that \(S_1 \cap (G S_2) = \emptyset\).
Step 4:
Let us suppose that there was a \(g \in (G S_1) \cap (G S_2)\).
\(g \in G S_1\), so, \(g = g_1 s_1\) where \(s_1 \in S_1\) and \(g_1 \in G\).
\(g \in G S_2\), so, \(g = g_2 s_2\) where \(s_2 \in S_2\) and \(g_2 \in G\).
\(g_1 s_1 = g = g_2 s_2\), so, \(s_1 = {g_1}^{-1} g_2 s_2\).
But as \(G\) was a subgroup, \({g_1}^{-1} g_2 \in G\).
So, \(s_1 \in S_1 \cap (G S_2)\), a contradiction against \(S_1 \cap (G S_2) = \emptyset\).
So, there is no \(g \in (G S_1) \cap (G S_2)\).
So, \((G S_1) \cap (G S_2) = \emptyset\).
