description/proof of that for group, subset, and subset that contains subset, inverse of 1st subset is contained in inverse of 2nd subset
Topics
About: group
The table of contents of this article
Starting Context
- The reader knows a definition of inverse of subset of group.
Target Context
- The reader will have a description and a proof of the proposition that for any group, any subset, and any subset that contains the 1st subset, the inverse of the 1st subset is contained in the inverse of the 2nd 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:
\(G\): \(\in \{\text{ the groups }\}\)
\(S_1\): \(\subseteq G\)
\(S_2\): \(\subseteq G\)
//
Statements:
\(S_1 \subseteq S_2\)
\(\implies\)
\({S_1}^{-1} \subseteq {S_2}^{-1}\)
//
2: Proof
Whole Strategy: Step 1: see that each element of \({S_1}^{-1}\) is in \({S_2}^{-1}\).
Step 1:
Let \({s_1}^{-1} \in {S_1}^{-1}\) be any.
\(s_1 \in S_1\).
So, \(s_1 \in S_2\).
So, \({s_1}^{-1} \in {S_2}^{-1}\).
So, \({S_1}^{-1} \subseteq {S_2}^{-1}\).
