description/proof of that for group and finite number of subsets, inverse of product of subsets is product of inverses of subsets in reverse order
Topics
About: group
The table of contents of this article
Starting Context
- The reader knows a definition of finite product of subsets of group.
- 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 and any finite number of subsets, the inverse of the product of the subsets in any order is the product of the inverses of the subsets in the reverse order.
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:
\(G\): \(\in \{\text{ the groups }\}\)
\(J\): \(\in \{\text{ the finite index sets }\}\), \(= \{j_1, ..., j_n\}\)
\(\{S_j \subseteq G \vert j \in J\}\):
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Statements:
\((S_{j_1} ... S_{j_n})^{-1} = {S_{j_n}}^{-1} ... {S_{j_1}}^{-1}\)
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2: Proof
Whole Strategy: Step 1: see that \((S_{j_1} ... S_{j_n})^{-1} \subseteq {S_{j_n}}^{-1} ... {S_{j_1}}^{-1}\); Step 2: see that \({S_{j_n}}^{-1} ... {S_{j_1}}^{-1} \subseteq (S_{j_1} ... S_{j_n})^{-1}\); Step 3: conclude the proposition.
Step 1:
Let \(p \in (S_{j_1} ... S_{j_n})^{-1}\) be any.
\(p = {p^{-1}}^{-1}\) where \(p^{-1} \in S_{j_1} ... S_{j_n}\).
\(p^{-1} = s_{j_1} ... s_{j_n}\) where \(s_{j_1} \in S_{j_1}, ..., s_{j_n} \in S_{j_n}\).
So, \(p = {p^{-1}}^{-1} = (s_{j_1} ... s_{j_n})^{-1} = {s_{j_n}}^{-1} ... {s_{j_1}}^{-1} \in {S_{j_n}}^{-1} ... {S_{j_1}}^{-1}\).
So, \((S_{j_1} ... S_{j_n})^{-1} \subseteq {S_{j_n}}^{-1} ... {S_{j_1}}^{-1}\).
Step 2:
Let \(p \in {S_{j_n}}^{-1} ... {S_{j_1}}^{-1}\) be any.
\(p = {s_{j_n}}^{-1} ... {s_{j_1}}^{-1}\) where \(s_{j_1} \in S_{j_1}, ..., s_{j_n} \in S_{j_n}\).
\({s_{j_n}}^{-1} ... {s_{j_1}}^{-1} = (s_{j_1} ... s_{j_n})^{-1}\).
\(s_{j_1} ... s_{j_n} \in S_{j_1} ... S_{j_n}\).
So, \(p = (s_{j_1} ... s_{j_n})^{-1} \in (S_{j_1} ... S_{j_n})^{-1}\).
So, \({S_{j_n}}^{-1} ... {S_{j_1}}^{-1} \subseteq (S_{j_1} ... S_{j_n})^{-1}\).
Step 3:
So, \((S_{j_1} ... S_{j_n})^{-1} = {S_{j_n}}^{-1} ... {S_{j_1}}^{-1}\).
