description/proof of that for group, set of inverses of elements is group with no duplication
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 set of the inverses of the elements of the group is the group with no duplication.
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:
\(G\): \(\in \{\text{ the groups }\}\), \(= \{g_j \vert j \in J\}\), where \(J\) is any index set, with no duplication
\(G^{- 1}\): \(= \{{g_j}^{- 1} \vert j \in J\}\)
//
Statements:
\(G^{- 1} = G\)
\(\land\)
\(\forall j, l \in J \text{ such that } j \neq l ({g_j}^{- 1} \neq {g_l}^{- 1})\)
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2: Proof
Whole Strategy: Step 1: see that \(G^{- 1} \subseteq G\) and \(G \subseteq G^{- 1}\); Step 2: see that \(G^{- 1}\) has no duplication.
Step 1:
\(G^{- 1} \subseteq G\), because by the definition of group, the inverse of each element is contained in the group.
\(G \subseteq G^{- 1}\), because for each \(g \in G\), \(g = {g^{- 1}}^{- 1}\), and as \(g^{- 1} \in G\), \({g^{- 1}}^{- 1} \in G^{- 1}\).
So, \(G^{- 1} = G\).
Step 2:
For each \(j, l \in J\) such that \(j \neq l\), \({g_j}^{- 1} \neq {g_l}^{- 1}\), because if \({g_j}^{- 1} = {g_l}^{- 1}\), \({{g_j}^{- 1}}^{- 1} = {{g_l}^{- 1}}^{- 1}\), but the left hand side was \(g_j\) and the right hand side was \(g_l\), and so, \(g_j = g_l\), which would mean that \(j = l\), a contradiction.
That means that \(G^{- 1}\) has no duplication.
