union of a chain of groups as group
1. Proposition
2. Proof
2.2. inverse element
Let \(g_1 \in G\), then by assumption there exists a \(G_i\) such that \(g_1 \in G_i\) and hence \(g_1^{-1} \in G_i \subseteq G\)
Let \(g_1 \in G\), then by assumption there exists a \(G_i\) such that \(g_1 \in G_i\) and hence \(g_1^{-1} \in G_i \subseteq G\)
Date: nil
Created: 2024-10-11 Fr 22:28