A description/proof of that with respect to subgroup, coset by element of group equals coset iff element is member of latter coset
Topics
About: group
The table of contents of this article
Starting Context
- The reader knows a definition of left or right coset of subgroup by element of group.
Target Context
- The reader will have a description and a proof of the proposition that with respect to any subgroup, the coset by any element of the group equals a coset if and only if the element is a member of the latter coset, whether they are left cosets or right cosets.
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: Description
For any group, \(G_1\), any subgroup, \(G_2 \subseteq G_1\), and any element, \(p' \in G_1\), the left or right coset, \(p' G_2 = p G_2\) or \(G_2 p' = G_2 p\), if and only if \(p' \in p G_2\) or \(p' \in G_2 p\), respectively.
2: Proof
Let us suppose that \(p' \in p G_2\). \(p' = p q'\) for a \(q' \in G_2\). For any \(q \in G_2\), \(p' q = p q' q \in p G_2\), so, \(p' G_2 \subseteq p G_2\). As \(p = p' q'^{-1}\), \(p \in p' G_2\), so, likewise, \(p G_2 \subseteq p' G_2\). So, \(p G_2 = p' G_2\).
Let us suppose that \(p' \in G_2 p\). \(p' = q' p\) for a \(q' \in G_2\). For any \(q \in G_2\), \(q p' = q q' p \in G_2 p\), so, \(G_2 p' \subseteq G_2 p\). As \(p = q'^{-1} p'\), \(p \in G_2 p'\), so, likewise, \(G_2 p \subseteq G_2 p'\). So, \(G_2 p = G_2 p'\).
Let us suppose that \(p' G_2 = p G_2\). \(p' \in p G_2\), because \(e \in G_2\).
Let us suppose that \(G_2 p' = G_2 p\). \(p' \in G_2 p\), because \(e \in G_2\).
