description/proof of that for group, subgroup, and left or right coset of subgroup by element of group, conjugates of subgroup by elements or inverses of elements of coset are same
Topics
About: group
The table of contents of this article
Starting Context
Target Context
- The reader will have a description and a proof of the proposition that for any group, any subgroup, and the left or right coset of the subgroup by any element of the group, the conjugates of the subgroup by any elements or the inverses of any elements respectively of the coset are the same.
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:
\(G'\): \(\in \{\text{ the groups }\}\)
\(G\): \(\in \{\text{ the subgroups of } G'\}\)
\(g'\): \(\in G'\)
\(h_1\): \(\in g' G\)
\(h_2\): \(\in g' G\)
\(\widetilde{h_1}\): \(\in G g'\)
\(\widetilde{h_2}\): \(\in G g'\)
//
Statements:
\(h_1 G {h_1}^{-1} = h_2 G {h_2}^{-1}\)
\(\land\)
\(\widetilde{h_1}^{-1} G \widetilde{h_1} = \widetilde{h_2}^{-1} G \widetilde{h_2}\)
//
2: Proof
Whole Strategy: Step 1: express \(h_j\) as \(= g' g_j\); Step 2: see that \(h_j G {h_j}^{-1} = g' G {g'}^{-1}\) by the expressions of Step 1; Step 3: express \(\widetilde{h_j}\) as \(= g_j g'\); Step 4: see that \(\widetilde{h_j}^{-1} G \widetilde{h_j} = {g'}^{-1} G g'\) by the expressions of Step 3.
Step 1:
\(h_1 = g' g_1\) for a \(g_1 \in G\); \(h_2 = g' g_2\) for a \(g_2 \in G\).
Step 2:
\(h_j G {h_j}^{-1} = g' g_j G (g' g_j)^{-1} = g' g_j G {g_j}^{-1} {g'}^{-1} = g' (g_j G {g_j}^{-1}) {g'}^{-1} = g' G {g'}^{-1}\), by the proposition that for any group, the conjugation by any element is a 'groups - homomorphisms' isomorphism.
So, \(h_1 G {h_1}^{-1} = g' G {g'}^{-1} = h_2 G {h_2}^{-1}\).
Step 3:
\(\widetilde{h_1} = g_1 g'\) for a \(g_1 \in G\); \(\widetilde{h_2} = g_2 g'\) for a \(g_2 \in G\).
Step 4:
\(\widetilde{h_1}^{-1} G \widetilde{h_1} = (g_1 g')^{-1} G g_1 g' = g'^{-1} {g_1}^{-1} G g_1 g' = g'^{-1} ({g_1}^{-1} G g_1) g' = g'^{-1} G g'\), by the proposition that for any group, the conjugation by any element is a 'groups - homomorphisms' isomorphism.
So, \(\widetilde{h_1}^{-1} G \widetilde{h_1} = g'^{-1} G g' = \widetilde{h_2}^{-1} G \widetilde{h_2}\).
3: Note
As the logic demonstrates, using the inverses, \(\widetilde{h_1}^{-1}\) and \(\widetilde{h_2}^{-1}\), is crucial for the right coset.
As an immediate corollary, \(G\) has at most \([G' : G]\) conjugates: in order to create any conjugate, \(g'\) needs to be chosen from a left coset, but as each left coset creates only 1 conjugate, there can be at most \([G' : G]\) conjugates: some 2 left cosets may create the same conjugate (especially when \(G\) is a normal subgroup, all the left cosets create the same conjugate, \(G\)).
