939: 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

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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

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Topics

About: group

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Proof
• 3: Note

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Starting Context

• The reader knows a definition of left or right coset of subgroup by element of group.
• The reader knows a definition of conjugate subgroup of subgroup by element.
• The reader admits the proposition that for any group, the conjugation by any element is a 'groups - homomorphisms' isomorphism.

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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.

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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.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$G^{\prime }$: $\in \{\text{ the groups }\}$
$G$: $\in \{\text{ the subgroups of }{G}^{\prime }\}$
$g^{\prime }$: $\in G^{\prime }$
$h_1$: $\in g^{\prime }G$
$h_2$: $\in g^{\prime }G$
$\widetilde{h_1}$: $\in G{g}^{\prime }$
$\widetilde{h_2}$: $\in G{g}^{\prime }$
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Statements:
$h_{1}G{h_1}^{-1}=h_{2}G{h_2}^{-1}$
$\wedge$
${\widetilde{h_1}}^{-1}G\widetilde{h_1}={\widetilde{h_2}}^{-1}G\widetilde{h_2}$
//

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2: Proof

Whole Strategy: Step 1: express $h_j$ as $=g^{\prime }{g}_j$; Step 2: see that $h_{j}G{h_j}^{-1}=g^{\prime }G{g^{\prime }}^{-1}$ by the expressions of Step 1; Step 3: express $\widetilde{h_j}$ as $=g_j{g}^{\prime }$; Step 4: see that ${\widetilde{h_j}}^{-1}G\widetilde{h_j}={g^{\prime }}^{-1}G{g}^{\prime }$ by the expressions of Step 3.

Step 1:

$h_1=g^{\prime }{g}_1$ for a $g_1\in G$; $h_2=g^{\prime }{g}_2$ for a $g_2\in G$.

Step 2:

$h_{j}G{h_j}^{-1}=g^{\prime }{g}_{j}G(g^{\prime }{g}_j{)}^{-1}=g^{\prime }{g}_{j}G{g_j}^{-1}{g^{\prime }}^{-1}=g^{\prime }(g_{j}G{g_j}^{-1}){g^{\prime }}^{-1}=g^{\prime }G{g^{\prime }}^{-1}$, by the proposition that for any group, the conjugation by any element is a 'groups - homomorphisms' isomorphism.

So, $h_{1}G{h_1}^{-1}=g^{\prime }G{g^{\prime }}^{-1}=h_{2}G{h_2}^{-1}$.

Step 3:

$\widetilde{h_1}=g_1{g}^{\prime }$ for a $g_1\in G$; $\widetilde{h_2}=g_2{g}^{\prime }$ for a $g_2\in G$.

Step 4:

${\widetilde{h_1}}^{-1}G\widetilde{h_1}=(g_1{g}^{\prime }{)}^{-1}G{g}_1{g}^{\prime }=g^{\prime -1}{g_1}^{-1}G{g}_1{g}^{\prime }=g^{\prime -1}({g_1}^{-1}G{g}_1)g^{\prime }=g^{\prime -1}G{g}^{\prime }$, by the proposition that for any group, the conjugation by any element is a 'groups - homomorphisms' isomorphism.

So, ${\widetilde{h_1}}^{-1}G\widetilde{h_1}=g^{\prime -1}G{g}^{\prime }={\widetilde{h_2}}^{-1}G\widetilde{h_2}$.

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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 $[G^{\prime }:G]$ conjugates: in order to create any conjugate, $g^{\prime }$ needs to be chosen from a left coset, but as each left coset creates only 1 conjugate, there can be at most $[G^{\prime }:G]$ 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, $G$).

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References

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