description/proof of that for group and subgroup, conjugation for subgroup by group element is 'groups homomorphisms' isomorphism
Topics
About: group
The table of contents of this article
Starting Context
- The reader knows a definition of conjugate subgroup of subgroup by element.
- The reader knows a definition of %category name% isomorphism.
- The reader admits the proposition that any map between any groups that maps the product of any 2 elements to the product of the images of the elements is a group homomorphism.
- The reader admits the proposition that any bijective group homomorphism is a 'groups - homomorphisms' isomorphism.
Target Context
- The reader will have a description and a proof of the proposition that for any group and its any subgroup, the conjugation for the subgroup by any group element is a 'groups homomorphisms' isomorphism.
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:
\(G'\): \(\in \{\text{ the groups }\}\)
\(G\): \(\in \{\text{ the subgroups of } G'\}\)
\(g'\): \(\in G'\)
\(f_{g'}\): \(: G \to g' G g'^{-1}\), \(= \text{ the conjugation for } G \text{ by } g'\)
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Statements:
\(f_{g'} \in \{\text{ the 'groups - homomorphisms' isomorphisms }\}\)
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2: Note
Compare with the proposition that for any group, the conjugation by any element is a 'groups - homomorphisms' isomorphism, which is about the conjugation for the whole group, while this proposition is about the conjugation for a subgroup. This proposition is naturally expected from that proposition, but let us prove it for sure.
3: Proof
Whole Strategy: Step 1: see that \(g' G g'^{-1}\) is a group; Step 2: see that \(f_{g'}\) is a groups homomorphism; Step 3: see that \(f_{g'}\) is bijective; Step 4: conclude the proposition.
Step 1:
\(g' G g'^{-1}\) is indeed a group, because it is a conjugate subgroup of subgroup by element: see Note for the definition of conjugate subgroup of subgroup by element.
Step 2:
Let us see that \(f_{g'}\) is a group homomorphism.
For each \(g_1, g_2 \in G\), \(f_{g'} (g_1 g_2) = g' g_1 g_2 g'^{-1} = g' g_1 g'^{-1} g' g_2 g'^{-1} = f_{g'} (g_1) f_{g'} (g_2)\).
\(f_{g'}\) is a group homomorphism, by the proposition that any map between any groups that maps the product of any 2 elements to the product of the images of the elements is a group homomorphism.
Step 3:
Let us see that \(f_{g'}\) is bijective.
Let \(g_1, g_2 \in G\) be any such that \(g_1 \neq g_2\). Let us suppose that \(f_{g'} (g_1) = f_{g'} (g_2)\). \(g' g_1 g'^{-1} = g' g_2 g'^{-1}\), \(g_1 = g'^{-1} g' g_1 g'^{-1} g' = g'^{-1} g' g_2 g'^{-1} g' = g_2\), a contradiction. So, \(f_{g'} (g_1) \neq f_{g'} (g_2)\). So, \(f_{g'}\) is injective.
\(f_{g'}\) is surjective, because the codomain is determined to be the range.
Step 4:
\(f_{g'}\) is a 'groups - homomorphisms' isomorphism, by the proposition that any bijective group homomorphism is a 'groups - homomorphisms' isomorphism.
