description/proof of that for group, conjugation by element is 'groups - homomorphisms' isomorphism
Topics
About: group
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description
- 2: Natural Language Description
- 3: Proof
Starting Context
- The reader knows a definition of conjugation for group 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, the conjugation by any 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:
\(G\): \(\in \{\text{ the groups }\}\)
\(p\): \(\in G\)
\(f_p\): \(= \text{ the conjugation for } G \text{ by } p\)
//
Statements:
\(f_p \in \{\text{ the 'groups - homomorphisms' isomorphisms }\}\)
//
2: Natural Language Description
For any group, \(G\), any element, \(p \in G\), and the conjugation for \(G\) by \(p\), \(f_p\), \(f_p\) is a 'groups - homomorphisms' isomorphism.
3: Proof
Whole Strategy: Step 1: see that \(f_p\) is a group homomorphism; Step 2: see that \(f_p\) is bijective; Step 3: conclude that \(f_p\) is a 'groups - homomorphisms' isomorphism.
Step 1:
Let us see that \(f_p\) is a group homomorphism.
For each \(p_1, p_2 \in G\), \(f_p (p_1 p_2) = f (p_1) f (p_2)\)?
\(f_p (p_1 p_2) = p p_1 p_2 p^{-1} = p p_1 p^{-1} p p_2 p^{-1} = f_p (p_1) f_p (p_2)\).
\(f_p\) 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 2:
Let us see that \(f_p\) is bijective.
Let \(p_1, p_2 \in G\) be any such that \(p_1 \neq p_2\). Let us suppose that \(f_p (p_1) = f_p (p_2)\). \(p p_1 p^{-1} = p p_2 p^{-1}\), \(p_1 = p^{-1} p p_1 p^{-1} p = p^{-1} p p_2 p^{-1} p = p_2\), a contradiction. So, \(f_p (p_1) \neq f_p (p_2)\). So, \(f_p\) is injective.
Let \(p_2 \in G\) be any. Let \(p_1 = p^{-1} p_2 p \in G\). \(f_p (p_1) = f_p (p^{-1} p_2 p) = p p^{-1} p_2 p p^{-1} = p_2\). So, \(f_p\) is surjective.
So, \(f_p\) is bijective.
Step 3:
\(f_p\) is a 'groups - homomorphisms' isomorphism, by the proposition that any bijective group homomorphism is a 'groups - homomorphisms' isomorphism.
