description/proof of that for group, multiplication map with fixed element from left or right is bijection
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 group.
Target Context
- The reader will have a description and a proof of the proposition that for any group, the multiplication map with any fixed element from left or right is a bijection.
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:
\(G\): \(\in \{\text{ the groups }\}\)
\(p\): \(\in G\)
\(f_p\): \(: G \to G, p' \mapsto p p'\)
\(f'_p\): \(: G \to G, p' \mapsto p' p\)
//
Statements:
\(f_p \in \{\text{ the bijections }\}\)
\(\land\)
\(f'_p \in \{\text{ the bijections }\}\)
//
2: Natural Language Description
For any group, \(G\), and any element, \(p \in G\), the multiplication map for \(G\) with \(p\) from left, \(f_p: G \to G, p' \mapsto p p'\), or from right, \(f'_p: G \to G, p' \mapsto p' p\), is a bijection.
3: Proof
Whole Strategy: Step 1: see that \(f_p\) is injective by supposing some 2 distinct elements having the same image and finding a contradiction; Step 2: see that \(f_p\) is surjective by finding a domain element that maps to an arbitrary codomain element; Step 3: see that \(f'_p\) is injective by supposing some 2 distinct elements having the same image and finding a contradiction; Step 4: see that \(f'_p\) is surjective by finding a domain element that maps to an arbitrary codomain element.
Step 1:
Let us see that \(f_p\) is injective.
Let us suppose that there were some elements, \(p_1, p_2 \in G\), such that \(p_1 \neq p_2\) and \(f_p (p_1) = f_p (p_2)\).
\(f_p (p_1) = p p_1 = p p_2 = f_p (p_2)\). \(p_1 = p^{-1} p p_1 = p^{-1}p p_2 = p_2\), a contradiction.
So, for each \(p_1, p_2 \in G\) such that \(p_1 \neq p_2\), \(f_p (p_1) \neq f_p (p_2)\).
Step 2:
Let us see that \(f_p\) is surjective.
Let \(p' \in G\) be any. \(f_p (p^{-1} p') = p p^{-1} p' = p'\).
Step 3:
Let us see that \(f'_p\) is injective.
Let us suppose that there were some elements, \(p_1, p_2 \in G\), such that \(p_1 \neq p_2\) and \(f'_p (p_1) = f'_p (p_2)\).
\(f'_p (p_1) = p_1 p = p_2 p = f_p (p_2)\). \(p_1 = p_1 p p^{-1} = p_2 p p^{-1} = p_2\), a contradiction.
So, for each \(p_1, p_2 \in G\) such that \(p_1 \neq p_2\), \(f'_p (p_1) \neq f'_p (p_2)\).
Step 4:
Let us see that \(f'_p\) is surjective.
Let \(p' \in G\) be any. \(f'_p (p' p^{-1}) = p' p^{-1} p = p'\).
