description/proof of that map between groups that maps product of 2 elements to product of images of elements is group homomorphism
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: Note 1
- 4: Proof
- 5: Note 2
Starting Context
- The reader knows a definition of group.
- The reader knows a definition of %structure kind name% homomorphism.
Target Context
- The reader will have a description and a proof of 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.
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:
\(G_1\): \(\in \{\text{ the groups }\}\)
\(G_2\): \(\in \{\text{ the groups }\}\)
\(f\): \(: G_1 \to G_2\)
//
Statements:
\(\forall p_1, p_2 \in G_1 (f (p_1 p_2) = f (p_1) f (p_2))\)
\(\implies\)
\(f \in \{\text{ the group homomorphisms }\}\)
//
2: Natural Language Description
For any groups, \(G_1, G_2\), and any map, \(f: G_1 \to G_2\), if for each \(p_1, p_2 \in G_1\), \(f (p_1 p_2) = f (p_1) f (p_2)\), \(f\) is a group homomorphism.
3: Note 1
By a frequently seen definition, \(\forall p_1, p_2 \in G_1 (f (p_1 p_2) = f (p_1) f (p_2))\) is all that is required, but we do not take that stance: refer to Note for the definition of %structure kind name% homomorphism.
4: Proof
Whole Strategy: Step 1: see that \(f\) maps the identity to the identity; Step 2: see that \(f\) maps the inverse of each element to the inverse of the image of the element.
Step 1:
Let us see that \(f\) maps the identity to the identity.
\(f (1) = f (1 1) = f (1) f (1)\), so, \(1 = f (1) {f (1)}^{-1} = f (1) f (1) {f (1)}^{-1} = f (1)\).
Step 2:
Let us see that \(f\) maps the inverse of each element to the inverse of the image of the element.
For each \(p \in G_1\), \(1 = f (1) = f (p p^{-1}) = f (p) f (p^{-1})\) and \(1 = f (1) = f (p^{-1} p) = f (p^{-1}) f (p)\), which means that \(f (p^{-1}) = {f (p)}^{-1}\).
5: Note 2
We cannot prove that \(f\) maps the identity to the identity like this: "for each \(p \in G_1\), \(f (p) = f (1 p) = f (p 1) = f (1) f (p) = f (p) f (1)\)", because \(f (p)\) s do not necessarily cover the whole \(G_2\). In fact, that is the reason why even the frequently seen definition of ring homomorphism (refer to Note for the definition of %structure kind name% homomorphism) needs to have the requirement of mapping the multiplicative identity to the multiplicative identity in the definition.
